Class Notes · Fall 2026

Phil 147

Philosophy of Language

My lecture notes, cleaned up and organized by lecture, filled in from the handouts where the notes were thin or abbreviated. Green boxes and tags mark material added from the handout.

ReferenceDefinitions

Lec 1 · Locke

Iconic theory of meaning
The meaning of a linguistic item is a corresponding mental icon (an idea), or an operation on such an icon.
Being understood
Being made rationally intelligible: the hearer comes to know your mental states (beliefs, desires, goals) and how they rationalize each other and your action.
Empiricism
The view that all ideas are derived from experience.
Three touch-points
What a theory of language should explain: (1) language makes us rationally explicable to each other; (2) some of it is assessable as true or false; (3) meaning is systematic.
Functional vocabulary
Small words that don’t obviously refer to things in the world, e.g. the logical words and, or, not.
Compositionality
A language L is compositional iff for every complex expression e in L, the meaning of e is determined by the structure of e and the meanings of its constituents.
Semantics / syntax
Semantics: the study of meaning. Syntax: the study of which strings are grammatically well-formed in a language.

Lec 2 · Frege, Function & Concept

A priori / a posteriori
A belief is justified a priori iff its justification is completely independent of sense perception; a posteriori iff it depends at least partly on perception.
Analytic (Kant)
True by conceptual containment: the predicate is contained in the subject concept (“all bachelors are unmarried”).
Function
A rule associating each input (the argument) with an output (the function’s value for that argument).
Truth-functional connective
A word like not or and whose output truth value depends only on the truth values of the sentences it combines.
Content
How the world is according to a sentence; what can be true or false.
Force
What a speaker is doing with a content when they use a sentence: asserting it, wondering about it, hedging, etc.

Lec 3 · Frege, Sense & Reference

Reference (Bedeutung)
The thing an expression stands for. Both “Hesperus” and “Phosphorus” refer to Venus.
Sense (Sinn)
The mode of presentation of the referent, modeled on a descriptive condition. Sense determines reference.
Mode of presentation
The way an object is presented to a thinker, like one face of a mountain. Not a mental image; can be shared.
Cognitive significance
S and S′ differ in cognitive significance iff a rational thinker who understands both can believe S is true without believing S′ is true.
Frege’s Criterion
Expressions e₁ and e₂ have different senses iff there are sentences S, S′ that differ only by swapping e₁ for e₂ and that differ in cognitive significance.
Fregean Communication
Speakers communicate successfully with a sentence only if both grasp its sense.
Conception
All the information an individual associates with a term. Varies person to person; irrelevant to communication (unlike sense).

Lec 4 · Russell, On Denoting

Empty name / description
A name or description with no referent: “the present King of France,” “the largest natural number.”
Negative existential
A sentence denying existence, e.g. “The present King of France does not exist.”
Referential vs. quantificational NP
Referential noun phrases talk about particular things (“Mary”); quantificational ones express generalizations (“everyone,” “fifteen people”).
Theory of Descriptions
“The F is G” is true iff there is exactly one F and every F is G. Descriptions are quantifiers, not names.
Scope
The scope of an operator is what the operator applies to. Quantifiers (and descriptions) take scope; referring terms don’t.

Lec 5 · Russell, Acquaintance

Sense-data
Mental entities we are directly aware of: the mental counterparts of images, sounds, tastes. They account for what experience is like.
Acquaintance
Direct, infallible awareness, e.g. of sense-data: if it seems you’re acquainted with a sense-datum, you are.
Intentionality
Something has intentionality when it is about something else.
Proposition
The meaning of a sentence and the object of attitudes like belief. For Russell, an ordered structure of objects, properties, and relations: ⟨a, b, R⟩.
Logically proper name
A name that contributes an object itself to the proposition expressed.
Description Theory of Names
Every apparent name is really a covert definite description, except names for currently entertained sense-data.
Principle of Acquaintance
Every proposition we can understand must be composed wholly of constituents with which we are acquainted.

Lec 6 · Quine

Referential opacity
A context (like “believes that …”) where substituting co-referring terms can change truth value.
De dicto ascription
A belief ascription meant to identify both which object the subject is thinking about and how they think of it. Substitution invalid.
De re ascription
A belief ascription meant only to identify which object the subject is thinking about (“believes of x that …”). Substitution valid.
Notional / relational belief
Notional belief always involves a mode of presentation; relational belief would be a direct relation to an object with none.
Quantifying in
A quantifier outside an attitude verb binding a variable inside it: ∃x(Ralph believes that x is a spy).

Lec 7 · Perry

Indexical
An expression whose reference depends on the context of use: I, here, now, this, (some uses of) tall.
Essential indexical
An indexical in an explanation of action that can’t be replaced by a name or description without destroying the explanation.
De se belief
Self-locating belief, like Perry’s “I am making a mess,” tied to oneself in a way that connects directly to action.
Doctrine of Propositions
Propositions (1) are the objects of belief, (2) are true or false absolutely, (3) obey Frege’s criterion of difference of sense.
Relativized proposition
A proposition true or false only relative to a parameter (a time, a person, a world).

Lec 8 · Strawson

Meaning as use
The meaning of an expression is a rule for its use (not merely determined by use). Essentially social.
Type / token
A type is a class of more specific individuals; a token is a specific individual. An expression is a type; an utterance is a token.
Same use
Two utterances make the same use of an expression iff they refer to the same object (more generally, express the same truth conditions).
Entailment
S entails p iff whenever S is true p is true, and if S is not true p need not be.
Presupposition
Background taken for granted: p is implied by S and by every member of S’s family (its negation, conditional, possibility, question).
Attributive / referential use
Donnellan: “the φ” used about whoever is the φ (attributive) vs. used to pick out someone the speaker has in mind (referential).

Lec 9 · Kaplan

Downward entailment
A position is downward entailing if swapping in something more specific preserves truth (“I don’t have a pizza” ⇒ “I don’t have a pizza with anchovies”).
Character
Conventional meaning, what a competent speaker knows. A function from context to content.
Content
What is said: the proposition expressed in a context.
Context
A possible occasion of use, which fixes the interpretation of context-sensitive expressions. Formally, a tagged list (speaker, audience, place, time, demonstrata).
Circumstance
A possible situation at which a content is evaluated as true or false.
Metasemantics
The study of the facts in virtue of which expressions have their meanings (vs. semantics, the study of meaning).

Lec 10 · Lewis

Counterfactual
“If it had been that p, it would have been that q,” written p □→ q.
Material conditional
p ⊃ q is true iff p is false or q is true.
Strict conditional
□(p ⊃ q): the material conditional holds in every accessible world.
Sobel sequence
A sequence of counterfactuals, each adding to the previous antecedent and flipping the consequent (“…they’d topple over; but if they also used crutches, they wouldn’t”).
System of spheres ($)
Nested sets of worlds around the actual world, ranking worlds by similarity to it. The actual world is at the top; ties are allowed.
Modal realism
Other possible worlds are just as real as the actual world.
Ersatzism
Possible worlds are stand-ins (Ersatz), not further real worlds.

Lec 11 · Quine

Essential / contingent property
Essential: a property a thing couldn’t lose and still exist. Contingent: one it could have lacked.
Reconstruction
An analysis of a notion that sorts our talk using it into the unproblematic parts and the parts to be jettisoned.
Semantical predicate
A predicate true of sentences, like “contains two numerals” or (for Quine) “is provable.”
Objectual quantification
Quantification whose truth depends only on whether some object satisfies the open sentence, however that object is referred to.
Aristotelian essentialism
Objects have essential properties in themselves, independent of how they’re described.

Lec 12 · Kripke

Description Theory
A name’s reference is fixed by a cluster of descriptions the speaker associates with it (theses i–vi and C).
Direct reference
A name contributes its referent directly, not via a descriptive sense.
Rigid designator
An expression that refers to the same object in every possible world in which it refers at all.
Non-rigid designator
An expression that refers to different objects in at least two possible worlds.
Psychological essentialism
Our (especially children’s) strong tendency to ascribe essences to things, including things that don’t have them.

Lecture 1Locke and the iconic theory of meaning

Locke, An Essay Concerning Human Understanding (1689), III.ii

Words mark ideas

For Locke, the goal of language is to be understood, which here means being made rationally intelligible. Not in a very deep sense: just “I had a goal, constraints, opportunities, so I acted as I did.” To understand someone is to know their mental states (beliefs, desires, goals, reasoning) and how those states rationalize each other and their behavior. Rationality matters a lot for Locke.

So words “mark ideas.” Ideas are the things we reason with: pictures before the mind that we manipulate in order to reason. Locke is an empiricist: all ideas come from experience.

  • Worry: it seems very hard to share ideas themselves. But perhaps you give me an image of your having the image.
  • Nice feature: it’s easy to see what it would mean for an idea to be right or wrong; we can point to false beliefs. So there are two ways to assess someone’s mental states: by the relations between them (do they rationalize each other?), and by their truth or falsity.
The Iconic Theory

The meaning of a linguistic item is a corresponding mental icon, or an operation on such an icon.

We can compose meaning-bearing elements to understand new sentences: different parts are different ideas, and some parts are operations on those ideas. To understand “Taylor Swift is watching a game at Harvard Stadium,” take your image of Taylor Swift and your image of the stadium during a game and smush them together.

Three touch-points for a theory of language

A philosophical theory of language should answer to three things, and Locke’s covers all three:

  1. Rational explicability. We use language to make ourselves intelligible to each other.
  2. Truth. Some of language is properly assessed as true or false.
  3. Systematicity. We understand novel sentences from the meanings of their parts and how they’re put together.

Locke’s project is about the rational, the semantic, and the systematic. It sets aside, e.g., the way poetry makes us feel.

Objections: Berkeley’s triangle and the word “not”

  • Berkeley. Locke has to say reference is fixed by resemblance. But a mental picture of a triangle is always of some specific triangle, and we can still reason about triangles in general.
  • Functional vocabulary (small words that don’t obviously stand for things, like and, or, not). And is easy: for “Mary and Jane ate a cookie,” combine the two images. Not is hard. What image corresponds to “Taylor Swift is not walking”?
    • Not a picture of her walking, since then it would mean the same as “Taylor Swift is walking.”
    • Not a picture of her walking with a bar through it. HandoutThat image doesn’t resemble any situation in the world.
    • Not a picture of her doing something else, because asserting the sentence doesn’t commit you to her doing anything in particular.
  • Or looks just as hard. HandoutSo do quantifiers (“fifteen,” “some,” “none”) and tense.

Frege’s alternative: start from systematicity

Locke took nouns and verbs as the paradigm and then had to fit everything else in. Frege starts from the other end: from systematicity (“an infinite use of finite means”) and from functional vocabulary like and and not. He asks what nouns and verbs must mean for that to work. Truth comes first; rationality is dealt with later.

Compositionality

A language L is compositional iff for every complex expression e in L, the meaning of e in L is determined by the structure of e in L and the meanings of the constituents of e in L.

Handout

Why “structure” is in the definition: “Iago loves Desdemona” and “Desdemona loves Iago” have the same parts but different meanings. Also defined: semantics (the study of meaning) and syntax (the study of which strings are grammatically well-formed).

Lecture 2Frege: function and concept

Frege, “Function and Concept” (1891)

Review: the three touch-points

  • Rational explanation is key for Locke: rational reaction as opposed to pure causation.
  • For language to be assessed as true or false, it has to be about the world. (It’s hard for music to be about something.)
  • Language must be systematic: an infinite use of finite means.

Locke started with rational explanation. Frege starts with systematicity.

Motivation: is arithmetic analytic?

For Kant, some a priori knowledge is unpuzzling because it’s just containment: “all bachelors are unmarried” is analytic. But math is weird. The concept “7 + 5” doesn’t seem to contain “12,” yet “7 + 5 = 12” is clearly a priori. So Kant calls arithmetic synthetic a priori.

Frege wants to show that arithmetic is analytic after all: its first principles can be proved from definitions using logical laws alone. But proofs written in natural language might smuggle in unnoticed assumptions. So he builds the Begriffsschrift (“concept-script”), a formal (higher-order) logical language that lets nothing through that wasn’t expressly assumed. That’s why systematicity matters so much to him.

Numbers vs. numerals. A numeral is the symbol we write; a number is the abstract object it refers to. So numerals stand to numbers as names stand to people.

Connectives as truth functions

A function is a rule taking an argument to a value. Negation flips a truth value, so it looks a lot like a function, and so does and:

not(x) = True if x = False; False if x = True
and(x, y) = True iff x = True and y = True

Take “Mary ran and Jane did not run.” If and is a function from two truth values to one, then the sentences it combines must refer to truth values. Sentences refer to the True or the False.

Predicates as functions, names as arguments

If “Mary runs” refers to the True, what are “Mary” and “runs” doing? Something has to output the True. Frege: “runs” is the function (from objects to truth values, giving True exactly for things that run) and “Mary” is the argument. In lecture this was presented as a free choice.

The same pattern shows up everywhere:

  • the capital of ( ) countries → cities; Germany ↦ Berlin
  • the husband of ( ) people → people

Force vs. content

  • Content: how the world is according to the sentence; what can be true or false.
  • Force: what the speaker is doing with that content, such as asserting it (committing to its truth), wondering about it (“I wonder… maybe…”), etc.

Frege wants to keep these apart, because force doesn’t pass through every operator. Asserting “Mary ate a cookie and Jane ran” commits you to “Mary ate a cookie”; asserting “Mary ate a cookie or Jane ran,” or “it’s not the case that Mary ate a cookie,” doesn’t. The same content appears without being asserted.

This is where the iconic theory fails badly. It’s fine with and but fails on not, because on the empiricist picture, having an image before the mind is believing it.

What Frege wants is a semantics you can plug and chug through to get truth values. Open question: how does any of this give us rational explanation?

Handout

Substitution and the President puzzle (sets up Lec 3). A function cares only about what its argument is, not how it’s picked out, so swapping co-referring terms should preserve truth. That works for “The President is in the White House; Trump is the President; so Trump is in the White House.” It fails for “The President is elected every four years; Trump is the President; so Trump is elected every four years.”

Lecture 3Frege: sense and reference

Frege, “On Sense and Reference” (1892)

Recall that Frege’s semantics is built from functions and the objects they take as arguments, so names are central. (On papers: the question should be central, with a clear criterion for an answer.)

Hesperus and Phosphorus: informative identities

Take an identity statement “a = b.” The “is” here is the is of identity, not the copula of “I am good.”

  • Hesperus is Hesperus.
  • Hesperus is Phosphorus.

If a name’s contribution is just its referent (the Reference View), both say the same thing about Venus. But the first is trivial and the second was an astronomical discovery. So what makes one analytic and the other empirical?

The Name View answers that “a = b” is about the names. Frege rejects it (two arguments in the reading).

Cognitive significance

Compare “Andy is a bachelor” with “Andy is an unmarried man.” The two have different associations but are analytically equivalent. Frege cares only about what bears on our rational capacities, our reasons for believing and acting, and not about associations.

Cognitive significance

S and S′ differ in cognitive significance iff a rational thinker who understands all the expressions in both can believe S is true while not believing S′ is true.

Sense as mode of presentation

Frege’s solution is sense: the mode of presentation of the referent, the way it is presented to you. Think of one view of a mountain as opposed to the mountain itself. HandoutHiking: “this mountain is this mountain” vs. “this mountain is the mountain I saw this morning.”

You can approximate a mode of presentation as an image, but modes of presentation are not mental images. The pill example:

  • Draw twice from a pile of indistinguishable pills, getting the same pill both times. The images are identical, but you have two modes of presentation, since you don’t know it’s the same pill.
  • Now follow one pill continuously with your eyes. The image may change, but there’s only one mode of presentation.

Modes of presentation can also be shared: two people can see the same view of a mountain. Images can’t. HandoutFrege’s telescope: the image in the lens is public, the image on each observer’s retina is private. On the (controversial) Frege–Schlick view, we can’t even ask whether the contents of two people’s minds are similar.

How sense determines reference

Generalize the model: a sense is like a descriptive criterion. “Hesperus” has a sense like “the first heavenly body visible in the evening sky”; Handout“Phosphorus,” “the last heavenly body visible in the morning sky.”

How does language attach to the world? A word is conventionally tied to a sense, the sense is a descriptive condition, and the referent is whatever satisfies it.

  • “H = H” is trivial: take the same sense on both sides and follow it out into the world. Of course it’s true.
  • “H = P” is informative: the criteria are different, so you have to check empirically that they pick out the same thing (they do).
Frege’s Criterion

Expressions e₁ and e₂ have different senses iff there are sentences S₁ and S₂ such that (1) they differ only in that e₁ is replaced by e₂, and (2) they differ in cognitive significance.

Sense and successful communication

Fregean Communication

Speakers communicate successfully using a sentence only if they both grasp the sense of that sentence.

  • Co-reference isn’t enough. Paderewski was a famous pianist who later became Polish prime minister. You say “Paderewski is a great musician”; I’ve only met him as a politician and tell him “your music is wonderful!” Same referent, yet communication failed. HandoutSimilarly, Hannah Smith: I form a true belief from your true claim, but about a different Hannah Smith.
  • Sharing all beliefs isn’t required. HandoutA conception is everything one person associates with a term; it varies person to person and is irrelevant to communication.

So sense is the middle ground: co-reference is necessary but not sufficient, and sharing all your beliefs is sufficient but not necessary. Frege does admit that senses vary between people (the “Aristotle” footnote) and calls this an imperfection of language. HandoutHe says it can be tolerated as long as the referent is the same, which sits uneasily with the communication principle.

Lecture 4Russell: on denoting

Russell, “On Denoting” (1905)

Handout

I have no notes for this lecture, so this section is written from the handout. The last part is from my Section 2 notes, which reviewed it.

Empty names and negative existentials

Some names and descriptions are empty: “the planet farthest from the Sun,” “the largest natural number,” or a name a journalist invents to attribute fabricated quotes. “The planet farthest from the Sun is larger than Pluto” seems like a whole thought, but if the description refers to nothing it’s semantically just “     is larger than Pluto.” Frege offers two answers: such sentences really are empty, so neither true nor false (Sense and Reference), or the description gets a dummy referent (Function and Concept).

The sharpest form of the problem is the negative existential:

  1. A subject–predicate sentence with a referring subject is true iff the subject refers to something that has the predicated property.
  2. “The present King of France does not exist” is such a sentence.
  3. It’s true.
  4. So “the present King of France” refers to something.

Meinong bites the bullet: the King “subsists” without existing. Russell wants to avoid this.

Referring vs. quantifying noun phrases

Referential noun phrases talk about particular things (“Mary,” “the professor for PHIL 147”). Quantificational ones express generalizations (“someone,” “fifteen people,” “everyone in this course,” “many students”). For Russell, “everything,” “nothing,” and “something” are defined from the primitive “C(x) is always true.” Quantification can’t be reduced to reference: “Every person in the room wears glasses” equals “Andre, Barbara, and Clyde wear glasses” only given the extra fact that they’re all and only the people in the room.

Quantified sentences have the shape Quantifier · Restrictor · Nuclear scope (“Every F G”), and “Some F G” can be defined as “not: all F not-G.”

The Theory of Descriptions

“The TF for PHIL 147 teaches a couple of sections” implies existence and uniqueness: there is exactly one TF, and that TF teaches a couple of sections.

Russell’s Theory of Descriptions

“The F is G” is true iff there is exactly one F and every F is G.

∃x (Fx ∧ ∀y (Fy → y = x) ∧ Gx)

This solves the puzzle. “The present King of France does not exist” means it is not the case that there is exactly one present King of France:

¬∃x (Kx ∧ ∀y (Ky → y = x))

There is no subject term left to refer to anything, so there’s no need for Meinong.

Scope: “The King of France is not bald”

The scope of an operator is what it applies to: compare 2(3) + 2 = 8 with 2(3 + 2) = 10, or “I saw the woman with binoculars.” Quantificational expressions enter scope relations; referential ones don’t. Since descriptions are quantifiers, “The present King of France is not bald” has two readings:

  • It’s not the case that the present King of France is bald. true: there is no such king
  • The present King of France is such that he is not bald. false

Compare “All that glitters is not gold.” On “The present King of France is bald,” Frege says it’s neither true nor false; Russell says it’s false.

Section 2 review

  • For Frege, “the King of France does not exist” comes out neither true nor false, but it’s obviously true. For Russell it is true, as the negation of an existence-and-uniqueness claim. (Does Frege have a response here?)
  • For Russell, quantifiers don’t have reference: “everyone” doesn’t refer to specific people. (Could quantifiers refer to the whole world, as though the world had the property of existing?)
  • Russell likes his view because it’s better ontologically (no subsisting non-existents), and perhaps epistemically.
  • Correspondence theory of truth: “S” is true iff S is among the true things. Open questions: what is Russell’s theory of what makes something true, and of how communication and mutual understanding happen, i.e. how words attach to objects (see Lec 5)?
Handout

Two problems for Russell to think about. Uniqueness: “I’m the daughter of a farmer” doesn’t imply you’re an only child. Desires: “Mary wants to squash the rat in her attic” becomes “Mary wants there to be exactly one rat in her attic and to squash it,” which implies she wants a rat in her attic.

Lecture 5Russell: acquaintance and description

Russell, “Knowledge by Acquaintance and Knowledge by Description” (1910)

Sense-data and acquaintance

Start with colour (or X-) constancy: when the light changes, my perception of my pencil and the underlying sense-data change, but perception at a higher level doesn’t really.

For Russell, what I’m most immediately aware of is sense-data. HandoutThese are mental entities, the counterparts of images, sounds, and tastes, that account for what experience is like. We are acquainted with sense-data, and acquaintance is infallible: seeing a pencil can go wrong, but there’s no appearance/reality gap for sense-data themselves.

That’s a substantive claim, though. The speckled hen: I see a hen with lots of speckles in the distance, and it seems to have an indeterminate number of them. Does the sense-datum have an indeterminate number of speckles?

Three accounts of intentionality

Something has intentionality when it’s about something. Three accounts so far:

  • Iconic theory: resemblance. The problem is that an image must resemble its object in some respects, or else everything resembles everything. Specifying which respects requires interpreting the image, and that defeats the purpose, since the theory was meant to explain what an idea is about.
  • Frege: descriptive criteria (senses).
  • Russell: descriptive and causal. A thought is about a thing because that thing causes the sense-data I’m aware of and I think descriptive thoughts about it. Issue: many things are causally responsible for my current sense-data, so how do we get the right description? (To think more about.)

Russellian propositions

Propositions are the meanings of sentences beyond the words: one proposition can be expressed in many languages, which makes them a good way to pin down what a disagreement is about. HandoutThey’re also the objects of belief. For Russell, a proposition contains the objects it’s about plus the properties and relations predicated of them, as an ordered structure:

⟨Sun, Moon, occludes⟩ ≠ ⟨Moon, Sun, occludes⟩

Order matters, since “the Sun blocks the Moon” ≠ “the Moon blocks the Sun.”

Names as disguised descriptions

What about fake names? A con artist introduces “Joe Smith, who’s excited about this opportunity.” How is the victim’s thought meaningful? Their proposition isn’t about Joe Smith but about “the man introduced to me as a backer.” The name is short for a description, so it doesn’t refer.

  • A logically proper name contributes an object itself to the proposition.
  • The test: if you can suppose the subject doesn’t exist and the proposition is still meaningful, the subject is a description, not a name.
  • Since we can always build Cartesian skeptical scenarios, this applies to everything outside our current sense-data.
Description Theory of Names

Every apparent name is really a covert definite description, except names for currently entertained sense-data.

What drives Russell: when it seems to me that I’m entertaining a proposition, I really am. I can be wrong about whether it’s true, but not about whether I’m thinking a contentful thought. And propositions contain what they’re about, so their constituents must be things whose existence is guaranteed, which means sense-data.

Principle of Acquaintance (handout)

Every proposition we can understand must be composed wholly of constituents with which we are acquainted.

Handout

Strictly, Russell thinks we are also acquainted with properties and relations (universals), so descriptions are built from sense-data plus universals. The description behind a name also varies between people and over time. That raises the communication puzzle: if each of us only entertains propositions about our own sense-data, two speakers never grasp the same proposition, so what makes communication succeed?

Lecture 6Quine: quantifiers and propositional attitudes

Quine, “Quantifiers and Propositional Attitudes” (1956)

Recall the three touch-points: rational explanation, systematicity, and aboutness. Quine is interested in two ways of ascribing belief, and in whether that implies a metaphysical difference (two kinds of belief).

Ralph and Ortcutt: substitution in belief reports

Ralph suspects the man in the brown hat is a spy. He thinks the man he saw at the beach (a pillar of the community) is not. Unbeknownst to Ralph, they are the same man, Bernard J. Ortcutt.

  1. Ralph believes that the man in the brown hat is a spy.
  2. The man in the brown hat is the man on the beach.
  3. ✗So Ralph believes that the man on the beach is a spy.

The inference fails, which is weird if functions don’t care about the way their argument is denoted. This is a failure of substitution, due to perspective (like sense). HandoutSuch a context is referentially opaque.

  • Frege’s solution: in belief reports, words refer to their customary sense (their “indirect reference”). The two descriptions have different senses, so the substitution never applied. This is intuitive: what motivates action isn’t what we’re referring to but how we think of it. Belief reports are about modes of presentation, not referents.
  • Russell’s solution: rewrite descriptions as quantifiers (“there’s exactly one man in a brown hat and he’s a spy”), and that doesn’t entail the beach version. Issue: since names are descriptions built from the believer’s own sense-data, we can’t characterize how Ralph is thinking in shareable terms. HandoutFregean senses, by contrast, are public.

De dicto vs. de re ascriptions

“Ralph believes that someone is a spy” can mean:

  • (a) just on general principles: there are spies, but he can’t point to one. Ralph believes ∃x(x is a spy)
  • (b) he has someone specific in mind. ∃x(Ralph believes x is a spy)

These motivate, and justify, different behavior. Compare “Ernest is hunting a lion”: no particular lion vs. “there is a lion such that Ernest is hunting it.”

De dicto ascription

A belief ascription is de dicto iff it’s intended to identify which object the subject is thinking about and how they’re thinking of it. “L believes that S can F.”

De re ascription

Only intended to identify which object the subject is thinking about. “L believes of S that he can F.”

The difference is whether the referent is introduced inside or outside the belief report. Co-referring terms can be substituted in de re ascriptions and cannot in de dicto ones.

Quine’s worry about de re belief

Can we say Ralph believes of Ortcutt that he is a spy? It seems so: if he believes that the man in the brown hat is a spy, he believes of that man that he’s a spy, and since de re allows substitution, he believes of Ortcutt that he’s a spy. By the same route from the beach, he believes of Ortcutt that he’s not a spy. So Ralph both does and doesn’t believe of Ortcutt that he’s a spy.

Sometimes we engage the world under a mode of presentation (notional belief). Quine also posits relational belief, a direct relation to the object with no mode of presentation (as “seeing” seems to allow substitution).

Handout

Why the incoherence depends on relational belief. The incoherence only bites if de re ascriptions report a distinct, purely relational belief. Then Ralph would stand in, and not stand in, one and the same direct relation to Ortcutt. If all belief is notional, the two de re ascriptions just involve two different modes of presentation, and there’s no contradiction. HandoutQuine also notes a second problem: de re ascriptions come apart from what Ralph would sincerely assert.

Kaplan: reducing de re to de dicto

Suppose all belief goes through a mode of presentation. Then a de re ascription just quantifies over modes of presentation (Kaplan 1968):

  • There is some MoP m such that Ralph believes, under m, that Ortcutt is a spy;
  • and there is some MoP m′ such that Ralph believes, under m′, that Ortcutt is not a spy.

We divorce the mode of presentation in the reporter’s words from the one quantified over and leave open which one Ralph is using. No contradiction. So there are still two kinds of ascription, which differ semantically, but only one kind of belief: notional.

Moral: a semantic difference doesn’t necessarily mean a metaphysical one. De dicto and de re are kinds of belief ascription; to claim they’re different kinds of belief takes a separate metaphysical argument.

Lecture 7Perry: the essential indexical

Perry, “The Problem of the Essential Indexical” (1979)

Self-locating belief

The paper is from 1979, close in time to Jackson’s Mary case, and the two are quite similar: in the shopper case you can be completely clear about all the facts and still be missing something. What you learn is a purely perspectival fact. What is that new knowledge, and is it an extra kind of fact? This is the challenge of self-locating belief.

De se belief

Belief connected to oneself in a way that is directly connected to action, the kind of belief Perry acquires in the shopper case.

The messy shopper and the Doctrine of Propositions

Perry follows a trail of sugar around a supermarket counter, looking for the shopper with the torn sack, and then realizes: “I was the shopper I was trying to catch.” What did he learn?

  • It clearly concerns Perry, but it’s distinct from “the shopper with the leaking bag is making a mess,” which he believed all along (Fregean distinctness).
  • So some special mode of presentation by which Perry is presented to himself is doing the work.
  • It can’t simply be the sense of “I am making a mess,” because that sentence means something different in someone else’s mouth. It isn’t absolutely true or false.

Perry’s target is the Doctrine of Propositions. Propositions:

  1. are the objects of belief;
  2. are true or false absolutely;
  3. obey Frege’s criterion of difference of sense.

We’d like a proposition that explains the change in behavior and satisfies all three. How could that work for indexicals?

Why no description can replace “I”

Try another sense for “Perry”: “the only bearded philosopher in a Safeway west of the Mississippi is making a mess.” That still doesn’t explain why he changes what he does. You need “…and that’s me!”, which is an indexical again. This is the essential indexical: replacing it with any other term destroys the explanation.

An indexical is an expression whose reference depends on context: pronouns, demonstratives, adjectives like “tall,” etc.

Nickel: linguistic vs. conceptual indexicals

Perry’s argument slips between different notions of “indexical.” Consider:

  • Every girl turned in her homework. “her” isn’t one person, but it isn’t indexical either
  • Perry learned that he was making a mess.
  • Perry learned that he, Perry, was making a mess.
Handout

What the examples show. Perry’s thesis is about a concept, the “missing conceptual ingredient” in the belief that explains action. But he argues for it using a linguistic category, indexical words. The examples pull these apart. The “he” in the second sentence is a bound pronoun, like “her” in the first, not an indexical. “He, Perry” isn’t indexical at all. Yet both correctly report what Perry learned and explain why he stopped. So the linguistic category isn’t what’s essential, and tying a debate about mental states to it is odd.

Frege: a private sense of oneself

Frege would say there is a special mode of presentation, a thought that leads immediately to action. You can’t grasp my version of it, since your thought can’t move my leg. Each person is presented to themselves in a special, primitive way that only they can grasp. HandoutFrege’s example is Dr. Lauben, who has to use “I” in a public sense (“he who is speaking to you”) to communicate.

Perry grants he has no knock-down argument against these private, limited-access propositions, but he rejects them anyway: he believes “only in a common actual world.” The metaphysical cost is too high.

Relativized propositions

When two people each think “I am making a mess,” they seem to learn the same thing. For Frege that can’t be right: different referents mean different senses and different thoughts.

The fix is to give up Absoluteness. Believing “Garber is President” seems to cause the same action at different times, so treat it as one proposition whose truth value changes over time, and index propositions to times. The proposal is to do the same for people: take the common structure (a relativized proposition with no particular person built in) and evaluate its truth against a particular context. HandoutA Fregean can already relativize to worlds, so this isn’t a deep departure.

From section: imagine two omniscient gods in different places. They know every proposition about the world, yet neither knows which god it is. Only something that just one of them could know would tell them apart.

Perry’s objection and Nickel’s reply

Perry: both Perry and another shopper watching him can believe that “I am making a mess” is true-for-Perry, but they act very differently. The context of belief need not be the context of evaluation.

Nickel: this assumes that believing p is the same as believing that p is true for me. But believing a proposition is different from believing it’s true relative to some parameter. The other shopper believes that Perry’s proposition is true relative to Perry; he doesn’t believe the proposition himself. So the objection doesn’t land.

Handout

Perry’s own positive view: the two shoppers are in the same belief state (which explains their similar behavior) but believe different things.

Lecture 8Strawson: on referring

Strawson, “On Referring” (1950); Donnellan, “Reference and Definite Descriptions” (1966)

Meaning as a rule of use

Strawson rejects representational theories of meaning, on which you understand an expression or sentence by looking at how it represents the world as being. No single theory explains all three touch-points plus motivating action: sense can’t do it all. He also thinks systematicity has been oversold.

Meaning as use

The meaning of an expression is a rule for its use. Not that meaning is determined by use: meaning and the rule of use are the same thing.

  • This is a social theory. All of Frege’s problems and points would make sense for a solitary individual; Strawson’s definition is essentially social.
  • Example: the meaning of “I” is the rule that it’s “used by someone to refer to themselves in speaking or writing.” It would make no sense to define it by pointing at someone.
  • HandoutSo rules of use are not Fregean senses. A sense determines a referent, but one rule for “I” yields a different referent for every speaker.

Expression, utterance, use

  • An expression is a type (the expression itself); an utterance is a token (a particular person saying it on a particular occasion).
  • A use is in between: two utterances make the same use of an expression iff they refer to the same object (or express the same truth conditions).

Against Russell: meaningful vs. true-or-false

Words don’t refer; people do. Take “The present King of France is bald.” Strawson says Russell conflates two questions:

  1. Is the sentence meaningful?
  2. Can it be used to express a true-or-false proposition when there’s no King of France?

“Meaningful” just tests whether there’s a coherent rule of use, i.e. whether the rules for the parts fit together, and this sentence has perfectly clear rules. Even so, a particular utterance of a meaningful sentence can fail to make a true-or-false assertion. (Strawson slides between “meaningful” and “significant.”)

Another problem: Russell’s theory applied to “the table is covered with stuff” says there’s exactly one table in the world, which is weird. HandoutThe speaker means the table that’s salient in context; rules for descriptions are partial and lean on context.

Presupposition vs. entailment

Two kinds of things are implied by (true if) a sentence S:

  1. entailments of S;
  2. presuppositions: background taken for granted.

Handout“It is Mary who won the race” asserts that Mary won and presupposes that someone won. Frege had seen this: “Whoever discovered the elliptic orbits died in misery” presupposes that someone discovered them.

Family-of-sentences test

p is a presupposition of S if p is implied by S and also by each of: It’s not the case that S; If S, then S′; It’s possible that S; Is it the case that S?

The idea is that p is implied every time you even put the sentence forward, without committing to its truth. Every member of the “present King of France” family commits you to there being a unique present King of France. That makes presupposition a good point against Russell, but not against Frege.

Donnellan: attributive vs. referential uses

  1. Attributive: if nothing is the φ, nothing has been said to be ψ. A detective standing over Smith’s mutilated body says “Smith’s murderer is insane,” meaning whoever did this.
  2. Referential: even if nothing is the φ, something has still been said to be ψ. At the trial, looking at Jones, someone says “Smith’s murderer is insane” and means Jones, guilty or not.

In the referential case, reference succeeds through the speaker’s intentions and the context, even if nothing is the φ. That something is the φ is a presumption, not an entailment. Handout“The man drinking a martini seems to be enjoying himself” can be true even if he’s drinking water from a martini glass and nobody at the party has a martini.

Lecture 9Kaplan: demonstratives

Kaplan, “Demonstratives” (1989)

Does natural language have a logic? The case of “any”

Russell and Frege thought sentences have truth values and so stand in logical relations. Strawson says natural language can’t be reduced to a formal language with well-defined truth conditions (“no exact logic”). Chomsky thinks you can have formal logic for syntax but not for semantics. E.g. a description doesn’t always have to be satisfied for you to use it successfully (the courtroom murderer from Lec 8).

Evidence that logic is built into language comes from the word “any”:

  • I don’t have any money. ✓
  • *I have any money. ✗

We’ve internalized some rule, and it’s tied to downward entailment. Positive contexts are upward-entailing; negative ones are downward-entailing (“I don’t have a pizza” ⇒ “I don’t have a pizza with anchovies”), and “any” is allowed in downward-entailing positions. With “every,” the restrictor is downward-entailing and the nuclear scope is upward-entailing: “Everyone who completes any assignment passes” is fine, but “Everyone who takes this class will get any grade” isn’t. So grammaticality is sensitive to logical properties, and those only make sense if sentences have truth conditions.

Compositionality as a problem for Strawson

Recall that a language is compositional iff the meaning of every complex expression is determined by its structure and the meanings of its constituents. Strawson struggles here, because two expressions can have the same rule of use but contribute differently to larger sentences. Handout“I” and “the speaker” are both used to refer to the speaker, yet “If I were silent and Snowball were speaking, I would be a grad student” and “…, the speaker would be a grad student” differ in truth conditions.

Character and content

Everyone agrees meaning splits in two. For Kaplan:

Character

A function from context to content. It’s set by linguistic convention, determines the content in each context, and is what a competent speaker knows. It’s close to the everyday notion of “meaning.”

Content

The proposition expressed by the sentence in a context. Its subject is the actual person, not a name or description.

  • Agreement and disagreement need the same content. A: “I’m a skeptic.” B: “No, I’m not a skeptic.” They failed to disagree: same character, different content.
  • Content is what gets composed, so it’s what makes compositionality work.

Context, circumstance, and possible worlds

Use possible-worlds semantics: a proposition is true in some worlds and false in others, and asserting a sentence narrows down the worlds.

  • The context fixes the interpretation of the sentence: who “I” is, what “here” is, and so on.
  • The circumstance is the situation (world) where we evaluate the resulting content as true or false.

The grad-student examples: “I am a graduate student” and “The speaker is a graduate student” agree at the actual world but come apart at other worlds. Donkeys: “If we called a tail a leg, how many legs would a donkey have?” Four, because we evaluate what we say at that circumstance, holding our meanings fixed. But “…what would the correct answer be, if someone asked there?” is five. HandoutLikewise, “If we all spoke French, how many legs would a donkey have?” Four, not “quatre.” So the proposition itself is purely objective: what we say is separate from the words we use to say it.

“I am here now”: logically true but contingent

Traditionally, a sentence is valid iff it can be derived from the empty set of premises, i.e. it’s true without investigation and necessarily true (“P or not P”). But “I am here now” is true whenever you say it, because the context of utterance is also the circumstance of evaluation. Yet it isn’t necessary: I could have stayed home today. So it’s logically true but not necessary. The proposition expressed could easily have been false.

What is a context? Semantics vs. metasemantics

Strawson says context is messy; Kaplan says it’s just “the input to character.”

Kaplan’s Principle 1: the referent of a pure indexical depends on the context, and the referent of a demonstrative depends on the associated demonstration. But this isn’t automatic. You can say “that” without pointing, and for “here” the size of the area (“I live here”) depends heavily on context. So Principle 1 is at best incomplete.

Handout

A Kaplanian context is a very partial description of a situation, a tagged list (speaker, audience, place, time, demonstrata). Kaplan doesn’t try to say how we get that neat list from a messy situation. He sets that aside as metasemantics (the study of the facts by which expressions have meaning), as opposed to semantics (the study of meaning).

Kaplan’s four components

Context— character →Content— at a →Circumstance→Truth value

Lecture 10Lewis: counterfactuals

Lewis, Counterfactuals (1973), ch. 1

Modality and possible worlds

Originally “modality” just meant a way (mode) of being or doing; the lingo comes from Kant. “Mary must keep her promise” is put forward apodeictically, “Mary keeps her promise” categorically, and “Mary may keep her promise” problematically.

Kaplan already treated propositions as implicitly modal. Putting one forward divides the possible worlds by whether it’s true there and says we’re on the true side, so it’s a claim about all worlds. That’s why “BN” and “I” have to be treated as referring expressions, not descriptions: descriptions behave very differently under modal operators (see Lec 12).

Why counterfactuals?

Counterfactuals (“if it had been that p, it would have been that q,” written p □→ q) matter for:

  1. the analysis of causation, since a causal relation might just be counterfactual dependence;
  2. decision theory, which compares actions by what would happen under each.

Lewis’s theoretical ambitions differ from everyone we’ve read so far:

  • The logical project: how can natural language make for valid arguments? Which inference rules are correct for counterfactuals? Some counterfactual inferences are clearly well supported; others just seem faulty.
  • The metaphysical project: what are counterfactuals really about? If they’re claims about other worlds, what kind of evidence could we have for them?

The two projects are closely intertwined.

From the material conditional to spheres of similarity

  • Material conditional: p ⊃ q is true iff p is false or q is true. It validates modus ponens, but it makes every counterfactual with a false antecedent true. Kangaroos do have tails, so “if kangaroos didn’t have tails, they’d have wings” comes out true.
  • Strict conditional: □(p ⊃ q), the material conditional holding in every accessible world. This rules out accidental connections, but it fails on Sobel sequences: “If kangaroos had no tails, they’d topple over; but if they had no tails and used crutches, they wouldn’t.” No single set of accessible worlds makes both halves true non-vacuously. Equivalently, strict conditionals validate strengthening the antecedent, and counterfactuals don’t.

Lewis’s fix uses two pieces of machinery: possible worlds and a similarity ranking. A system of spheres $ nests worlds around the actual world @, with more similar worlds in smaller spheres. Every counterfactual is evaluated against the same ranking, but each one looks at a different region of it.

Lewis’s semantics

p □→ q is true iff either there are no accessible p-worlds (vacuous truth), or q is true in all the closest p-worlds (the p-worlds in the smallest sphere that contains any). The “might” counterfactual ◇→ gets a different analysis.

Features of the ranking show up as logical properties:

  • Ties are allowed. Take “If Bizet and Verdi had been compatriots, they’d have been French” and “…they’d have been Italian.” The closest compatriot worlds are tied, some French and some Italian, so both come out false, as we intuitively judge.
  • @ is at the top. That’s what licenses modus ponens with counterfactuals. From p □→ q, every closest p-world is a q-world. From p, @ is a p-world, and since it’s at the top it’s one of the closest. So @ is a q-world, i.e. q.
  • No transitivity. Indicative conditionals chain; counterfactuals don’t:
    • If Harvard is in Cambridge, it’s in MA; if it’s in MA, it’s in the US; so if it’s in Cambridge, it’s in the US. valid
    • If Hoover had been born Russian, he’d have been a communist; if he’d been a communist, he’d have been a traitor; so if he’d been born Russian, he’d have been a traitor. invalid
    (The Hoover inference seems valid to me, though? Maybe it turns on “traitor to whom.”)

The metaphysical project: modal realism

Lewis is a modal realist: other possible worlds are just as real as ours, and “actual” just means this one. Objections:

  • The incredulous stare. This is the most common reaction. But everyone has to take some things for granted, and this objection is just people being content to presuppose the explanandum.
  • Epistemology. If other worlds are causally isolated from us, how can we know anything about them? Evaluating counterfactuals seems to require contingent truths about other worlds, but we learn contingent truths only through perception, which is causal.
  • Ethics. If every possible world is real, then whatever we do, the bad things still happen somewhere. Are we just moving harm and benefit around?

Ersatzism (Ersatz = substitute) is an alternative. The semantics doesn’t need real worlds, only some points of evaluation, a $, and a similarity notion. So you can keep Lewis’s logic and give worlds a different foundation, e.g. as stand-ins (Stalnaker 1976).

Is similarity the right notion?

Similarity is super weird: every world is similar to ours in some ways and dissimilar in others. Ordinarily we have to specify a respect of comparison, but Lewis fixes his spheres by overall similarity, without one.

  • Caesar: “If Caesar had fought in Vietnam, he’d have used catapults” vs. “…he’d have dropped an atomic bomb.” Each preserves a different invariant. So maybe different counterfactuals need different similarity notions, set by context?
  • Nixon and the button: Nixon is told the USSR has launched a nuclear attack and must decide whether to push “the button”; then it turns out to be a false alarm. “If Nixon had pushed the button, there’d have been a nuclear holocaust” seems true. But isn’t the most similar world one where the button just malfunctions, since the rest of history then stays the same?

Lecture 11Quine: three grades of modal involvement

Quine, “Three Grades of Modal Involvement” (1953)

Quine’s suspicion of modality

Some features of a thing are contingent: Alan Garber is President of Harvard, but he might not have been. Others are supposedly essential: features it couldn’t lose and still exist. Many philosophers treat possibility and necessity as basic concepts and explain other things in terms of them.

Quine prefers a “desert landscape” to the “jungle”: the absolute minimum. That preference comes from a sense of futility. Every few hundred years people reject metaphysics as making no progress.

  • Is rationality essential to humans? You construct an edge case, revise, find more cases, and so on. After all that, are we getting anything interesting? Coming up with ever more complicated theories seems degenerating, unlike in physics.
  • Hume went all the way: a book with neither abstract reasoning about quantity or number nor experimental reasoning about matters of fact should be committed to the flames as sophistry and illusion.

But Hume is too extreme. We do need modal talk: natural science needs it to tell a law from a mere correlation. (I don’t like this: can’t you do all of natural science without a grounding theory of modality??)

So Quine wants a reconstruction of necessity: an analysis that keeps the parts of our modal talk that commit us to the right things and jettisons the rest. Compare “nations”: we ascribe some really weird properties to them, and we can reconstruct the notion by analyzing it from the ground up. Quine’s verdict on the three grades:

  1. necessity as a predicate of sentences: good;
  2. necessity as an operator on statements: reducible to the first, but tempting;
  3. quantifying into modal contexts: bad, because it leads back into the jungle of Aristotelian essentialism.

Kripke’s Naming and Necessity is aimed squarely at this (see Lec 12).

First grade: necessity as a semantical predicate

Start with quotation. “Cicero was a Roman orator” is about the man. “‘Cicero’ has six letters” is about the name, and it’s true regardless of anything about the man. So substitution fails inside quotes.

Substitution fails for necessity in exactly the same way:

Necessity

  • It’s necessary that 9 > 5.
  • The number of planets is 9.
  • ✗So it’s necessary that the number of planets > 5.

Quotation

  • “9 is greater than 5” contains two numerals.
  • The number of planets is 9.
  • ✗So “the number of planets is greater than 5” contains two numerals.

(The necessity conclusion is arguably ambiguous, though. “Necessarily, however many planets there are, there are more than 5” is false. “The number that is in fact the number of planets (9) is necessarily > 5” is true.)

So Quine’s idea is that necessity is a property of sentences: provability in a logical system, i.e. axioms plus rules for proving further theorems. Read “it is necessary that…” as “this sentence is provable: …”. That makes necessity a semantical predicate, true of sentences just as “contains two numerals” is.

  • A standard system that can prove arithmetic proves “9 > 5” but can’t even process “planets.” That gives the right truth values.
  • Is Alan Garber essentially rational? “Garber is rational” isn’t provable, but “The rational President of Harvard is rational” is. So the question only makes sense relative to a way of speaking about him.

Problems for the first grade

Nickel: there are no essences here at all. On Quine’s view, if you refer to Garber with a term that includes rationality, he’s “necessarily” rational; otherwise he isn’t. So the terms have these features, and the features don’t carry over to the person. It’s like an Anglicized name: having an English name doesn’t make someone English. So it’s better to say that on the first grade there’s no such thing as essential properties.

Which system? Provability is relative to a system, and even laws of nature get relativized: Newton’s laws are provable in some systems and not in others. We can even build systems where it’s not necessary that 9 > 5. Why not build a system with the axiom “x is human ⊃ □(x is rational)”? (Can’t we construct whatever system we need for every claim? Then necessity basically doesn’t exist.) So how do we judge one system better than another? Quine rejects the distinction between what we stipulate to be true and what we have good reason to believe. Is this a metaphysical question? Quine doesn’t want those.

Third grade: quantifying into modal contexts

“Somebody stole my cookies” is true iff some object satisfies “x stole my cookies,” regardless of how we refer to it. If Cookie Monster is the blue muppet, “Cookie Monster stole my cookies” and “The blue muppet stole my cookies” must agree in truth value. Quine calls this quantification objectual.

Now quantify into necessity: “Someone is such that it is necessary that they are rational.” That’s true iff some object is necessarily rational, however we refer to it. So it assumes objects have essential properties in themselves: Aristotelian essentialism. (But why must these properties be necessary?)

The same route takes you from a = b to □(a = b):

  1. ∀x □(x = x).
  2. Hesperus = Phosphorus, so by substitution, □(H = P).
  3. But then there’s no world where H ≠ P. That seems wrong, since we learned H = P, which seemed to narrow down the range of worlds.

Lecture 12Kripke: naming and necessity I

Kripke, Naming and Necessity (1972), Lecture I

Why Kripke was revolutionary

Naming and Necessity has three themes:

  1. the return of metaphysics, against Quine’s “Three Grades”;
  2. the rejection of descriptivist theories of reference (Frege, Russell) in favor of direct reference;
  3. the causal theory of reference: what we say depends on how we’re embedded in the social and natural world.

It attacks descriptivism with three arguments: modal, epistemic, and semantic. Lecture I is the modal one. But the result runs into questions. With no role for anything like Fregean sense, what about informative identities, belief ascriptions, and communication?

The Description Theory

When I talk about Aristotle, what makes my intention about the philosopher and not the shipping magnate? The natural answer: I have descriptive information in mind that identifies him, and if you asked for clarification I could give it. Indeed, if I can’t answer, it seems I’m not really talking about anyone. Suppose “wrote the Odyssey and Iliad” is the only description I have of Homer. Then if I say “maybe Homer didn’t write the Odyssey and Iliad,” I don’t seem to be speaking about anyone.

Description Theory (handout)
  1. (i) A name “X” has a corresponding cluster of properties: the φ such that the speaker believes “φX.”
  2. (ii) One or more of them are believed to pick out a unique individual.
  3. (iii) If most (or a weighted most) of the φs are satisfied by a unique object y, y is the referent.
  4. (iv) If there’s no unique object, “X” doesn’t refer.
  5. (v) “If X exists, X has most of the φs” is known a priori.
  6. (vi) …and it expresses a necessary truth.
  7. (C) The account can’t be circular: the properties can’t themselves involve reference ineliminably.
  • (v) is the epistemic side. If I know “Aristotle” carries the description “philosopher,” then I know a priori that Aristotle is a philosopher.
  • (vi) is the modal side, and both Kaplan-style and Quinean theorists end up committed to it:
    • Comfortable with possible worlds: “Necessarily, if there’s a last great philosopher of antiquity, the LGPoA is a philosopher” is true. If “Aristotle” abbreviates that description (Russell), then “Necessarily, if Aristotle exists, Aristotle is a philosopher” is true as well.
    • Quine: “If Aristotle exists, Aristotle is a philosopher” just is the description sentence, which is provable, and provability is all necessity is.

The modal argument

  • (7) It is necessary that the last great philosopher of antiquity was a philosopher. true
  • (8) It is necessary that Aristotle was a philosopher. false

So they can’t mean the same thing. Done! “Necessary” works like universal quantification over worlds: every world must conform, and in some world Aristotle never went into philosophy. “The LGPoA” picks out different people in different worlds; “Aristotle” doesn’t.

Rigid designator

An expression e is a rigid designator iff e refers to the same object in every possible world in which e refers at all. It is non-rigid iff it refers to different objects in at least two possible worlds.

The argument:

  1. Ordinary proper names are rigid.
  2. The descriptions a descriptivist could plausibly associate with them are non-rigid.
  3. So “necessarily, X is the φ” is false, and (vi) is false.

Why “plausibly”? Descriptions that pick out an object by its essential properties are rigid, like “the natural number between 2 and 4.” But we rarely know enough about essences to pick anything out that way.

Responses: limiting the damage

  1. Wide scope. (7) might be false too, read as “the LGPoA, you got that guy? It’s necessary that he was a philosopher.” It has two readings, just like “the number of planets” in Lec 11:
    • It’s necessary that the LGPoA, whoever that may be, was a philosopher. true
    • The LGPoA is such that it’s necessary that they were a philosopher. false
    The descriptivist says a name’s description obligatorily takes wide scope over modal operators, so (8) gets only the false reading. Kripke’s reply (Preface): the argument doesn’t need □ at all. The embedded sentences “The LGPoA was a philosopher” and “Aristotle was a philosopher” differ in truth value at some worlds, so they can’t mean the same thing.
  2. Actualized descriptions. Adding “actually” (“the person who was actually the LGPoA”) makes the description rigid.

Both responses concede that (vi) is false; they just limit the damage.

Where does this leave Quine?

Quine would say that talk of essences is just talk, and reflects a mistaken view of the world. Kripke’s point is that people can make sense of “that guy might have lost” (pointing at Nixon) without specifying which description they’re using. Someone who replies “only if you describe him as ‘Nixon,’ not as ‘the winner’” is the one with a philosophical theory. So there’s intuitive content to objects having modal properties, and the burden of proof shifts: Quine can’t treat his view as the default.

Nickel thinks this undersells Quine and oversells Kripke.

  • Psychological essentialism (Gelman 2003): there’s a developmental stage where children think lots of things have essential properties (“why are mountains high? so they’re for climbing!”). We have a strong tendency to ascribe essences to things that clearly don’t have them; some adults ascribe essences to race. That doesn’t mean they’re right. Our intuitions can systematically misfire.
  • There’s no tabula rasa: like Goodman, we need to take some (but not all) things for granted. But an intuition is a presumption, not at all conclusive. We keep intuitions until they’re rejected.

Math 131

Topology

Topic 1Topological spaces and bases

Munkres §12–13 · Lecture 1

Topology is the study of shape, and of the minimum structure needed to talk about shape.

Topologies and open sets

Topology

Let $X$ be a set. A topology on $X$ is a collection $\mathcal U$ of subsets of $X$ (its elements are called open sets) such that:

  1. $\varnothing, X \in \mathcal U$;
  2. any union of open sets is open;
  3. finite intersections of open sets are open.

A topological space is a set together with a topology on it.

Neighborhood

A neighborhood of $x \in X$ is an open set $U$ containing $x$.

Finer, coarser, comparable

Topologies $\mathcal U_1, \mathcal U_2$ on the same $X$ are comparable if $\mathcal U_1 \subset \mathcal U_2$ or $\mathcal U_2 \subset \mathcal U_1$. The larger one is finer, the smaller one coarser. They are equal if both containments hold.

Examples
  • Standard topology on $\mathbb R$: the open sets are unions of open intervals $(a,b) = \{x \in \mathbb R : a \lt x \lt b\}$.
  • Discrete topology on any set $X$: $\mathcal U = \mathcal P(X)$, so all subsets are open. Points are isolated: $\{x\}$ is a neighborhood of $x$ by itself.
  • Finite-complement topology $\mathbb R_{fc}$: the open sets are $\varnothing$ and complements of finite sets. Axiom 2 holds because a union just “fills in holes”; axiom 3 holds because a finite intersection just creates finitely many more holes.
  • WS 1Indiscrete (trivial) topology on any set: only $\varnothing$ and $X$ are open.
  • WS 1Countable-complement topology: the open sets are $\varnothing$ and the sets whose complement is countable.
  • Comparisons: discrete is finer than every other topology; finite-complement is coarser than standard.
  • WS 1On $X = \{a,b,c\}$, $\{\varnothing, \{b\}, \{a,b\}, \{b,c\}, X\}$ is a topology, but $\{\varnothing, \{a\}, \{b\}, X\}$ is not: the union $\{a,b\}$ is missing. Topologies needn’t be comparable: $\{\varnothing, \{a\}, X\}$ and $\{\varnothing, \{b\}, X\}$ aren’t.

Bases and subbases

Usually we define a topology by giving a basis.

Basis

A basis for a topology on $X$ is a collection $\mathcal B$ of subsets of $X$ such that:

  1. (B1) $\bigcup_{B \in \mathcal B} B = X$ (the basis elements cover $X$);
  2. (B2) if $B_1, B_2 \in \mathcal B$ and $x \in B_1 \cap B_2$, then there is $B_3 \in \mathcal B$ with $x \in B_3 \subset B_1 \cap B_2$.

The topology generated by $\mathcal B$, $\mathcal U_{\mathcal B}$, is the collection of all unions of basis elements.

Propositionproof ↓

A basis generates a topology. If $\mathcal B$ is a basis, then $\mathcal U_{\mathcal B}$ is a topology on $X$.

Subbasis

A subbasis for a topology on $X$ is any collection $\mathcal S$ of subsets of $X$ whose union is $X$. It generates a topology by taking all finite intersections of elements of $\mathcal S$, then all unions of those.

Examples
  • The singletons $\{x\}$ form a basis for the discrete topology.
  • Any topology is a basis for itself.
  • The open intervals $(a,b)$ form a basis for the standard topology on $\mathbb R$.

Comparing topologies; two more topologies on ℝ

Lemma 13.3AddedWS 2proof ↓

Comparing via bases. Let $\mathcal B, \mathcal B'$ be bases for topologies $\mathcal U, \mathcal U'$ on $X$. Then $\mathcal U'$ is finer than $\mathcal U$ iff for each $x$ and each $B \in \mathcal B$ containing $x$, there is $B' \in \mathcal B'$ with $x \in B' \subset B$. In particular, it suffices to check that every element of $\mathcal B$ is open in $\mathcal U'$.

Lower limit topology $\mathbb R_\ell$WS 2

The topology on $\mathbb R$ with basis all half-open intervals $[a,b) = \{x : a \le x \lt b\}$.

K-topology $\mathbb R_K$WS 3

Let $K = \{1/n : n \in \mathbb N\}$. The topology on $\mathbb R$ with basis all open intervals $(a,b)$ and all sets $(a,b) - K$.

Lemma 13.4AddedWS 2WS 3proof ↓

$\mathbb R_\ell$ and $\mathbb R_K$ vs. standard. $\mathbb R_\ell$ and $\mathbb R_K$ are each strictly finer than the standard topology, but they are not comparable with each other.

Topic 2Metric and subspace topologies

Munkres §16, §20 · Lectures 1–2

Metric spaces

Metric space

A set $X$ is a metric space if there is a distance function $d : X \times X \to \mathbb R_{\ge 0}$ such that for all $x,y,z \in X$:

  • $d(x,y) = d(y,x)$;
  • $d(x,y) \ge 0$, and $d(x,y) = 0 \iff x = y$;
  • $d(x,y) \le d(x,z) + d(z,y)$ (triangle inequality).
Metric topology

The metric topology on $(X,d)$ is the topology generated by the open balls $B(x,\varepsilon) = \{y \in X : d(x,y) \lt \varepsilon\}$.

Examples
  • Added$\mathbb R$ with $d(x,y) = |x-y|$: the balls are the open intervals, so the metric topology is the standard topology.
  • On $\mathbb R^2$ the Euclidean metric topology equals the product topology (see Topic 6).
  • WS 1WS 2Not every topology comes from a metric. Metric topologies are Hausdorff (Topic 4), so $\mathbb R_{fc}$ and the indiscrete topology on two or more points aren’t metric. On a finite set, a metric topology is discrete: take $\varepsilon$ below the smallest distance.
  • WS 1The discrete metric, $d(x,y) = 1$ for $x \ne y$ and $d(x,x) = 0$, gives the discrete topology: $B(x, \tfrac12) = \{x\}$.

Subspaces

Subspace topology

Let $X$ be a topological space with topology $\mathcal U_X$, and let $S \subset X$ be any subset. The subspace topology on $S$ is $$\mathcal U_{S} = \{\, U \cap S : U \in \mathcal U_X \,\}.$$

Examples
  • AddedIn $Y = [0,1] \subset \mathbb R$, the set $[0,\tfrac12) = (-1,\tfrac12) \cap Y$ is open in $Y$, though it isn’t open in $\mathbb R$.
  • $X = \mathbb R^2$ and $S$ = the Sierpiński triangle: the subspace topology gives something interesting.
  • WS 2In $\mathbb R_{fc} \times \mathbb R_{fc}$, the line $y = -x$ gets the finite-complement topology.
  • WS 2In $\mathbb R_\ell \times \mathbb R_\ell$, the line $y = -x$ gets the discrete topology, since $[x, x+\varepsilon) \times [-x, -x+\varepsilon)$ meets it only at $(x,-x)$. The line $y = x$ gets the lower limit topology.
  • WS 2The line $L = \{y = -x\} \subset \mathbb R^2$ gets the standard topology of $\mathbb R$ via $x \mapsto (x,-x)$: a ball in $\mathbb R^2$ meets $L$ in an open segment, and every open segment of $L$ arises this way.

Topic 3Closed sets, closure, and limit points

Munkres §17 · Lecture 3

Closed sets

Closed set

A subset $A$ of a topological space $X$ is closed if $X - A$ is open.

Theorem 17.1proof ↓

Let $X$ be a topological space. Then:

  1. $\varnothing$ and $X$ are closed;
  2. arbitrary intersections of closed sets are closed;
  3. finite unions of closed sets are closed.
Theorem 17.2proof ↓

Let $Y$ be a subspace of $X$. Then $A$ is closed in $Y$ iff $A = C \cap Y$ for some closed set $C$ of $X$.

Theorem 17.3proof ↓

Let $Y$ be a subspace of $X$. If $A$ is closed in $Y$ and $Y$ is closed in $X$, then $A$ is closed in $X$.

Interior and closure

Interior and closure

The interior $\operatorname{int} A$ is the union of all open sets contained in $A$: the largest open set inside $A$.

The closure $\bar A$ is the intersection of all closed sets containing $A$: the smallest closed set containing $A$.

So $\operatorname{int} A \subset A \subset \bar A$.

Theorem 17.4proof ↓

Let $Y$ be a subspace of $X$, let $A \subset Y$, and let $\bar A$ be the closure of $A$ in $X$. Then the closure of $A$ in $Y$ equals $\bar A \cap Y$.

Theorem 17.5proof ↓

Let $A$ be a subset of a topological space $X$.

  1. $x \in \bar A$ iff every open set $U$ containing $x$ intersects $A$.
  2. If the topology on $X$ is given by a basis, then $x \in \bar A$ iff every basis element $B$ containing $x$ intersects $A$.

Limit points

Limit point

If $A \subset X$ and $x \in X$, then $x$ is a limit point of $A$ if every neighborhood of $x$ intersects $A$ in some point other than $x$.

Theorem 17.6proof ↓

Let $A \subset X$ and let $A'$ be the set of all limit points of $A$. Then $\bar A = A \cup A'$.

Corollary 17.7Addedproof ↓

A subset of a topological space is closed iff it contains all its limit points.

FactWS 3proof ↓

Closures of unions. For subsets $A_\alpha \subset X$, $\;\bigcup_\alpha \bar A_\alpha \subset \overline{\bigcup_\alpha A_\alpha}$. Equality holds for finite unions, but can fail for infinite ones.

ExamplesAdded
  • In $\mathbb R$: $\overline{\mathbb Q} = \mathbb R$ and $\operatorname{int}\mathbb Q = \varnothing$.
  • In $\mathbb R$, for $A = \{1/n : n \in \mathbb N\}$: $A' = \{0\}$, so $\bar A = A \cup \{0\}$ and $A$ is not closed.
  • In a discrete space every set is both open and closed, and no point is a limit point of anything.
  • WS 3$K = \{1/n\}$ in four topologies on $\mathbb R$. Standard: neither open nor closed, $\bar K = K \cup \{0\}$. $\mathbb R_K$: closed, not open, $\bar K = K$. $\mathbb R_\ell$: neither, $\bar K = K \cup \{0\}$ (every $[0,\varepsilon)$ meets $K$). $\mathbb R_{fc}$: neither, $\bar K = \mathbb R$ (the only infinite closed set is $\mathbb R$).

Topic 4Hausdorff spaces and convergence

Munkres §17 · Lecture 3

Definitions

Hausdorff space

A topological space $X$ is Hausdorff if for each pair of distinct points $x_1 \ne x_2$ there are neighborhoods $U_1$ of $x_1$ and $U_2$ of $x_2$ that are disjoint.

Convergence

A sequence $x_1, x_2, \ldots \in X$ converges to $x \in X$ if for each neighborhood $U$ of $x$ there is $N$ such that $x_n \in U$ for all $n \ge N$.

$T_1$ axiom

$X$ satisfies the $T_1$ axiom if every finite point set is closed.

FactWS 3proof ↓

$T_1$ means points are closed. $X$ is $T_1$ iff every single point is closed, iff for all $x \ne y$ there is an open set containing $x$ but not $y$.

Results

Theorem 17.8proof ↓

Every finite point set in a Hausdorff space is closed. (So Hausdorff spaces are $T_1$.)

Theorem 17.9proof ↓

Let $X$ be $T_1$ and $A \subset X$. Then $x$ is a limit point of $A$ iff every neighborhood of $x$ contains infinitely many points of $A$.

Theorem 17.10proof ↓

If $X$ is Hausdorff, then a sequence of points of $X$ converges to at most one point of $X$.

Theorem 17.11Addedproof ↓

The product of two Hausdorff spaces is Hausdorff, and a subspace of a Hausdorff space is Hausdorff. (Munkres also includes: every simply ordered set is Hausdorff in the order topology.)

FactPS 2

Hausdorff via the diagonal. $X$ is Hausdorff iff the diagonal $\Delta = \{(x,x) : x \in X\}$ is closed in $X \times X$. (Munkres §17 #13)

ExamplesAdded
  • Every metric space is Hausdorff: if $d(x,y) = r \gt 0$, the balls $B(x, r/2)$ and $B(y, r/2)$ are disjoint.
  • $\mathbb R_{fc}$ is $T_1$ but not Hausdorff: any two nonempty open sets intersect. In it, the sequence $1, 2, 3, \ldots$ converges to every point. PS 2So does $x_n = 1/n$ (§17 #14).
  • WS 3$\mathbb R_\ell$ and $\mathbb R_K$ are Hausdorff: any topology finer than a Hausdorff one is Hausdorff.

Topic 5Continuous functions and homeomorphisms

Munkres §18

Continuity

Continuous function

Let $X, Y$ be topological spaces. A function $f : X \to Y$ is continuous if for each open subset $V$ of $Y$, the set $f^{-1}(V)$ is an open subset of $X$. (Note that $f^{-1}$ means preimage here, not an inverse function.)

Checking on a basis or subbasis. If $Y$ has basis $\mathcal B$, it suffices to show that the preimage of every basis element is open. If $Y$ has subbasis $\mathcal S$, it suffices to show that the preimage of every subbasis element is open.

Theorem 18.1proof ↓

Let $X, Y$ be topological spaces and $f : X \to Y$. The following are equivalent:

  1. $f$ is continuous;
  2. for every $A \subset X$, $f(\bar A) \subset \overline{f(A)}$;
  3. for every closed set $B$ of $Y$, $f^{-1}(B)$ is closed in $X$;
  4. for each $x \in X$ and each neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ with $f(U) \subset V$.
FactPS 2

Maps out of $\mathbb R_{fc}$. If $X \subset \mathbb R_{fc}$ is infinite (subspace topology), every continuous $f : X \to \mathbb R$ is constant.

Homeomorphisms and embeddings

Homeomorphism

If $f : X \to Y$ is a bijection, $f$ is continuous, and $f^{-1}$ is also continuous, then $f$ is a homeomorphism. When a homeomorphism from $X$ to $Y$ exists, $X$ and $Y$ are homeomorphic.

This means $V$ is open iff $f^{-1}(V)$ is open, so $f$ gives a one-to-one correspondence between the open sets of $X$ and those of $Y$. It is the key notion of equivalence in topology.

Embedding

If $f : X \to Z$ is continuous and injective, and $f$ is a homeomorphism from $X$ onto $Y = f(X) \subset Z$ with the subspace topology, then $f$ is an embedding of $X$ into $Z$.

It’s much harder to show two spaces are not homeomorphic: you have to rule out every possible map.

Examples
  • $\mathbb R$ (standard) and $(0,1) \subset \mathbb R$ (subspace) are homeomorphic, via something like the tangent function: $x \mapsto \tan\big(\pi(x - \tfrac12)\big)$ from $(0,1)$ to $\mathbb R$.
  • PS 2Continuous maps needn’t preserve limit points: if $x$ is a limit point of $A$, $f(x)$ need not be a limit point of $f(A)$ (§18 #2).
  • WS 4$\mathbb R^2$ is homeomorphic to the open unit disk via $x \mapsto x / (1 + |x|)$, with inverse $y \mapsto y / (1 - |y|)$.
  • Non-example. $f : \mathbb R \to \mathbb R_K$, $x \mapsto x$, is not a homeomorphism, because $K = \{1/n : n \in \mathbb N\}$ is closed in $\mathbb R_K$ but not in $\mathbb R$. This only rules out this one map, not every map. Added$\mathbb R_K$ has as basis all open intervals $(a,b)$ and all sets $(a,b) - K$.
  • Non-example. Let $f : \mathbb R \to \mathbb R^2$ be continuous and injective, with image a loop whose far end comes back and gets infinitely close to the crossing point $f(x_0) = (0,0)$ (it “slows down”). Give the image $Y$ the subspace topology. Then $f^{-1} : Y \to \mathbb R$ is not continuous. The preimage of a small neighborhood of $(0,0)$ in $Y$ is an interval around $x_0$ plus a tail $(t, \infty)$, so a small interval around $x_0$ alone is not the preimage of any open set of $Y$. So $f$ is not an embedding. WS 4The image $Y$ is still Hausdorff, as is any subspace of $\mathbb R^2$.
  • Let $X = \{a,b,c\}$ with the discrete topology. Sending $a, b, c$ to any three distinct points embeds $X$ in $\mathbb R^2$.
  • $\mathbb R_{fc}$ and $\mathbb R^2_{fc}$ are homeomorphic: any bijection $f : \mathbb R \to \mathbb R^2$ works, and one exists since the two sets have the same cardinality. If $U \ne \varnothing$ is open in $\mathbb R_{fc}$, then $U = \mathbb R - F$ with $F$ finite, so $f(U) = \mathbb R^2 - f(F)$ is open too. The same argument works the other way.
  • $\mathbb Z \subset \mathbb R$ and $\mathbb Q \subset \mathbb R$ (subspace topologies) are not homeomorphic. $\mathbb Z$ is discrete, so every point is isolated, but no point of $\mathbb Q$ is isolated: every open set in $\mathbb Q$ contains infinitely many points. If $g : \mathbb Z \to \mathbb Q$ were a homeomorphism, $\{0\}$ is open in $\mathbb Z$ but $g(\{0\})$ is not open in $\mathbb Q$.
  • $\mathbb Q$ vs. $\mathbb Q \cup \{\sqrt2\}$ (left without a conclusion in my notes). AddedThese are homeomorphic. The isolated-point argument can’t tell them apart, and by Sierpiński’s theorem every countable metric space with no isolated points is homeomorphic to $\mathbb Q$.

Topic 6The product topology

Munkres §15, §19 · Lecture 2 and later

Finite products

Product topology on $X \times Y$

Let $X, Y$ be topological spaces with topologies $\mathcal U_X, \mathcal U_Y$. The product topology on $X \times Y$ is generated (as a basis) by the sets $U \times V$ with $U \in \mathcal U_X$, $V \in \mathcal U_Y$, which are called open rectangles.

Propositionproof ↓

Metric = product on $\mathbb R^2$. The metric topology and the product topology on $\mathbb R^2$ are the same. This is the standard topology on $\mathbb R^2$.

Examples
  • In $\mathbb R^2 = \mathbb R \times \mathbb R$, the open sets literally are unions of rectangles (without their edges).
  • WS 2$\mathbb R_{fc} \times \mathbb R_{fc}$ is strictly finer than $\mathbb R^2_{fc}$. Complements of points are product-open, but $(\mathbb R - \{0\}) \times \mathbb R$ is product-open and its complement, a line, is infinite. Neither topology comes from a metric, since neither is Hausdorff.

Infinite products

Product topology on $\prod X_\alpha$

Let $X = \prod_{\alpha \in A} X_\alpha$, where $A$ is an index set. An element $x \in X$ is a function $x : A \to \bigcup_\alpha X_\alpha$ with $x(\alpha) \in X_\alpha$, written $x = (x_\alpha)_{\alpha \in A}$.

The product topology on $X$ is generated by the basis of sets $$\prod_{i \in I} U_i \times \prod_{\alpha \in A - I} X_\alpha, \qquad U_i \subset X_i \text{ open},\ I \text{ a finite set of indices}.$$ So all but finitely many factors must be the whole space.

Box topology

The box topology on $\prod_{\alpha \in A} X_\alpha$ has basis all products of open sets: $\big\{\prod_{\alpha \in A} U_\alpha : U_\alpha \subset X_\alpha \text{ open}\big\}$. For finite products it agrees with the product topology.

Projections are continuous. For each $\beta \in A$, the projection $\pi_\beta : \prod_\alpha X_\alpha \to X_\beta$, $(x_\alpha)_{\alpha \in A} \mapsto x_\beta$, is continuous.

Theorem 19.6proof ↓

Maps into products. Let $Z$ be a topological space and $f : Z \to \prod_{\alpha \in A} X_\alpha$ a function, with coordinates $f_\alpha : Z \to X_\alpha$. Then $f$ is continuous in the product topology iff every $f_\alpha$ is continuous. This is false for the box topology.

FactPS 3

Convergence in products. In $\prod X_\alpha$ with the product topology, $x_n \to x$ iff $\pi_\alpha(x_n) \to \pi_\alpha(x)$ for every $\alpha$. This fails in the box topology. (Munkres §19 #6)

Examples
  • $\mathbb R^\omega = \mathbb R^{\mathbb N} = \prod_{n \in \mathbb N} X_n$ with each $X_n = \mathbb R$ (a countable product). The infinite open cube $(0,1)^{\mathbb N}$ is not open in the product topology, but $\prod_{n \le N} (0,1) \times \prod_{n \gt N} \mathbb R$ is. In the box topology, $(0,1)^{\mathbb N}$ is open.
  • WS 5$\{0,1\}^{\mathbb N}$ with discrete factors: the product topology is not discrete, since basic open sets fix only finitely many coordinates, so no point is open. The box topology is discrete: $\{x\} = \prod_n \{x_n\}$.
  • PS 3Let $\mathbb R^\infty \subset \mathbb R^\omega$ be the sequences that are eventually zero. It is dense in the product topology but closed in the box topology (§19 #7).
  • AddedWhy 19.6 fails for the box topology: $f : \mathbb R \to \mathbb R^\omega$, $f(t) = (t, t, t, \ldots)$ has continuous coordinates, but $f^{-1}\big(\prod_n (-\tfrac1n, \tfrac1n)\big) = \{0\}$ is not open.

Topic 7Quotient spaces

Munkres §22

Quotient topology and quotient maps

Quotient topology

Let $X$ be a topological space, $S$ a set, and $q : X \to S$ surjective. The quotient topology on $S$ from $q$ is $$\{\, U \subset S : q^{-1}(U) \text{ is open in } X \,\},$$ and $q$ is called a quotient map. Think of this as pushing the topology forward from $X$ to $S$.

Quotient map

If $f : X \to Y$ is a map between topological spaces, $f$ is a quotient map if:

  1. it is surjective;
  2. it is continuous;
  3. the topology on $Y$ coincides with the quotient topology from $f$. That is, $V \subset Y$ is open iff $f^{-1}(V)$ is open.
Saturated set

$A \subset X$ is saturated (for $q$) if $A = q^{-1}(q(A))$, i.e. $A$ is itself the full preimage of a set, not just a piece of one.

Quotient maps via saturated sets. A continuous surjection $f : X \to Y$ is a quotient map iff $f(U)$ is open in $Y$ for every saturated open set $U \subset X$. This is the same as condition 3 above.

Theorem 22.2AddedPS 3proof ↓

Maps out of a quotient. Let $p : X \to Y$ be a quotient map, and let $g : X \to Z$ be constant on each set $p^{-1}(y)$. Then $g$ induces $f : Y \to Z$ with $f \circ p = g$. Moreover, $f$ is continuous iff $g$ is, and $f$ is a quotient map iff $g$ is.

Corollary 22.3AddedPS 3proof ↓

Recognizing a quotient. Let $g : X \to Z$ be a continuous surjection, and let $X^* = \{g^{-1}(z) : z \in Z\}$ with the quotient topology. Then $g$ induces a continuous bijection $g^* : X^* \to Z$, which is a homeomorphism iff $g$ is a quotient map.

FactPS 3

Sections and retractions give quotient maps. If $p : X \to Y$ is continuous and there is a continuous $f : Y \to X$ with $p \circ f = \mathrm{id}_Y$, then $p$ is a quotient map. In particular, a retraction $r : X \to A$ (continuous, with $r(a) = a$ for all $a \in A$) is a quotient map. (Munkres §22 #2)

Examples
  • $f : \mathbb R \to [0,\infty)$, $f(x) = x^2$, with the subspace topology on $[0,\infty)$. Consider two choices of domain topology: (1) $\mathbb R$ standard, (2) $\mathbb R$ discrete. In both cases $f$ is continuous and surjective. For continuity in (1), the preimage of $(a,b)$ with $0 \le a \lt b$ is $(-\sqrt b, -\sqrt a) \cup (\sqrt a, \sqrt b)$. But (1) is a quotient map and (2) is not. In (2), $\{-1, 1\}$ is saturated and open, but $f(\{-1,1\}) = \{1\}$ is not open.

Quotients by equivalence relations

Quotient space $X/{\sim}$

Let $X$ be a topological space and $\sim$ an equivalence relation on $X$: $x \sim x$; $x \sim y \iff y \sim x$; and $x \sim y,\ y \sim z \Rightarrow x \sim z$. Let $[x] = \{y \in X : y \sim x\}$ be the equivalence class of $x$.

The quotient space $Y = X/{\sim}$ is the set of all equivalence classes with the quotient topology from $q : X \to X/{\sim}$, $q(x) = [x]$. Its open sets are $\{U \subset X/{\sim} : q^{-1}(U) \text{ open in } X\}$, i.e. sets of equivalence classes whose union is open in $X$.

Propositionproof ↓

$\mathbb R/\mathbb Z$ is the circle. On $X = \mathbb R$, define $x \sim y$ iff $x - y \in \mathbb Z$, so $[x] = \{x + n : n \in \mathbb Z\}$. Then $Y = X/{\sim}$ is homeomorphic to $S^1 = \{(x,y) \in \mathbb R^2 : x^2 + y^2 = 1\}$ (subspace topology), via the map induced by $f : \mathbb R \to \mathbb R^2$, $f(t) = (\cos 2\pi t, \sin 2\pi t)$.

Examples
  • In $\mathbb R/\mathbb Z$: $3.75 \sim 0.75$. Every equivalence class has a representative in $[0,1)$, and $0 \sim 1$, so it’s an interval with its ends glued: a circle.
  • PS 3$\mathbb R/\mathbb Z \cong S^1$ is an instance of Corollary 22.3.

Topic 8Connectedness

Munkres §23–25 · Lecture 8 and earlier

Connected spaces

Separation

Let $X$ be a topological space. A separation of $X$ is a pair of nonempty open sets $U, V$ with $U \cup V = X$ and $U \cap V = \varnothing$.

Connected

$X$ is connected if it has no separation. A subset $S \subset X$ is connected if it is connected in the subspace topology.

Clopen criterion. $X$ is connected iff the only subsets of $X$ that are both open and closed (“clopen”) are $\varnothing$ and $X$.

Propositionproof ↓

$[0,1]$ is connected. $[0,1] \subset \mathbb R$ is connected.

Lemma 23.2Addedproof ↓

Connected sets stay on one side. If $C, D$ form a separation of $X$ and $Y \subset X$ is connected, then $Y \subset C$ or $Y \subset D$.

Examples
  • $X = [0,1] \subset \mathbb R$ is connected.
  • $Y = [0,1] \cup [2,3]$ is not connected. Take $U = (-\tfrac12, \tfrac32) \cap Y$ and $V = (\tfrac32, \tfrac72) \cap Y$: they are nonempty, disjoint, open in $Y$, and span $Y$, so they form a separation.

Building connected sets

Theorem 23.3Addedproof ↓

Unions with a common point. The union of a collection of connected subspaces of $X$ that have a point in common is connected.

FactPS 4

Chains of connected sets. If each $A_n \subset X$ ($n \in \mathbb N$) is connected and, for each $n \gt 1$, $A_n \cap A_m \ne \varnothing$ for some $m \lt n$, then $\bigcup_n A_n$ is connected.

Theorem 23.4WS 7proof ↓

Closures of connected sets. If $A$ is connected and $A \subset B \subset \bar A$, then $B$ is connected. In particular, the closure of a connected set is connected.

Theorem 23.5WS 7proof ↓

Continuous images. If $f : X \to Y$ is continuous and $A \subset X$ is connected, then $f(A)$ is connected.

Theorem 23.6WS 8proof ↓

Products. A finite product of connected spaces is connected. In fact any product of connected spaces is connected in the product topology (§23 #10), but not necessarily in the box topology.

Examples
  • WS 7Preimages of connected sets needn’t be connected: for $f(x) = x^2$ on $\mathbb R$, $f^{-1}((1,4)) = (-2,-1) \cup (1,2)$.
  • WS 8$\mathbb R^\omega$ with the product topology is path connected: $\gamma(t) = (1-t)x + ty$ is continuous because each coordinate is (Theorem 19.6). It is also locally connected, since basic open sets $\prod_{i \in I} (a_i, b_i) \times \prod \mathbb R$ are path connected in the same way.
  • WS 8$\mathbb R^\omega$ with the box topology is not connected: the bounded sequences and the unbounded sequences are both open, since the box $\prod_n (x_n - 1, x_n + 1)$ keeps you bounded or unbounded.

Connected subsets of ℝ

FactAddedproof ↓

Connected subsets of $\mathbb R$. A subset of $\mathbb R$ is connected iff it is an interval (or a point).

Theorem 24.3Addedproof ↓

Intermediate value theorem. Let $f : X \to \mathbb R$ be continuous with $X$ connected. If $a, b \in X$ and $r$ lies between $f(a)$ and $f(b)$, then $f(c) = r$ for some $c \in X$.

PropositionPS 4WS 4

Removing points. $(0,1)$, $(0,1]$ and $[0,1]$ are pairwise non-homeomorphic, and $\mathbb R$ is not homeomorphic to $\mathbb R^n$ for $n \gt 1$. (Munkres §24 #1)

FactPS 4

Connected metric spaces are big. A connected metric space with at least two points is uncountable.

Examples
  • PS 4$\mathbb R^2 - A$ is path connected for any countable $A$ (§24 #9).

Components and total disconnectedness

Connected components

The connected components of $X$ are its maximal nonempty connected subsets. The component of $x \in X$ is the union of all connected $C \subset X$ with $x \in C$.

Totally disconnected

$X$ is totally disconnected if its connected components are single points.

Caution: components are not necessarily both open and closed. AddedThey are always closed (by the closure fact above), but not always open. In $\mathbb Q$, for example, the components are single points, which are not open.

Examples
  • Any set with the discrete topology is totally disconnected.
  • The Cantor set $\mathcal C$ is totally disconnected. Intuitively, $\mathcal C = \bigcap_n C_n$, where each $C_n$ is a union of $2^n$ disjoint closed intervals, each of length $1/3^n$. To see that $\mathcal C$ is not connected, use $C_1 = [0,\tfrac13] \cup [\tfrac23, 1]$: $\mathcal C$ meets both pieces, so the two pieces cut down to $\mathcal C$ give a separation.
  • PS 4Subsets of $\mathbb R_{fc}$: a finite subset with at least two points is disconnected, and every infinite subset is connected.
  • WS 7$\mathbb R_\ell$ is totally disconnected: each $[a,b)$ is clopen, so $\{x \lt c\}$ and $\{x \ge c\}$ separate any set containing points on both sides of $c$. $\mathbb R_{fc}$ is connected: in an infinite space with the finite-complement topology, any two nonempty open sets intersect.

Path connectedness

Path connected

$X$ is path connected if any two points $x, y \in X$ can be joined by a path, i.e. there is a continuous $\gamma : [0,1] \to X$ with $\gamma(0) = x$ and $\gamma(1) = y$.

Propositionproof ↓

Path connected ⇒ connected. If a space is path connected, then it is connected. The converse is false, so path connectedness is the stronger condition.

Propositionproof ↓

$\mathbb R^n$ is path connected. $\mathbb R^n$ is path connected (hence connected).

Examples
  • Topologist’s sine curve (connected but not path connected). Let $\Gamma = \{(x, \sin(1/x)) : x \gt 0\}$ and $X = \bar\Gamma = \Gamma \cup (\{0\} \times [-1,1]) \subset \mathbb R^2$. $X$ is connected, since $\Gamma$ is connected and taking the closure keeps it connected. $X$ is not path connected: you can’t connect a point on the limit segment to anything else.

Local connectedness

Locally connected

$X$ is locally connected if it has a basis for its topology consisting of open connected sets. That is, for any neighborhood $V$ of $x$ there is an open connected neighborhood $U$ of $x$ with $U \subset V$.

Examples
  • $X = \{0, 1\} \subset \mathbb R$ is locally connected but not connected.
  • The topologist’s sine curve is not locally connected near the segment $\{0\} \times [-1,1]$: small neighborhoods there look like infinitely many separate strands.
  • Infinite comb (path connected and connected, but not locally connected): $$E = \big([0,1] \times \{0\}\big) \cup \big(\{0\} \times [0,1]\big) \cup \bigcup_{n \in \mathbb N} \{\tfrac1n\} \times [0,1].$$ It’s path connected because any two points can be joined by going down to the base, so it’s also connected. It’s not locally connected, because a small neighborhood of $p = (0,1)$ looks like infinitely many separate vertical pieces.

Topic 9Compactness

Munkres §26–27 · Lectures 9–10

Open covers and compactness

Open cover, subcover

Let $X$ be a topological space. A collection $\mathcal V$ of open sets in $X$ is an open cover if $\bigcup_{V \in \mathcal V} V = X$. A subcover is a subcollection $\mathcal V' \subset \mathcal V$ that also covers $X$. You can remove sets, but you can’t change them.

Compact

$X$ is compact if every open cover of $X$ has a finite subcover, i.e. there are $V_1, \ldots, V_n \in \mathcal V$ with $V_1 \cup \cdots \cup V_n = X$.

A subset $Y \subset X$ is compact if it is compact as a subspace. Equivalently, if $\mathcal V$ is a collection of sets open in $X$ with $\bigcup_{V \in \mathcal V} V \supset Y$, then there is a finite subcollection $V_1 \cup \cdots \cup V_n \supset Y$.

Examples
  • $\mathbb R$ is not compact: one open cover without a finite subcover is enough. Take $\mathcal V = \{(a, a+1) : a \in \mathbb R\}$.
  • $\mathbb R_{fc}$ is compact.
  • AddedIn $\mathbb R_{fc}$ every subset is compact (same argument). So $\mathbb N \subset \mathbb R_{fc}$ is compact but not closed.

Compact sets, closed sets, and Hausdorff spaces

Closed in compact is compact. A closed subset $Y$ of a compact space $X$ is compact.

Propositionproof ↓

Compact in Hausdorff is closed. Suppose $X$ is Hausdorff. If $Y \subset X$ is compact, then $Y$ is closed. In a general topological space, compact does not imply closed (see the $\mathbb R_{fc}$ example).

FactWS 9proof ↓

Separating compact sets. In a Hausdorff space, disjoint compact sets $A$ and $B$ have disjoint open neighborhoods $U \supset A$ and $V \supset B$. In particular, this holds for a point and a compact set.

FactAddedproof ↓

Nested closed sets. If $C_1 \supset C_2 \supset \cdots$ are nonempty closed sets in a compact space, then $\bigcap_n C_n \ne \varnothing$.

Theorem 27.3proof ↓

Heine–Borel. A subset $Y \subset \mathbb R^n$ (standard topology) is compact iff $Y$ is closed and bounded, i.e. there is $R \gt 0$ with $Y \subset B(0, R)$. “Bounded” is only defined in metric spaces.

Compactness and continuous maps

Theorems 26.5, 26.6proof ↓

Suppose $f : X \to Y$ is continuous and $X$ is compact.

  1. The image $f(X)$ is compact.
  2. If also $f$ is injective and $Y$ is Hausdorff, then $f$ is an embedding.
Theorem 27.7WS 10proof ↓

Perfect compact spaces are uncountable. If $X$ is a nonempty compact Hausdorff space with no isolated points, then $X$ is uncountable.

Theorem

Brouwer, 1910. If $X$ is a nonempty, compact, totally disconnected metric space with no isolated points, then $X$ is homeomorphic to $\mathcal C$, the standard middle-thirds Cantor set in $[0,1] \subset \mathbb R$. Recall that $\mathcal C \cong \{0,1\}^{\mathbb N}$.

Unpacking the hypotheses: “no isolated points” means no singletons are open, and “totally disconnected” means the connected components are singletons. Its proof won’t be tested.

ReferenceProofs

Proofs of the results above, in the same order. Where my notes only sketched a proof, or said to read it in the book, the rest is filled in and tagged.

Topologies and bases

A basis generates a topology. If $\mathcal B$ is a basis, then $\mathcal U_{\mathcal B}$ (all unions of basis elements) is a topology.

(1) $\varnothing \in \mathcal U_{\mathcal B}$ as the union over nothing. $X \in \mathcal U_{\mathcal B}$ by (B1).

(2) Suppose $\mathcal V \subset \mathcal U_{\mathcal B}$ is any subcollection. For each $V \in \mathcal V$ there is $\mathcal B_V \subset \mathcal B$ with $V = \bigcup_{B \in \mathcal B_V} B$. So $\bigcup_{V \in \mathcal V} V = \bigcup_{V \in \mathcal V} \bigcup_{B \in \mathcal B_V} B$ is a union of basis elements, and therefore in $\mathcal U_{\mathcal B}$.

(3) Added(My notes say to read the book’s proof, using (B2).) It suffices to do two sets, by induction. Let $U_1, U_2 \in \mathcal U_{\mathcal B}$ and $x \in U_1 \cap U_2$. Pick basis elements $B_1 \subset U_1$ and $B_2 \subset U_2$ containing $x$. By (B2) there is $B_x \in \mathcal B$ with $x \in B_x \subset B_1 \cap B_2 \subset U_1 \cap U_2$. Then $U_1 \cap U_2 = \bigcup_{x} B_x$ is a union of basis elements. ∎

ProofAdded↑ statement

Lemma 13.3. (⇐) Let $U \in \mathcal U$ and $x \in U$. There is $B \in \mathcal B$ with $x \in B \subset U$, and then $B' \in \mathcal B'$ with $x \in B' \subset B \subset U$. So $U$ is a union of elements of $\mathcal B'$, hence in $\mathcal U'$. (⇒) If $x \in B \in \mathcal B$, then $B \in \mathcal U \subset \mathcal U'$, so some $B' \in \mathcal B'$ has $x \in B' \subset B$. For the last claim: if every $B \in \mathcal B$ is $\mathcal U'$-open, every $\mathcal U$-open set, being a union of such $B$, is $\mathcal U'$-open. ∎

ProofAdded↑ statement

Lemma 13.4. $\mathbb R_\ell$: $(a,b) = \bigcup_{a \lt x \lt b} [x, b)$ is open in $\mathbb R_\ell$, so $\mathbb R_\ell$ is finer (Lemma 13.3). It is strictly finer, since $[a,b)$ contains no open interval around $a$. $\mathbb R_K$: every standard basis element is an $\mathbb R_K$ basis element, so $\mathbb R_K$ is finer. It is strictly finer, since no open interval around $0$ lies inside $(-1,1) - K$. Not comparable: $[10, 11)$ is open in $\mathbb R_\ell$ but not in $\mathbb R_K$, because every $\mathbb R_K$ basis element containing $10$ contains points left of $10$. And $(-1,1) - K$ is open in $\mathbb R_K$ but not in $\mathbb R_\ell$, because every $[0, \varepsilon)$ meets $K$. ∎

Closed sets, closure, and limit points

ProofAdded↑ statement

Theorem 17.1. $X - \varnothing = X$ and $X - X = \varnothing$ are open. By De Morgan, $X - \bigcap_\alpha A_\alpha = \bigcup_\alpha (X - A_\alpha)$, a union of open sets, and $X - \bigcup_{i=1}^n A_i = \bigcap_{i=1}^n (X - A_i)$, a finite intersection of open sets. Both are open. ∎

ProofAdded↑ statement

Theorem 17.2. (⇐) If $A = C \cap Y$ with $C$ closed in $X$, then $X - C$ is open in $X$, so $(X - C) \cap Y = Y - A$ is open in $Y$. Hence $A$ is closed in $Y$.

(⇒) If $A$ is closed in $Y$, then $Y - A$ is open in $Y$, so $Y - A = U \cap Y$ for some $U$ open in $X$. Then $C = X - U$ is closed in $X$ and $A = C \cap Y$. ∎

ProofAdded↑ statement

Theorem 17.3. By 17.2, $A = C \cap Y$ with $C$ closed in $X$. Since $Y$ is closed in $X$, $A$ is an intersection of two closed sets of $X$, so it is closed. ∎

ProofAdded↑ statement

Theorem 17.4. Let $B$ be the closure of $A$ in $Y$. $\bar A$ is closed in $X$, so $\bar A \cap Y$ is closed in $Y$ (17.2) and contains $A$. Hence $B \subset \bar A \cap Y$.

Conversely, $B$ is closed in $Y$, so $B = C \cap Y$ with $C$ closed in $X$ (17.2). Since $C \supset A$, we get $C \supset \bar A$, so $\bar A \cap Y \subset C \cap Y = B$. ∎

ProofAdded↑ statement

Theorem 17.5. (a) We prove the contrapositive: $x \notin \bar A$ iff some open $U \ni x$ misses $A$. If $x \notin \bar A$, then $U = X - \bar A$ is open, contains $x$, and misses $A$. If such a $U$ exists, $X - U$ is closed and contains $A$, so $\bar A \subset X - U$ and $x \notin \bar A$.

(b) Every open set containing $x$ contains a basis element containing $x$, and basis elements are open. So “every open set meets $A$” and “every basis element meets $A$” are equivalent. ∎

ProofAdded↑ statement

Theorem 17.6. (⊃) If $x \in A'$, every neighborhood of $x$ meets $A$, so $x \in \bar A$ by 17.5. Also $A \subset \bar A$.

(⊂) Let $x \in \bar A$. If $x \in A$ we’re done. If $x \notin A$, every neighborhood $U$ of $x$ meets $A$ (17.5), and since $x \notin A$, it meets $A$ in a point other than $x$. So $x \in A'$. ∎

ProofAdded↑ statement

Corollary 17.7. $A$ is closed iff $A = \bar A = A \cup A'$ (17.6), iff $A' \subset A$. ∎

ProofWS 3↑ statement

Closures of unions. (⊂) Each $A_\beta \subset \bigcup A_\alpha \subset \overline{\bigcup A_\alpha}$, which is closed, so $\bar A_\beta \subset \overline{\bigcup A_\alpha}$. Finite unions: $\bar A_1 \cup \cdots \cup \bar A_n$ is closed (finite union) and contains $\bigcup A_i$, so it contains $\overline{\bigcup A_i}$. Infinite unions: in $\mathbb R$ take $A_n = \{1/n\}$. Each is closed, so $\bigcup \bar A_n = K$, but $\overline{\bigcup A_n} = K \cup \{0\}$. This is also the mistake in Worksheet 3’s fake proof of the reverse inclusion: every neighborhood $U$ of $x$ meets some $A_\alpha$, but which $\alpha$ depends on $U$, so no single $A_\alpha$ need be met by every neighborhood. ∎

Hausdorff spaces and convergence

ProofAdded↑ statement

Theorem 17.8. Finite unions of closed sets are closed, so it suffices to show $\{x_0\}$ is closed. If $x \ne x_0$, Hausdorff gives disjoint neighborhoods $U \ni x$ and $V \ni x_0$. Then $U$ misses $\{x_0\}$, so $x \notin \overline{\{x_0\}}$ (17.5). Hence $\overline{\{x_0\}} = \{x_0\}$. ∎

ProofWS 3↑ statement

$T_1$ means points are closed. Finite unions of closed sets are closed, so “finite sets closed” is the same as “points closed.” (⇒) If $\{y\}$ is closed, then $X - \{y\}$ is open, contains $x$, and misses $y$. (⇐) For each $y \ne x$, pick an open $V_y \ni y$ missing $x$. Then $X - \{x\} = \bigcup_{y \ne x} V_y$ is open, so $\{x\}$ is closed. ∎

ProofAdded↑ statement

Theorem 17.9. (⇐) If every neighborhood of $x$ contains infinitely many points of $A$, it contains one other than $x$.

(⇒) Suppose $x \in A'$ but some neighborhood $U$ of $x$ meets $A - \{x\}$ in only finitely many points $x_1, \ldots, x_m$. Since $X$ is $T_1$, $X - \{x_1, \ldots, x_m\}$ is open. So $U \cap (X - \{x_1, \ldots, x_m\})$ is a neighborhood of $x$ that misses $A - \{x\}$, which contradicts $x \in A'$. ∎

ProofAdded↑ statement

Theorem 17.10. Suppose $x_n \to x$ and $y \ne x$. Take disjoint neighborhoods $U \ni x$ and $V \ni y$. $U$ contains $x_n$ for all $n \ge N$, so $V$ contains only finitely many $x_n$. Hence $x_n \not\to y$. ∎

ProofAdded↑ statement

Theorem 17.11. Products. If $(x_1,y_1) \ne (x_2,y_2)$, they differ in some coordinate, say $x_1 \ne x_2$. Take disjoint open $U_1 \ni x_1$ and $U_2 \ni x_2$ in $X$. Then $U_1 \times Y$ and $U_2 \times Y$ are disjoint and open.

Subspaces. For $x \ne y$ in $Y \subset X$, take disjoint neighborhoods $U, V$ in $X$. Then $U \cap Y$ and $V \cap Y$ are disjoint neighborhoods in $Y$. ∎

Continuity

ProofAdded↑ statement

Checking on a basis or subbasis. Any open $V \subset Y$ is a union of basis elements, $V = \bigcup_\alpha B_\alpha$, so $f^{-1}(V) = \bigcup_\alpha f^{-1}(B_\alpha)$ is a union of open sets. For a subbasis, every basis element is a finite intersection $S_1 \cap \cdots \cap S_n$, and $f^{-1}(S_1 \cap \cdots \cap S_n) = \bigcap_i f^{-1}(S_i)$ is open. ∎

ProofAdded↑ statement

Theorem 18.1. (1 ⇒ 2) Let $x \in \bar A$ and let $V$ be a neighborhood of $f(x)$. Then $f^{-1}(V)$ is open and contains $x$, so it meets $A$ (17.5), say at $y$. Then $f(y) \in V \cap f(A)$. Every neighborhood of $f(x)$ meets $f(A)$, so $f(x) \in \overline{f(A)}$.

(2 ⇒ 3) Let $B$ be closed in $Y$ and $A = f^{-1}(B)$. If $x \in \bar A$, then $f(x) \in f(\bar A) \subset \overline{f(A)} \subset \bar B = B$, so $x \in A$. Thus $\bar A = A$ is closed.

(3 ⇒ 1) If $V$ is open in $Y$, then $Y - V$ is closed, so $f^{-1}(Y - V) = X - f^{-1}(V)$ is closed, and $f^{-1}(V)$ is open.

(1 ⇒ 4) Take $U = f^{-1}(V)$.

(4 ⇒ 1) Let $V$ be open in $Y$. For each $x \in f^{-1}(V)$, $V$ is a neighborhood of $f(x)$, so there is a neighborhood $U_x$ of $x$ with $f(U_x) \subset V$, i.e. $U_x \subset f^{-1}(V)$. So $f^{-1}(V) = \bigcup_x U_x$ is open. ∎

Product topology

Metric = product on $\mathbb R^2$. Metric-open ⇒ product-open. Let $U \in \mathcal U_{\text{met}}$, so $U$ is a union of balls. It suffices to show each ball $B((x_0,y_0), \varepsilon)$ is a union of rectangles. Take any $(x,y)$ in the ball and let $d = d\big((x,y),(x_0,y_0)\big) \lt \varepsilon$. Choose $0 \lt \delta \lt (\varepsilon - d)/2$ and consider the rectangle $(x-\delta, x+\delta) \times (y-\delta, y+\delta)$. Each of its points is within $\sqrt2\,\delta \lt \varepsilon - d$ of $(x,y)$, so by the triangle inequality it lies in the ball. Since $(x,y)$ was arbitrary, there is a rectangle around every point of the ball, so the ball is a union of rectangles.

Product-open ⇒ metric-open. This is easy by the same argument. If $(x,y) \in U \times V$, pick $(x - r, x + r) \subset U$ and $(y - s, y + s) \subset V$. The ball of radius $\min(r,s)$ around $(x,y)$ lies in $U \times V$. ∎

Projections are continuous. For $U \subset X_\beta$ open, $\pi_\beta^{-1}(U) = U \times \prod_{\alpha \in A - \{\beta\}} X_\alpha$, which is a basis element (with $I = \{\beta\}$), so it is open. ∎

Theorem 19.6. (⇒) $f_\alpha = \pi_\alpha \circ f$, so $f_\alpha$ is a composition of continuous functions, which is continuous since $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$.

(⇐) Suppose each $f_\alpha : Z \to X_\alpha$ is continuous. Any open set of $\prod X_\alpha$ is a union of basis elements of the form $B = \prod_{i \in I} U_i \times \prod_{\alpha \in A - I} X_\alpha$, so it suffices to check that $f^{-1}(B)$ is open. Each $f_i$ is continuous, so $f_i^{-1}(U_i)$ is open in $Z$ for all $i \in I$. And straight from the definition, $f^{-1}(B) = \bigcap_{i \in I} f_i^{-1}(U_i)$. This is a finite intersection of open sets, so it is open. In the box topology, the intersection would run over all $\alpha$, and that last step fails. ∎

Quotient spaces

ProofAdded↑ statement

Quotient maps via saturated sets. (⇒) If $U$ is saturated and open, then $f^{-1}(f(U)) = U$ is open, so $f(U)$ is open by condition 3.

(⇐) If $f^{-1}(V)$ is open, it is saturated, and $V = f(f^{-1}(V))$ by surjectivity, so $V$ is open. Together with continuity, this gives “$V$ open iff $f^{-1}(V)$ open.” ∎

ProofAdded↑ statement

Theorem 22.2. $f(y) = g(x)$ for any $x \in p^{-1}(y)$ is well defined since $g$ is constant on $p^{-1}(y)$, and $f \circ p = g$. For $V \subset Z$ open, $g^{-1}(V) = p^{-1}(f^{-1}(V))$. Since $p$ is a quotient map, $f^{-1}(V)$ is open iff $g^{-1}(V)$ is, so $f$ is continuous iff $g$ is. Quotient: if $f$ is a quotient map, so is the composite $g = f \circ p$. Conversely, if $g$ is a quotient map, then $f$ is surjective and continuous, and if $f^{-1}(V)$ is open, then $g^{-1}(V) = p^{-1}(f^{-1}(V))$ is open, so $V$ is open. ∎

ProofAdded↑ statement

Corollary 22.3. The quotient map $p : X \to X^*$ sends $x$ to $g^{-1}(g(x))$, and $g$ is constant on each fiber. So Theorem 22.2 gives a continuous $g^*$ with $g^* \circ p = g$, and $g^*$ is a bijection by construction. A continuous bijection is a homeomorphism iff it is a quotient map, and by 22.2 $g^*$ is a quotient map iff $g$ is. ∎

$x \mapsto x^2$ example. (2) is not a quotient map. In the discrete topology on $\mathbb R$, $\{-1,1\} = f^{-1}(\{1\})$ is saturated and open, but $f(\{-1,1\}) = \{1\}$ is not open in $[0,\infty)$.

(1) is a quotient map. AddedSuppose $f^{-1}(U)$ is open in $\mathbb R$. Let $g : [0,\infty) \to \mathbb R$, $g(y) = \sqrt y$, which is continuous. Since $f(g(y)) = y$, we get $g^{-1}(f^{-1}(U)) = U$, so $U$ is open in $[0,\infty)$. ∎

From my notes: the image of $f(t) = (\cos 2\pi t, \sin 2\pi t)$ is $S^1$, $f$ is surjective onto it, and $f$ is continuous. The rest (via Corollary 22.3) was Problem Set 3 #5.

Connectedness

Clopen criterion. If $U, V$ is a separation, then $V = X - U$ with $U$ open, so $V$ is closed as well as open. And $V \ne \varnothing, X$, so $V$ is a nontrivial clopen set. Conversely, if $A$ is clopen with $\varnothing \ne A \ne X$, then $A$ and $X - A$ are disjoint, nonempty, open, and cover $X$, so they form a separation. ∎

$[0,1]$ is connected. Suppose there is a separation: $U, V$ nonempty, open in $X = [0,1]$, $U \cup V = X$, $U \cap V = \varnothing$. Say $0 \in U$, and let $y_0 \in V$. Consider $x_0 = \sup\big(U \cap [0, y_0]\big)$.

  • $x_0 \notin V$. Otherwise, since $V$ is open, an interval $(x_0 - \varepsilon, x_0]$ lies in $V$, so $x_0 - \varepsilon$ would be a smaller upper bound for $U \cap [0, y_0]$. In particular $x_0 \ne y_0$, so $x_0 \lt y_0$. Also $x_0 \gt 0$, since $0 \in U$ and $U$ is open.
  • $x_0 \notin U$. Otherwise, since $U$ is open and $x_0 \lt y_0$, we can always go further to the right: some $[x_0, x_0 + \varepsilon) \subset U \cap [0,y_0]$, contradicting that $x_0$ is the sup.

But $U \cup V = X$, which is a contradiction. ∎

ProofAdded↑ statement

Lemma 23.2. $C \cap Y$ and $D \cap Y$ are open in $Y$, disjoint, and cover $Y$. If both were nonempty they would separate $Y$, so one is empty. ∎

ProofAdded↑ statement

Theorem 23.3. Let $p$ be the common point and $Y = \bigcup A_\alpha$. Suppose $C, D$ separate $Y$, with $p \in C$. Each $A_\alpha$ lies in $C$ or $D$ (23.2) and contains $p$, so it lies in $C$. Then $Y \subset C$ and $D = \varnothing$, which is a contradiction. ∎

ProofAdded↑ statement

Closures of connected sets. Suppose $C, D$ is a separation of $B$. Then $C \cap A$ and $D \cap A$ are open in $A$ and disjoint, so since $A$ is connected one of them is empty: say $A \subset C$. $C = B - D$ is closed in $B$, so the closure of $A$ in $B$, which is $\bar A \cap B = B$ (17.4), lies in $C$. Then $D = \varnothing$, which is a contradiction. ∎

ProofWS 7↑ statement

Theorem 23.5. Suppose $C, D$ separate $f(A)$. The restriction $f|_A : A \to f(A)$ is continuous, so $f|_A^{-1}(C)$ and $f|_A^{-1}(D)$ are open in $A$, disjoint, and cover $A$. They are nonempty because $C$ and $D$ lie in the image. So they separate $A$, which is a contradiction. ∎

ProofWS 8↑ statement

Theorem 23.6. Two factors: fix $(a,b) \in X \times Y$. For each $x$, $T_x = (X \times \{b\}) \cup (\{x\} \times Y)$ is connected: it is a union of copies of $X$ and $Y$ meeting at $(x,b)$ (23.3). All the $T_x$ contain $(a,b)$, so $X \times Y = \bigcup_x T_x$ is connected. Induct for finitely many factors. Any product (WS 8): fix $x \in X = \prod X_\alpha$. For finite $I \subset A$, $X_I = \prod_{i \in I} X_i \times \prod_{\alpha \notin I} \{x_\alpha\}$ is homeomorphic to a finite product, hence connected. All the $X_I$ contain $x$, so $B = \bigcup_I X_I$ is connected. A basic open set restricts only finitely many coordinates $I$, so it meets $X_I$. Hence $\bar B = X$, which is connected by 23.4. ∎

ProofAdded↑ statement

Connected subsets of $\mathbb R$. Intervals are connected: the sup argument used for $[0,1]$ works for any interval. Conversely: if $S \subset \mathbb R$ is not an interval, there are $a \lt c \lt b$ with $a, b \in S$ and $c \notin S$. Then $S \cap (-\infty, c)$ and $S \cap (c, \infty)$ separate $S$. ∎

ProofAdded↑ statement

Intermediate value theorem. If no $c$ has $f(c) = r$, then $f(X) \cap (-\infty, r)$ and $f(X) \cap (r, \infty)$ are disjoint, open in $f(X)$, cover it, and are nonempty (they contain $f(a)$ and $f(b)$). This separates $f(X)$, contradicting 23.5. ∎

Discrete spaces are totally disconnected. If $C \subset X$ has two points $x \ne y$, then $\{x\}$ and $C - \{x\}$ are open in $C$ (every subset is open), disjoint, nonempty, and cover $C$. So $C$ is not connected, and the only connected sets are single points. ∎

The Cantor set is totally disconnected. Take $x \ne y \in \mathcal C$. We want to show they can’t lie in the same connected component. Choose $n$ with $|x - y| \gt 1/3^n$. The intervals of $C_n$ each have length $1/3^n$, so $x$ and $y$ lie in different intervals of $C_n$. Let $I$ be the interval containing $x$. Then $I \cap \mathcal C$ and $\mathcal C - I$ (the rest of $C_n$, cut down to $\mathcal C$) are open in $\mathcal C$, disjoint, and cover $\mathcal C$, so they separate $x$ from $y$. ∎

Path connected ⇒ connected. Let $X$ be path connected, and suppose it has a separation $U, V$. Choose $x \in U$, $y \in V$, and a path $\gamma : [0,1] \to X$ with $\gamma(0) = x$, $\gamma(1) = y$. Consider $\gamma^{-1}(U), \gamma^{-1}(V) \subset [0,1]$. They are open (since $\gamma$ is continuous), disjoint, nonempty ($0$ and $1$), and cover $[0,1]$. So they form a separation of $[0,1]$, which is impossible since $[0,1]$ is connected. ∎

$\mathbb R^n$ is path connected. For any $x = (x_1, \ldots, x_n)$ and $y = (y_1, \ldots, y_n)$, take the convex combination $\gamma : [0,1] \to \mathbb R^n$, $\gamma(t) = (1-t)\,x + t\,y$. It is continuous, with $\gamma(0) = x$ and $\gamma(1) = y$. ∎

Compactness

$\mathbb R$ is not compact. $\mathcal V = \{(a, a+1) : a \in \mathbb R\}$ is a cover: every $y \in \mathbb R$ lies in some such interval. But it has no finite subcover. If $V_i = (a_i, a_i + 1)$ for $i = 1, \ldots, n$, then $x = \max_i a_i + 1$ is not in their union. ∎

$\mathbb R_{fc}$ is compact. Let $\mathcal V$ be any open cover and take any nonempty $V_0 \in \mathcal V$. Then $V_0 = \mathbb R - \{x_1, \ldots, x_n\}$. Since $\mathcal V$ is a cover, for each $i$ there is $V_i \in \mathcal V$ with $x_i \in V_i$. Then $V_0 \cup V_1 \cup \cdots \cup V_n = \mathbb R$ is a finite subcover. ∎

Closed in compact is compact. (From the worksheet.) Let $\mathcal V$ cover $Y$ by sets open in $X$. Since $Y$ is closed, $X - Y$ is open, so $\mathcal V \cup \{X - Y\}$ is an open cover of $X$. It has a finite subcover $V_1, \ldots, V_n$ (possibly together with $X - Y$). Since $X - Y$ misses $Y$, $V_1 \cup \cdots \cup V_n \supset Y$. ∎

Compact in Hausdorff is closed. We want to show $X - Y$ is open. Fix $x \in X - Y$. For each $y \in Y$, by Hausdorff find disjoint open $U_y \ni x$ and $V_y \ni y$. Then $\mathcal V = \{V_y : y \in Y\}$ is an open cover of $Y$. $Y$ is compact, so there is a finite subcover $V_{y_1}, \ldots, V_{y_m}$. Then $U = U_{y_1} \cap \cdots \cap U_{y_m}$ is open (we can take this intersection because it’s finite), contains $x$, and $U \cap Y = \varnothing$. We can do this for every $x \in X - Y$, so $X - Y$ is open and $Y$ is closed. ∎

ProofWS 9↑ statement

Separating compact sets. Point and compact set: let $x \notin B$. As in the previous proof, choose disjoint open $U_b \ni x$ and $V_b \ni b$ for each $b \in B$, and take a finite subcover $V_{b_1}, \ldots, V_{b_m}$ of $B$. Then $U = \bigcap U_{b_i}$ and $V = \bigcup V_{b_i}$ are disjoint. Two compact sets: for each $a \in A$, get disjoint open $U_a \ni a$ and $V_a \supset B$. Finitely many $U_{a_j}$ cover $A$, and then $U = \bigcup_j U_{a_j}$ and $V = \bigcap_j V_{a_j}$ work. ∎

ProofAdded↑ statement

Nested closed sets. If $\bigcap C_n = \varnothing$, the open sets $X - C_n$ cover $X$. They increase with $n$, so a finite subcover means $X = X - C_N$ for some $N$, i.e. $C_N = \varnothing$. That is a contradiction. ∎

ProofAdded↑ statement

Heine–Borel. (My notes say to read this proof in Munkres. Here is an outline.)

(⇒) $\mathbb R^n$ is Hausdorff, so compact $Y$ is closed. The balls $B(0,m)$, $m = 1, 2, \ldots$, cover $Y$. A finite subcover gives $Y \subset B(0, M)$, so $Y$ is bounded.

(⇐) If $Y \subset B(0,R)$, then $Y \subset [-R,R]^n$. Closed intervals $[a,b]$ are compact (Munkres 27.1, from the least upper bound property), and finite products of compact spaces are compact (Munkres 26.7), so $[-R,R]^n$ is compact. $Y$ is closed in it, so $Y$ is compact. ∎

Theorems 26.5, 26.6. (1) AddedLet $\mathcal V$ cover $f(X)$ by sets open in $Y$. Then $\{f^{-1}(V) : V \in \mathcal V\}$ is an open cover of $X$, so it has a finite subcover $f^{-1}(V_1), \ldots, f^{-1}(V_n)$, and $V_1, \ldots, V_n$ cover $f(X)$.

(2) Assume $X$ is compact, $f$ is injective and continuous, and $Y$ is Hausdorff. We want to show $X$ is homeomorphic to $f(X)$ (subspace topology) via $f$. Indeed, $f$ is bijective onto its image (injective, restricted to the image) and continuous. For continuity of the inverse, take $U \subset X$ open. Then $X - U$ is closed, hence compact (closed in compact), so $f(X - U) \subset Y$ is compact (part 1). A compact subset of a Hausdorff space is closed, so $f(X - U)$ is closed and $Y - f(X - U)$ is open. Using injectivity, $f(U) \cap f(X - U) = \varnothing$, so $f(U) = f(X) \cap \big(Y - f(X - U)\big)$, which is open in the subspace topology. ∎

ProofWS 10↑ statement

Theorem 27.7. Step 1: for any nonempty open $U$ and any point $x$, there is a nonempty open $V \subset U$ with $x \notin \bar V$. Pick $y \in U$ with $y \ne x$; this is possible since $U \ne \{x\}$, as $x$ isn’t isolated. Take disjoint open $W_1 \ni x$ and $W_2 \ni y$, and let $V = W_2 \cap U$. Then $V \subset X - W_1$, which is closed, so $\bar V$ misses $x$. Step 2: let $f : \mathbb N \to X$ be any function and $x_n = f(n)$. By step 1, get nonempty open $X \supset V_1 \supset V_2 \supset \cdots$ with $x_n \notin \bar V_n$. The sets $\bar V_n$ are nested, nonempty, and closed, so $\bigcap \bar V_n \ne \varnothing$. A point in the intersection is no $x_n$, so $f$ is not surjective and $X$ is uncountable. ∎