Ultrafilters in topology

Remark: To forestall empty set difficulties, whenever I talk about arbitrary sets in this post they will be non-empty.

We continue our exploration of ultrafilters from the previous post. Recall that a (proper) filter on a poset Image may be NSFW.
Clik here to view. is a non-empty subset Image may be NSFW.
Clik here to view. such that

For every Image may be NSFW.
Clik here to view. $x, y \in F$ , there is some Image may be NSFW.
Clik here to view. $z \in F$ such that Image may be NSFW.
Clik here to view. $z \le x, z \le y$ .
For every Image may be NSFW.
Clik here to view. $x \in F$ , if Image may be NSFW.
Clik here to view. $x \le y$ then Image may be NSFW.
Clik here to view. $y \in F$ .
Image may be NSFW.
Clik here to view. is not the whole set Image may be NSFW.
Clik here to view..

If Image may be NSFW.
Clik here to view. has finite infima (meets), then the first condition, given the second, can be replaced with the condition that if Image may be NSFW.
Clik here to view. $x, y \in F$ then Image may be NSFW.
Clik here to view. $x \wedge y \in F$ . (This holds in particular if Image may be NSFW.
Clik here to view. is the poset structure on a Boolean ring, in which case Image may be NSFW.
Clik here to view. $x \wedge y = xy$ .) A filter is an ultrafilter if in addition it is maximal under inclusion among (proper) filters. For Boolean rings, an equivalent condition is that for every Image may be NSFW.
Clik here to view. $x \in B$ either Image may be NSFW.
Clik here to view. $x \in F$ or Image may be NSFW.
Clik here to view. $1 - x \in F$ , but not both. Recall that this condition tells us that ultrafilters are precisely complements of maximal ideals, and can be identified with morphisms in Image may be NSFW.
Clik here to view. $\text{Hom}_{\text{BRing}}(B, \mathbb{F}_2)$ . If Image may be NSFW.
Clik here to view. $B = \text{Hom}(S, \mathbb{F}_2)$ for some set Image may be NSFW.
Clik here to view., then we will sometimes call an ultrafilter on Image may be NSFW.
Clik here to view. an ultrafilter on Image may be NSFW.
Clik here to view. (for example, this is what people usually mean by “an ultrafilter on Image may be NSFW.
Clik here to view. $\mathbb{N}$ “).

Today we will meander towards an ultrafilter point of view on topology. This point of view provides, among other things, a short, elegant proof of Tychonoff’s theorem.

Some intuitions

There are several ways to get a more intuitive grasp on what it is, exactly, that ultrafilters are. We saw one of these intuitions already: an ultrafilter on Image may be NSFW.
Clik here to view. is a maximal consistent deductively closed set of propositions in Image may be NSFW.
Clik here to view.. If Image may be NSFW.
Clik here to view. is a set of propositions in Image may be NSFW.
Clik here to view., then ultrafilters on Image may be NSFW.
Clik here to view. containing Image may be NSFW.
Clik here to view. can be identified with “models of Image may be NSFW.
Clik here to view.,” that is, ways to make the statements of Image may be NSFW.
Clik here to view. all true. (We do not actually need to construct these models because the only thing we care about is which statements are true or false in a model, which is precisely what the ultrafilter tells us!) According to this intuition, the ultrafilter axioms are just familiar properties of propositional logic:

Conjunction: if Image may be NSFW.
Clik here to view. and Image may be NSFW.
Clik here to view., then Image may be NSFW.
Clik here to view. $p \wedge q$ (Image may be NSFW.
Clik here to view. and Image may be NSFW.
Clik here to view.).
Modus ponens: if Image may be NSFW.
Clik here to view. and Image may be NSFW.
Clik here to view. $p \rightarrow q$ (that is, Image may be NSFW.
Clik here to view. $p \le q$ , or Image may be NSFW.
Clik here to view. implies Image may be NSFW.
Clik here to view.), then Image may be NSFW.
Clik here to view..
Excluded middle: Image may be NSFW.
Clik here to view. if and only if Image may be NSFW.
Clik here to view. $\neg (\neg p)$ (not not Image may be NSFW.
Clik here to view.).

In his post on ultrafilters, Terence Tao also uses the closely related “voting” intuition. Here we only consider ultrafilters on sets. We think of the set Image may be NSFW.
Clik here to view. as a collection of voters, and a collection of votes as a subset of Image may be NSFW.
Clik here to view. (the subset of people who voted “yes” on something). An ultrafilter Image may be NSFW.
Clik here to view. on Image may be NSFW.
Clik here to view. is then a consistent way of deciding elections: given a collection of votes on some proposition, the outcome of the election is “yes” if the set of “yes” votes is in Image may be NSFW.
Clik here to view. and “no” otherwise. According to this intuition, the ultrafilter axioms then say that this voting system does not produce voting paradoxes: the explanations are essentially the same as the ones above if one substitutes “Image may be NSFW.
Clik here to view. is true” with “the electorate voted for Image may be NSFW.
Clik here to view..”

Terence Tao also uses a third probabilistic, or measure-theoretic, intuition: an ultrafilter Image may be NSFW.
Clik here to view. on Image may be NSFW.
Clik here to view. can be thought of as defining a “measure” Image may be NSFW.
Clik here to view. $\mu$ which assigns to each subset of Image may be NSFW.
Clik here to view. measure Image may be NSFW.
Clik here to view. if the subset is in Image may be NSFW.
Clik here to view. (“almost surely true”) and Image may be NSFW.
Clik here to view. otherwise (“almost surely false”). Informally, this is a kind of zero-one law. Image may be NSFW.
Clik here to view. $\mu$ is precisely the morphism Image may be NSFW.
Clik here to view. $\phi : \text{Hom}(S, \mathbb{F}_2) \to \mathbb{F}_2$ associated to the maximal ideal associated to Image may be NSFW.
Clik here to view., but here we are thinking of its codomain in a somewhat different way. According to this intuition, the ultrafilter axioms set out several natural properties a “measure” like Image may be NSFW.
Clik here to view. $\mu$ should have:

If Image may be NSFW.
Clik here to view. $\mu(A) = 1$ and Image may be NSFW.
Clik here to view. $\mu(B) = 1$ , then Image may be NSFW.
Clik here to view. $\mu(A \cap B) = 1$ .
If Image may be NSFW.
Clik here to view. $\mu(A) = 1$ and Image may be NSFW.
Clik here to view. $A \subseteq B$ , then Image may be NSFW.
Clik here to view. $\mu(B) = 1$ .
Image may be NSFW.
Clik here to view. $\mu(A) = 1$ if and only if Image may be NSFW.
Clik here to view. $\mu(A^c) = 0$ .

Together with the other axioms, the third axiom is actually equivalent to Image may be NSFW.
Clik here to view. $\mu$ being finitely additive (and equal to Image may be NSFW.
Clik here to view. on the entire set): if Image may be NSFW.
Clik here to view. A_1, ... A_n are disjoint subsets, then either they all have measure Image may be NSFW.
Clik here to view. (hence so does their union) or some Image may be NSFW.
Clik here to view. A_i has measure Image may be NSFW.
Clik here to view., hence its complement (hence all of the other subsets) has measure Image may be NSFW.
Clik here to view.. This is not as strong as countable additivity, so Image may be NSFW.
Clik here to view. $\mu$ does not actually form a measure.

Nevertheless, finite additivity is a convenient way to think about how ultrafilters work: if Image may be NSFW.
Clik here to view. $X : S \to \mathbb{R}$ is a “random variable” which takes on finitely many values, it breaks Image may be NSFW.
Clik here to view. up into a disjoint union of finitely many sets, exactly one of which can have measure Image may be NSFW.
Clik here to view., hence Image may be NSFW.
Clik here to view. takes that value “almost surely” (so that value is its “expected value”). In voter-theoretic terms, the electorate can hold elections with any finite number of options, rather than just two.

Non-principal ultrafilters

Given a set Image may be NSFW.
Clik here to view., there are some obvious ultrafilters we can write down on Image may be NSFW.
Clik here to view.. These are the ultrafilters consisting of the set of all subsets containing some Image may be NSFW.
Clik here to view. $s \in S$ , or the principal ultrafilters. It’s not hard to see that this condition is equivalent to the condition that the ultrafilter has a least element (the intersection of all of its elements), since this least element must be a singleton. Algebraically, principal ultrafilters correspond to principal ideals generated by elements of Image may be NSFW.
Clik here to view. $B = \text{Hom}(S, \mathbb{F}_2)$ vanishing at one point Image may be NSFW.
Clik here to view. and not vanishing anywhere else, and the corresponding morphism Image may be NSFW.
Clik here to view. $B \to \mathbb{F}_2$ is the obvious projection to the Image may be NSFW.
Clik here to view. $i^{th}$ coordinate. In voting terms these are “dictatorial” voting systems, in measure-theoretic terms these are Dirac measures supported on points, and in probabilistic terms these are deterministic random variables taking a single value.

When Image may be NSFW.
Clik here to view. is finite, all ultrafilters are principal. However, when Image may be NSFW.
Clik here to view. is infinite, we know that Image may be NSFW.
Clik here to view. $\text{Spec } B$ is compact, Hausdorff, and totally disconnected. But it’s not hard to see that Image may be NSFW.
Clik here to view. $S \subset \text{Spec } B$ has the discrete topology, so there must be more points than the elements of Image may be NSFW.
Clik here to view.; in other words, there must be non-principal ultrafilters (hence morphisms Image may be NSFW.
Clik here to view. $B \to \mathbb{F}_2$ other than the obvious ones) which, as points in the spectrum, give a compactification of the discrete space Image may be NSFW.
Clik here to view..

Proposition: An ultrafilter on an infinite set Image may be NSFW.
Clik here to view. is non-principal if and only if it contains the Fréchet filter of cofinite subsets of Image may be NSFW.
Clik here to view..

Proof. An ultrafilter contains the Fréchet filter if and only if its complement contains all finite subsets of Image may be NSFW.
Clik here to view.. Since principal ultrafilters contain singletons, no principal ultrafilter can contain the Fréchet filter, and conversely if a principal ultrafilter contains the Fréchet filter it cannot contain any singletons.

At this point, the statements I’ve been making about compactification are probably worth justifying. Recall that given a space Image may be NSFW.
Clik here to view., a space Image may be NSFW.
Clik here to view. equipped with a continuous injection Image may be NSFW.
Clik here to view. $T \to C$ is said to be a compactification of Image may be NSFW.
Clik here to view. if Image may be NSFW.
Clik here to view. is compact Hausdorff and the image of Image may be NSFW.
Clik here to view. is dense. In this case, Image may be NSFW.
Clik here to view. $C = \text{Spec } B$ has the property that its topology is determined by the continuous functions Image may be NSFW.
Clik here to view. $C \to \mathbb{F}_2$ , so to show that Image may be NSFW.
Clik here to view. T = S is dense it suffices to show that a function is determined by its values on Image may be NSFW.
Clik here to view.. But this is true by definition. So the space of ultrafilters on Image may be NSFW.
Clik here to view. is in fact a compactification of the discrete space Image may be NSFW.
Clik here to view., which sits inside it as the subspace of principal ultrafilters.

In voting terms, if a principal ultrafilter is like having a dictator, a non-principal ultrafilter is like having a “virtual” or “ideal” dictator. This perspective was explained to me, in different words, by Todd Trimble in the comments to the previous post. Given an ultrafilter Image may be NSFW.
Clik here to view., imagine playing the following game (which we might call “Find the Dictator”): repeatedly ask yes or no questions of the electorate. Then you know that the set of voters who voted against the outcome of any particular election have no say in any of them; in probabilistic terms, they have measure Image may be NSFW.
Clik here to view.. So the next question you ask, you only look at the set of voters who do have some say (measure Image may be NSFW.
Clik here to view.), and ask them some question. Continuing in this way you end up looking at a smaller and smaller set of voters who have some say in the elections. (In topological terms, these smaller and smaller sets of voters are precisely intersections of neighborhoods of the ultrafilter Image may be NSFW.
Clik here to view. with the discrete subspace of principal ultrafilters; by density, this essentially determines the neighborhoods.) If Image may be NSFW.
Clik here to view. is principal, then if you get lucky with the questions you’ll eventually find the dictator. But if Image may be NSFW.
Clik here to view. is non-principal, there isn’t one! So your chain of questions is attempting to converge to a dictator which isn’t actually there, but which is mysteriously somehow still influencing the election.

Convergence and Tychonoff’s theorem

The remarks about convergence above suggest the following definition.

Definition: Let Image may be NSFW.
Clik here to view. be a topological space and let Image may be NSFW.
Clik here to view. be a filter on Image may be NSFW.
Clik here to view.. We say that Image may be NSFW.
Clik here to view. converges to Image may be NSFW.
Clik here to view. $x \in X$ (written Image may be NSFW.
Clik here to view. $F \to x$ ) if for every open set Image may be NSFW.
Clik here to view. containing Image may be NSFW.
Clik here to view., there exists Image may be NSFW.
Clik here to view. $f \in F$ such that Image may be NSFW.
Clik here to view. $f \subseteq U$ .

Filter convergence is a generalization of convergence of sequences which is better adapted to spaces which are far from being metric spaces (for example, spaces which aren’t first-countable). For a thorough explanation of the issues here, I recommend Pete Clark’s notes on convergence.

It is useful to extend the definition of filter convergence and clustering to filter bases, or prefilters, which are nonempty collections of nonempty sets closed under finite intersection. The collection of all supersets of the elements of a prefilter forms a filter, the filter generated by the prefilter, and a prefilter converges to a point Image may be NSFW.
Clik here to view. if and only if the filter it generates converges to Image may be NSFW.
Clik here to view..

We can recover convergence of sequences as follows. If Image may be NSFW.
Clik here to view. $a : X \to Y$ is a continuous function and Image may be NSFW.
Clik here to view. a prefilter on Image may be NSFW.
Clik here to view., then Image may be NSFW.
Clik here to view. $a(F) = \{ a(f) : f \in F \}$ is a prefilter on Image may be NSFW.
Clik here to view. (the image or pushforward prefilter). If Image may be NSFW.
Clik here to view. $X = \mathbb{N}$ and Image may be NSFW.
Clik here to view. is the Fréchet filter, then Image may be NSFW.
Clik here to view. a(F) converges to Image may be NSFW.
Clik here to view. precisely when the sequence defined by Image may be NSFW.
Clik here to view. converges to Image may be NSFW.
Clik here to view. in the usual sense. If we want to stick to filters, we can instead take the filter generated by Image may be NSFW.
Clik here to view. a(F) .

In a metric space, the closure of a subset is precisely the set of limits of sequences of elements in the subset. This fails for more general topological spaces, but the corresponding statement is true for filters.

Proposition: Let Image may be NSFW.
Clik here to view. be a topological space, Image may be NSFW.
Clik here to view. $S \subset X$ , and Image may be NSFW.
Clik here to view. $x \in X$ . Then the following are equivalent:

Image may be NSFW.
Clik here to view. $x \in \bar{S}$ .
There exists a prefilter on Image may be NSFW.
Clik here to view. which converges to Image may be NSFW.
Clik here to view..
There exists a filter containing Image may be NSFW.
Clik here to view. which converges to Image may be NSFW.
Clik here to view..
There exists an ultrafilter containing Image may be NSFW.
Clik here to view. which converges to Image may be NSFW.
Clik here to view..

Proof. Let Image may be NSFW.
Clik here to view. N_x denote the collection of all neighborhoods of Image may be NSFW.
Clik here to view.. This is the neighborhood filter at Image may be NSFW.
Clik here to view., and by definition, Image may be NSFW.
Clik here to view. $x \in \bar{S}$ if and only if Image may be NSFW.
Clik here to view. $N_x \cap S$ converges to Image may be NSFW.
Clik here to view.. This gives Image may be NSFW.
Clik here to view. $1 \Rightarrow 2$ . That Image may be NSFW.
Clik here to view. $2 \Rightarrow 3$ follows by taking the filter generated by the prefilter in question, and that Image may be NSFW.
Clik here to view. $3 \Rightarrow 4$ follows by taking any ultrafilter containing the filter in question, so it remains to prove that Image may be NSFW.
Clik here to view. $4 \Rightarrow 1$ .

Let Image may be NSFW.
Clik here to view. be an ultrafilter containing Image may be NSFW.
Clik here to view. and converging to Image may be NSFW.
Clik here to view., hence for every open set Image may be NSFW.
Clik here to view. containing Image may be NSFW.
Clik here to view. there exists Image may be NSFW.
Clik here to view. $f \in F$ such that Image may be NSFW.
Clik here to view. $f \subset U$ . Then Image may be NSFW.
Clik here to view. does not contain any subset of the complement of Image may be NSFW.
Clik here to view., so for any Image may be NSFW.
Clik here to view. $f \in F$ the intersection Image may be NSFW.
Clik here to view. $S \cap f \in F$ is nonempty. Moreover, Image may be NSFW.
Clik here to view. $f \subset U$ implies that Image may be NSFW.
Clik here to view. $S \cap f \subset U$ . It follows that every open set Image may be NSFW.
Clik here to view. containing Image may be NSFW.
Clik here to view. contains elements of Image may be NSFW.
Clik here to view., so Image may be NSFW.
Clik here to view. $x \in \bar{S}$ as desired.

In voting terms, an ultrafilter containing Image may be NSFW.
Clik here to view. is an electorate where the voters in Image may be NSFW.
Clik here to view. are the ones which have power, and it converges to Image may be NSFW.
Clik here to view. if and only if any neighborhood of Image may be NSFW.
Clik here to view. contains voters in power. In probabilistic terms, an ultrafilter containing Image may be NSFW.
Clik here to view. is a “random variable” which is in Image may be NSFW.
Clik here to view. almost surely, and it converges to Image may be NSFW.
Clik here to view. if and only if it is in any neighborhood of Image may be NSFW.
Clik here to view. almost surely.

Recall that the topology of a space is determined by the closure operator Image may be NSFW.
Clik here to view. $S \mapsto \bar{S}$ , and that conversely any such operator satisfying the Kuratowski closure axioms defines a topology. It follows that the topology of a space is determined by (pre/ultra)filter convergence. In particular, a function Image may be NSFW.
Clik here to view. $f : X \to Y$ is continuous if and only if it preserves convergence of (pre/ultra)filters. As the following proposition shows, this is a convenient way to rephrase certain topological properties.

Proposition: Let Image may be NSFW.
Clik here to view. be a topological space.

Image may be NSFW.
Clik here to view. is Hausdorff if and only if every ultrafilter on Image may be NSFW.
Clik here to view. converges to at most one point.
Image may be NSFW.
Clik here to view. is compact if and only if every ultrafilter on Image may be NSFW.
Clik here to view. converges to at least one point.

Proof.

Image may be NSFW.
Clik here to view. $\Rightarrow$ : Let Image may be NSFW.
Clik here to view. be an ultrafilter converging to Image may be NSFW.
Clik here to view., and let Image may be NSFW.
Clik here to view. $y \neq x$ . By assumption, there exist disjoint open sets Image may be NSFW.
Clik here to view. with Image may be NSFW.
Clik here to view. $x \in U, y \in V$ . Since Image may be NSFW.
Clik here to view. $F \to x$ , there is Image may be NSFW.
Clik here to view. $f \in F$ such that Image may be NSFW.
Clik here to view. $f \subset U$ . It follows that Image may be NSFW.
Clik here to view. $V \not \in F$ , so Image may be NSFW.
Clik here to view. cannot converge to Image may be NSFW.
Clik here to view.. Image may be NSFW.
Clik here to view. $\Leftarrow$ : By assumption, there exist Image may be NSFW.
Clik here to view. $x \neq y$ such that every neighborhood of one intersects a neighborhood of another. It follows that any ultrafilter containing the filter Image may be NSFW.
Clik here to view. $N_x \cap N_y$ (the collection of intersections of neighborhoods of Image may be NSFW.
Clik here to view. and neighborhoods of Image may be NSFW.
Clik here to view.) converges to both Image may be NSFW.
Clik here to view. and Image may be NSFW.
Clik here to view..
Image may be NSFW.
Clik here to view. $\Rightarrow$ : Given an ultrafilter Image may be NSFW.
Clik here to view., the collection of closures of elements of Image may be NSFW.
Clik here to view. satisfies the finite intersection property, so by assumption there exists Image may be NSFW.
Clik here to view. which is in the closure of every element of Image may be NSFW.
Clik here to view.. For any open set Image may be NSFW.
Clik here to view. containing Image may be NSFW.
Clik here to view., the closure of Image may be NSFW.
Clik here to view. does not contain Image may be NSFW.
Clik here to view., hence Image may be NSFW.
Clik here to view. cannot contain Image may be NSFW.
Clik here to view. and must contain Image may be NSFW.
Clik here to view.. It follows that Image may be NSFW.
Clik here to view. $F \to x$ . Image may be NSFW.
Clik here to view. $\Leftarrow$ : Any collection of closed sets Image may be NSFW.
Clik here to view. with the finite intersection property is contained in an ultrafilter Image may be NSFW.
Clik here to view.. (In order for a collection of sets to be contained in an ultrafilter it is necessary and sufficient that it have the finite intersection property.) By assumption this ultrafilter converges to some point Image may be NSFW.
Clik here to view.. Since Image may be NSFW.
Clik here to view. contains each Image may be NSFW.
Clik here to view., it cannot contain their complements Image may be NSFW.
Clik here to view. (or any subset thereof), so Image may be NSFW.
Clik here to view. cannot converge to any element of Image may be NSFW.
Clik here to view. $\bigcup C_i^c$ . It follows that Image may be NSFW.
Clik here to view. $x \in \bigcap C_i$ .

Note that unlike the case of sequential compactness of metric spaces, we don’t have to pass to a subsequence to find a limit; in some sense the ultrafilter has already done the passing to a subsequence for us. This is one reason why ultrafilters are so convenient in analysis.

In voting terms, an ultrafilter Image may be NSFW.
Clik here to view. converging to a set Image may be NSFW.
Clik here to view. of points is a voting system where the elements of Image may be NSFW.
Clik here to view. are “approximate dictators.” A Hausdorff space of voters is then one for which at most one approximate dictator can occupy the throne, and a compact space of voters is one for which at least one approximate dictator must occupy the throne. In probabilistic terms, an ultrafilter Image may be NSFW.
Clik here to view. converging to a set Image may be NSFW.
Clik here to view. of points is a “random variable” which is in a neighborhood of any Image may be NSFW.
Clik here to view. $s \in S$ almost surely. A Hausdorff space is then one for which such a random variable can be at most one such collection of neighborhoods almost surely, and a compact space is one for which such a random variable must be in at least one such collection of neighborhoods almost surely.

Since every ultrafilter on a finite set is principal, I think the above proposition is a elegant way to justify the intuition that compact spaces “behave like” finite discrete spaces. It also manifests a certain duality between the properties of being compact and being Hausdorff. Finally, it helps explains why compact Hausdorff spaces are so special: these are the ones for which every ultrafilter converges to exactly one point.
Thus giving the structure of a compact Hausdorff space on a set Image may be NSFW.
Clik here to view. is equivalent to giving a function Image may be NSFW.
Clik here to view. $\beta X \to X$ describing which ultrafilters converge to which points in a consistent manner. One can think of this as a kind of algebraic structure on Image may be NSFW.
Clik here to view. whose Image may be NSFW.
Clik here to view.-ary operations can be identified with ultrafilters on Image may be NSFW.
Clik here to view.; we’ll come back to this in a later post.

We will now give the ultrafilter proof of Tychonoff’s theorem. First recall that if Image may be NSFW.
Clik here to view. $X_i, i \in I$ are a collection of topological spaces, then the product topology on Image may be NSFW.
Clik here to view. $X = \prod X_i$ is the initial topology with respect to the projections Image may be NSFW.
Clik here to view. $\pi_i : X \to X_i$ . In ultrafilter terms, this means that an ultrafilter Image may be NSFW.
Clik here to view. on Image may be NSFW.
Clik here to view. converges to Image may be NSFW.
Clik here to view. $x \in X$ if and only if Image may be NSFW.
Clik here to view. $\pi_i(F) \to \pi_i(x)$ for all Image may be NSFW.
Clik here to view..

Theorem (Tychonoff): A product of compact spaces is compact.

Proof. Let Image may be NSFW.
Clik here to view. be an ultrafilter on Image may be NSFW.
Clik here to view. $X = \prod X_i$ . Since each space Image may be NSFW.
Clik here to view. X_i is compact, Image may be NSFW.
Clik here to view. $\pi_i(F)$ converges to some non-empty set of points Image may be NSFW.
Clik here to view. $S_i \subset X_i$ for each Image may be NSFW.
Clik here to view.. Then Image may be NSFW.
Clik here to view. converges to precisely the points in Image may be NSFW.
Clik here to view. $\prod S_i$ , and by the axiom of choice (!) this set is non-empty.

I find this proof much more elegant and easier to follow than the typical proof, which is some combination of the Alexander subbase theorem and messing around with a basis for the product topology. Munkres gave an alternate proof when I took his class, but I don’t remember any of the details. This proof, on the other hand, is perhaps maximally easy to reconstruct.

Note that if we assume in addition that each Image may be NSFW.
Clik here to view. X_i is Hausdorff, each Image may be NSFW.
Clik here to view. S_i is a singleton, so we do not need to use the axiom of choice to prove that Image may be NSFW.
Clik here to view. $\prod S_i$ is non-empty. We only need the ultrafilter lemma (to prove all of the equivalences above), so it follows that Tychonoff’s theorem for compact Hausdorff spaces follows from (hence is equivalent to) the ultrafilter lemma.

Topology and logic

Since ultrafilters are, at their heart, logical constructions, what we have described above is a tantalizing connection between topology and logic. There appear to be several such connections. There is, for example, this answer and this answer to Minhyong Kim’s MO question about open sets, which identifies topology with an idealization of measurement and with an idealization of computation, respectively, both of which appear to be related to what I’ve said above. There is also an interpretation of the open sets of a topological space as a Heyting algebra, which is related to intuitionistic logic. Perhaps this all fits into a cohesive picture which I’ll understand one day.

Ultrafilters in topology

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Microsoft Intune から展開しているアプリのアップデートについて

18-year-old girl was beaten for half an hour by two Northampton men in 'an...

Car crash in Dunton Bassett leaves driver in critical condition

Macky 2, Two Others In Road Accident

Application log 00000000000000089514: Could not convert queue DLVST90CLNT

Detroit mafia: D’Anna Brothers agree to plea deal

Delivery block field greyed out using VA02

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BREAKING NEWS: Diamond Platnumz Is Reported Dead After Ghastly Car Accident

FIAT 500 B0111 B0112