A product of two compact spaces is compact. The tube lemma did that, and induction stretches it to any finite number of factors. This page is about the jump to infinitely many factors, even uncountably many, where induction can't follow. Tychonoff's theorem says compactness survives the jump anyway.
The price is the axiom of choice. We'll need Zorn's lemma twice, and at the end we'll see that the theorem doesn't just use choice, it is choice in disguise. Along the way we prove the two facts about nets that the Nets page stated and skipped, give a second proof with no nets in it at all, and then spend the theorem on some strange compact spaces.
Start with what we already have.
On Compact Spaces, the tube lemma proved that $X \times Y$ is compact when $X$ and $Y$ are, and induction gave every finite product.
Induction proves a statement for $n = 1, 2, 3, \ldots$, one at a time. It never proves it for $\omega$ factors at once, let alone for $[0, 1]$ factors, one for every real number. Watch the finite proof climb and never arrive.
So we need a new idea. Before hunting for one, notice that the answer depends on which topology the product carries. The Product Topology page mentioned a rival in passing. Here it is properly.
Let $\{X_\alpha\}_{\alpha \in J}$ be an indexed family of spaces. The box topology on $\prod_{\alpha \in J} X_\alpha$ is the topology generated by the basis of all products $\prod_{\alpha \in J} U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$.
The product topology asks for $U_\alpha = X_\alpha$ for all but finitely many $\alpha$; the box topology drops that condition. With finitely many factors the two are the same. With infinitely many, the box topology has many more open sets, and that's bad news for compactness.
Give $\{0, 1\}$ the discrete topology. Is $\{0, 1\}^\omega$ compact in the box topology?
No. For a point $x = (x_1, x_2, \ldots)$, the set $\prod_n \{x_n\}$ is a product of open sets, so it's open in the box topology, and it's just $\{x\}$. Every point is open: the box topology on $\{0, 1\}^\omega$ is discrete.
The open cover by one-point sets has no finite subcover, because $\{0, 1\}^\omega$ is infinite (it's uncountable). So the space isn't compact.
In the product topology the same trick fails: a basic open set pins down only finitely many coordinates and leaves the rest free, so it always contains infinitely many points.
The box topology looked at $\{0, 1\}^\omega$ and gave every single point its own open set. Very generous. Completely useless.
The same thing happens inside the cube. Try it.
Show that $[0, 1]^\omega$ is not compact in the box topology.
Proof. Let $Z = \{0, 1\}^\omega \subseteq [0, 1]^\omega$, the points whose coordinates are all $0$ or $1$.
$Z$ is closed. If $y \notin Z$, some coordinate $y_n$ is strictly between $0$ and $1$, and $\pi_n^{-1}\big((0, 1)\big)$ is an open set containing $y$ that misses $Z$.
$Z$ is discrete. For $x \in Z$, the box $\prod_n \big((x_n - \frac12, x_n + \frac12) \cap [0, 1]\big)$ is open and meets $Z$ only in $x$.
If $[0, 1]^\omega$ were compact, its closed subset $Z$ would be compact. But $Z$ is an infinite discrete space, and the cover by one-point sets has no finite subcover. $\blacksquare$
So whatever proof we find had better use the "all but finitely many" clause somewhere. Both proofs below do, and we'll point at the exact line.
Both proofs need to pick something maximal: a collection of sets so big you can't add anything to it. Infinite sets don't come with biggest elements, so we need a tool that manufactures them. It runs on partial orders.
On Nets, a partial order $\preceq$ was a relation that is reflexive, antisymmetric, and transitive. Directed sets were partial orders where any two elements have something above both.
A partial order can have incomparable elements. The parts where everything is comparable are the ones you can climb.
Let $\preceq$ be a partial order on a set $P$. A chain in $P$ is a subset $C \subseteq P$ in which any two elements are comparable: for all $a, b \in C$, $a \preceq b$ or $b \preceq a$.
An upper bound is somewhere to stand above a whole chain at once. It has to live in $P$, but not necessarily in the chain.
Let $\preceq$ be a partial order on a set $P$ and let $C \subseteq P$. An element $u \in P$ is an upper bound of $C$ if $c \preceq u$ for every $c \in C$.
And maximal means nothing sits strictly above you. That's weaker than being above everything, and the difference matters.
Let $\preceq$ be a partial order on a set $P$. An element $m \in P$ is maximal if there is no $p \in P$ with $m \preceq p$ and $m \neq p$.
A small example shows all three at once.
Let $P$ be the set of proper subsets of $\{1, 2, 3\}$, ordered by $\subseteq$. Find a chain and an upper bound for it, and find every maximal element. Is there a largest element?
$\emptyset \subseteq \{1\} \subseteq \{1, 2\}$ is a chain, and $\{1, 2\}$ is an upper bound for it. The sets $\{1\}$ and $\{2\}$ aren't comparable, so $\{\{1\}, \{2\}\}$ is not a chain, but it still has the upper bound $\{1, 2\}$.
The maximal elements are $\{1, 2\}$, $\{1, 3\}$, and $\{2, 3\}$: the only set strictly bigger than any of them is $\{1, 2, 3\}$, which isn't in $P$. There is no largest element, since no proper subset contains all three of those. Three maximal elements, none of them on top.
In a finite poset, keep climbing and you hit a maximal element. In an infinite one you might climb forever. Zorn's lemma says that if every chain has somewhere to stand above it, you can't climb forever.
Let $P$ be a nonempty partially ordered set in which every chain has an upper bound. Then $P$ has a maximal element.
There's no proof to show. Given the other axioms of set theory, Zorn's lemma is equivalent to the axiom of choice, so it's an axiom wearing a lemma's name. Every "by Zorn's lemma" on this page is the axiom of choice at work.
What's yellow and equivalent to the axiom of choice? Zorn's lemon.
The hypothesis is the whole game. Here's a poset where it fails, and so does the conclusion.
Let $P$ be the set of finite subsets of $\mathbb{N}$, ordered by $\subseteq$. Does every chain have an upper bound? Does $P$ have a maximal element? What changes if $P$ is all subsets of $\mathbb{N}$?
The chain $\{1\} \subseteq \{1, 2\} \subseteq \{1, 2, 3\} \subseteq \cdots$ has no upper bound in $P$: an upper bound would contain every $n$, so it would be infinite. And there's no maximal element: for a finite $F$, the set $F \cup \{\max F + 1\}$ (or $\{1\}$ if $F = \emptyset$) is strictly bigger.
With all subsets of $\mathbb{N}$, the union of any chain is an upper bound, and $\mathbb{N}$ itself is maximal (in fact largest).
That union is the template for every Zorn argument you'll meet.
How a Zorn argument goes. (1) Pick the poset, usually "collections of sets with some property", ordered by ⊆. (2) Check that the union of a chain still has the property: that union is your upper bound. (3) Check the poset is nonempty. (4) Use maximality: "if we could add this set, we would have, so it's already in." Step (4) is where the real content lives. Munkres and the course notes sometimes use the Hausdorff maximal principle instead (every chain sits inside a maximal chain); it's equivalent to Zorn's lemma.
Before using it, test the hypothesis on a few posets. Watch for posets that have a maximal element even though Zorn's lemma doesn't apply.
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The first proof of Tychonoff runs on universal nets.
A net is universal if, for every subset $A$ of $X$, it is eventually in $A$ or eventually in $X \setminus A$. It is frequently in $A$ if for every $\alpha$ there's some $\beta \succeq \alpha$ with $\Gamma(\beta) \in A$.
A universal net has made up its mind about every subset of the space, and that's a lot of subsets. Sequences almost never manage it.
Is the sequence $x_n = \frac1n$ universal in $\mathbb{R}$? Is $y_n = 0$ for $n \ge 10$ (with any first nine terms)?
$(x_n)$ is not universal, even though it converges. Take $E = \{\frac12, \frac14, \frac16, \ldots\}$. The sequence lands in $E$ at every even $n$ and outside it at every odd $n$, so it is frequently in $E$ and frequently in $\mathbb{R} \setminus E$, and eventually in neither.
$(y_n)$ is universal. Given any $A$: if $0 \in A$, the sequence is in $A$ for all $n \ge 10$; if not, it is in $\mathbb{R} \setminus A$ for all $n \ge 10$. The first nine terms don't matter.
That's no accident. For sequences, being universal is the same as being eventually constant.
Show that a sequence $(x_n)$ in a set $X$ is universal if and only if it is eventually constant.
Proof. ($\impliedby$) If $x_n = c$ for $n \ge N$, then for any $A$ the sequence is eventually in $A$ (if $c \in A$) or eventually in $X \setminus A$ (if not).
($\implies$) Suppose $(x_n)$ is not eventually constant. If some value $c$ is taken for infinitely many $n$, let $A = \{c\}$: the sequence is frequently in $A$, and since it's not eventually constant, it is also frequently outside $A$. Otherwise every value is taken only finitely often, so we can pick $n_1 < n_2 < \cdots$ with the values $x_{n_k}$ all different. Let $A = \{x_{n_k} : k \text{ even}\}$. The sequence is in $A$ at every $n_k$ with $k$ even and outside $A$ at every $n_k$ with $k$ odd. Either way it is frequently in $A$ and frequently in $X \setminus A$, so it isn't universal. $\blacksquare$
So the universal subnets we're about to build are almost never subsequences. Subnets are allowed to repeat terms and to be indexed by huge directed sets, and the proof uses both freedoms. This is Theorem 6.13 in Bredon, and the Nets page skipped it.
Every net has a universal subnet.
Proof. Let $\Gamma\colon \mathcal{D} \to X$ be a net and write $x_\alpha = \Gamma(\alpha)$.
Step 1: a maximal collection. Consider collections $\mathcal{F}$ of subsets of $X$ such that
Order them by inclusion. $\{X\}$ is one, so the poset is nonempty. If $\mathcal{K}$ is a chain of such collections, its union satisfies (1), because each of its sets lies in some member of $\mathcal{K}$, and (2), because two of its sets lie in members of $\mathcal{K}$, one of which contains the other, so both sets and their intersection lie in that one. The union is an upper bound for $\mathcal{K}$, so by Zorn's lemma there is a maximal such collection $\mathcal{F}_0$.
$X \in \mathcal{F}_0$: adding $X$ to $\mathcal{F}_0$ keeps (1), since the net is always in $X$, and keeps (2), since $A \cap X = A$. By maximality, $X$ was already there.
Step 2: the subnet. Let $$\mathcal{D}_0 = \{(A, \alpha) \in \mathcal{F}_0 \times \mathcal{D} : x_\alpha \in A\},$$ ordered by $(A, \alpha) \preceq (B, \beta)$ if and only if $B \subseteq A$ and $\alpha \preceq \beta$. This is a partial order, and $\mathcal{D}_0$ is nonempty, since it contains $(X, \alpha)$ for every $\alpha$. It is directed: given $(A, \alpha)$ and $(B, \beta)$ in $\mathcal{D}_0$, choose $\gamma \in \mathcal{D}$ with $\alpha \preceq \gamma$ and $\beta \preceq \gamma$. By (2), $A \cap B \in \mathcal{F}_0$, so by (1) the net is frequently in $A \cap B$: there is $\delta \succeq \gamma$ with $x_\delta \in A \cap B$. Then $(A \cap B, \delta) \in \mathcal{D}_0$ lies above both.
Let $h\colon \mathcal{D}_0 \to \mathcal{D}$, $h(A, \alpha) = \alpha$. It is final: given $\alpha \in \mathcal{D}$, if $(X, \alpha) \preceq (B, \beta)$, then $\alpha \preceq \beta = h(B, \beta)$. So $\Gamma \circ h$, which sends $(A, \alpha)$ to $x_\alpha$, is a subnet.
Step 3: the subnet is universal. Let $S \subseteq X$ and suppose the subnet is frequently in $S$. First we show the original net is frequently in $S \cap A$ for every $A \in \mathcal{F}_0$. Fix $A \in \mathcal{F}_0$ and $\alpha \in \mathcal{D}$. By (1) there is $\alpha' \succeq \alpha$ with $x_{\alpha'} \in A$, so $(A, \alpha') \in \mathcal{D}_0$. Since the subnet is frequently in $S$, there is $(B, \beta) \succeq (A, \alpha')$ with $x_\beta \in S$. Then $\beta \succeq \alpha' \succeq \alpha$, and $x_\beta \in B \subseteq A$, so $x_\beta \in S \cap A$.
Now the collection $\mathcal{F}_0 \cup \{S \cap A : A \in \mathcal{F}_0\}$ satisfies (1), by what we just showed, and (2), since $(S \cap A) \cap (S \cap B) = S \cap (A \cap B)$ and $A \cap (S \cap B) = S \cap (A \cap B)$, with $A \cap B \in \mathcal{F}_0$. By maximality it equals $\mathcal{F}_0$, so $S = S \cap X \in \mathcal{F}_0$.
If the subnet were also frequently in $X \setminus S$, the same argument would put $X \setminus S$ in $\mathcal{F}_0$, and then $\emptyset = S \cap (X \setminus S) \in \mathcal{F}_0$ by (2). But the net is not frequently in $\emptyset$, contradicting (1). So the subnet is not frequently in $X \setminus S$, which means it is eventually in $S$.
Every $S$ has now been decided: either the subnet is frequently in $S$, and then eventually in $S$, or it is not, and then it is eventually in $X \setminus S$. The subnet is universal. $\blacksquare$
Read the proof like this. $\mathcal{F}_0$ is a maximal list of places the net keeps coming back to, closed under intersections. The subnet only walks inside sets on the list, deeper and deeper. Any set the subnet keeps visiting would have made the list already, so it can't keep visiting a set and its complement. Here's that picture on the bouncing sequence from the Nets page.
The proof's only real choice is which sets go on the list. You make those choices here: each time, say whether the subnet ends up inside the set or outside it, and watch which terms survive.
On a finite set every decision is easy, and the subnet ends up constant. The interesting universal nets live in infinite sets, and there's a catch.
Every universal net I can actually write down is eventually constant. The interesting ones provably exist, and nobody can write one down. Topology's cryptids.
If you've met ultrafilters before, you've met $\mathcal{F}_0$.
The maximal collection F₀ in the proof is an ultrafilter on X: it never contains the empty set, it's closed under finite intersections and under supersets, and for every S it contains S or its complement. Many books prove Tychonoff with ultrafilters instead of nets; it's the same proof in different clothes.
Two quick facts about universal nets, both used in a moment.
(a) Let $\Gamma$ be a universal net in $X$ and $f\colon X \to Y$ any function (not necessarily continuous). Show that $f \circ \Gamma$ is universal in $Y$. (b) Show that every subnet of a universal net is universal.
Proof. (a) Let $B \subseteq Y$. Since $\Gamma$ is universal, it is eventually in $f^{-1}(B)$ or eventually in $X \setminus f^{-1}(B) = f^{-1}(Y \setminus B)$. So $f \circ \Gamma$ is eventually in $B$ or eventually in $Y \setminus B$.
(b) Let $\Gamma\colon \mathcal{D} \to X$ be universal and $\Gamma \circ h$ a subnet, with $h\colon \tilde{\mathcal{D}} \to \mathcal{D}$ final. Given $A \subseteq X$, say $\Gamma$ is eventually in $A$: there is $\alpha_0$ with $\Gamma(\beta) \in A$ for all $\beta \succeq \alpha_0$. (The other case is the same with $X \setminus A$.) Since $h$ is final, there is $\tilde{\alpha}$ with $\alpha_0 \preceq h(\tilde{\beta})$ whenever $\tilde{\alpha} \preceq \tilde{\beta}$, and then $\Gamma(h(\tilde{\beta})) \in A$. So $\Gamma \circ h$ is eventually in $A$. $\blacksquare$
Now spot universal nets, and the convergent nets that only look universal.
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The second skipped fact. The Nets page listed four ways to say "compact" and proved none of them. Here's the one Tychonoff needs.
A space $X$ is compact if and only if every universal net in $X$ converges.
Proof. ($\implies$) Let $X$ be compact and $\Gamma\colon \mathcal{D} \to X$ universal, and suppose $\Gamma$ converges to no point. Then each $x \in X$ has an open neighborhood $U_x$ such that $\Gamma$ is not eventually in $U_x$. Since $\Gamma$ is universal, it is eventually in $X \setminus U_x$. By compactness, finitely many $U_{x_1}, \ldots, U_{x_n}$ cover $X$. Say $\Gamma$ is in $X \setminus U_{x_i}$ from $\alpha_i$ on. Since $\mathcal{D}$ is directed, some $\alpha$ lies above $\alpha_1, \ldots, \alpha_n$, and then $$\Gamma(\alpha) \in \bigcap_{i=1}^n (X \setminus U_{x_i}) = X \setminus \bigcup_{i=1}^n U_{x_i} = \emptyset,$$ which is impossible.
($\impliedby$) Suppose every universal net in $X$ converges. We use the closed-set form of compactness: let $\mathcal{A}$ be a collection of closed sets with the finite intersection property, and we show $\bigcap_{C \in \mathcal{A}} C \neq \emptyset$. Let $\mathcal{D}$ be the set of finite nonempty subcollections of $\mathcal{A}$, ordered by inclusion; it is directed, since $F \cup G$ lies above $F$ and $G$. For each $F \in \mathcal{D}$, choose $x_F \in \bigcap F$, which is nonempty by the finite intersection property.
By the universal subnet theorem, the net $(x_F)$ has a universal subnet, which converges to some $x$ by hypothesis. Fix $C \in \mathcal{A}$. The net is eventually in $C$: if $F \supseteq \{C\}$, then $x_F \in \bigcap F \subseteq C$. By finality, so is the subnet. If $x$ were not in $C$, the subnet would also be eventually in the neighborhood $X \setminus C$ of $x$, and some index would land in both $C$ and $X \setminus C$. So $x \in C$. This holds for every $C \in \mathcal{A}$, so $x \in \bigcap_{C \in \mathcal{A}} C$. $\blacksquare$
The forward direction is the honest one: a universal net has already decided about every neighborhood, so compactness can corner it. The backward direction quietly uses Zorn's lemma again, through the universal subnet. With this, (1) ⟺ (3) on the Nets page's list is proved. It also gives a quick way to see that a space is not compact.
Use the theorem to show that $(0, 1]$ is not compact.
The sequence $x_n = \frac1n$ has a universal subnet. In $\mathbb{R}$, every subnet of $(x_n)$ converges to $0$, and limits in $\mathbb{R}$ are unique because it's Hausdorff. So the subnet converges to no point of $(0, 1]$: a limit there would also be a limit in $\mathbb{R}$, different from $0$. A universal net in $(0, 1]$ that doesn't converge means $(0, 1]$ isn't compact.
The rest of the list follows quickly.
Show that $X$ is compact if and only if every net in $X$ has a convergent subnet.
Proof. ($\implies$) Every net has a universal subnet, which converges because $X$ is compact.
($\impliedby$) By the theorem, it's enough to show that every universal net $\Gamma$ converges. By hypothesis $\Gamma$ has a subnet $\Gamma \circ h$ converging to some $x$. Let $U$ be a neighborhood of $x$. If $\Gamma$ were not eventually in $U$, it would be eventually in $X \setminus U$ (it's universal), and by finality so would $\Gamma \circ h$. But $\Gamma \circ h$ is eventually in $U$, and it can't be eventually in both. So $\Gamma$ is eventually in every neighborhood of $x$: $\Gamma \to x$. $\blacksquare$
Everything is in place, and the proof from the course notes is three sentences. Take a universal net in $\prod X_\alpha$. Each coordinate net $\pi_\alpha \circ \Gamma$ is universal (part (a) of the exercise), so it converges in the compact space $X_\alpha$, to some $y_\alpha$. A net in a product converges exactly when every coordinate does, so $\Gamma$ converges to $y = (y_\alpha)$, and the product is compact by the theorem above.
That proof is written out in full, with the lemma on convergence coordinate by coordinate, on Stone–Čech Compactification, where it was first needed. Now every brick under it is proved too.
Here's that proof as a picture, with five coordinates standing in for all of them. Each axis settles down on its own, and the whole polyline settles with them.
The product topology is used in "converges exactly when every coordinate does". In the box topology that's false: $x_k = (\frac1k, \frac1k, \frac1k, \ldots)$ converges to $0$ in every coordinate, but the box $\prod_n (-\frac1n, \frac1n)$ around $0$ contains no $x_k$ at all. Choice shows up twice: once inside the universal subnet, and once in picking a limit $y_\alpha$ in every factor at the same time (limits needn't be unique if the factors aren't Hausdorff).
Now a second proof, Munkres's, with no nets anywhere. Instead of a net that has decided about every set, we build a collection of sets that can't get any bigger. The first step is Zorn again.
Let $X$ be a set and $\mathcal{A}$ a collection of subsets of $X$ with the finite intersection property. There is a collection $\mathcal{M} \supseteq \mathcal{A}$ of subsets of $X$ that has the finite intersection property and is maximal: no collection that properly contains $\mathcal{M}$ has the finite intersection property.
Proof. Let $P$ be the set of collections $\mathcal{B} \supseteq \mathcal{A}$ of subsets of $X$ with the finite intersection property, ordered by $\subseteq$. $P$ is nonempty, since $\mathcal{A} \in P$. Let $\mathcal{K}$ be a chain in $P$. If $\mathcal{K}$ is empty, $\mathcal{A}$ is an upper bound. Otherwise let $\mathcal{U}$ be the union of the collections in $\mathcal{K}$; it contains $\mathcal{A}$. Given $B_1, \ldots, B_n \in \mathcal{U}$, each $B_i$ lies in some $\mathcal{B}_i \in \mathcal{K}$, and since $\mathcal{K}$ is a chain, one of $\mathcal{B}_1, \ldots, \mathcal{B}_n$ contains the others. That one has the finite intersection property, so $B_1 \cap \cdots \cap B_n \neq \emptyset$. So $\mathcal{U} \in P$ is an upper bound for $\mathcal{K}$. By Zorn's lemma $P$ has a maximal element $\mathcal{M}$. $\blacksquare$
A maximal collection is greedy in two ways. It already holds every finite intersection of its sets, and it already holds every set that meets all of its sets.
Let $X$ be a set and let $\mathcal{M}$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property. Then:
Proof. (a) Let $B = M_1 \cap \cdots \cap M_n$ with each $M_i \in \mathcal{M}$. A finite intersection of elements of $\mathcal{M} \cup \{B\}$ is, after writing out $B$, a finite intersection of elements of $\mathcal{M}$, so it is nonempty. Thus $\mathcal{M} \cup \{B\}$ has the finite intersection property, and by maximality $B \in \mathcal{M}$.
(b) A finite intersection of elements of $\mathcal{M} \cup \{A\}$ that involves $A$ has the form $A \cap M_1 \cap \cdots \cap M_n = A \cap B$, where $B \in \mathcal{M}$ by (a). It is nonempty because $A$ meets every element of $\mathcal{M}$. So $\mathcal{M} \cup \{A\}$ has the finite intersection property, and by maximality $A \in \mathcal{M}$. $\blacksquare$
On a finite set you can see exactly what these collections are.
Let $X = \{1, 2, 3, 4\}$. Show that $\mathcal{M} = \{A \subseteq X : 3 \in A\}$ is maximal with the finite intersection property, and that every maximal collection on $X$ looks like this for some point.
Every set in $\mathcal{M}$ contains $3$, so every intersection does: $\mathcal{M}$ has the finite intersection property. If $A \notin \mathcal{M}$, then $3 \notin A$, so $A \cap \{3\} = \emptyset$ with $\{3\} \in \mathcal{M}$. Adding $A$ breaks the property, so $\mathcal{M}$ is maximal.
Conversely, let $\mathcal{M}$ be any maximal collection. It is finite, so by (a) the intersection $M$ of all its sets is in $\mathcal{M}$, and $M \neq \emptyset$. Pick $p \in M$. Every set of $\mathcal{M}$ contains $M$, hence $p$, so every set containing $p$ meets every set of $\mathcal{M}$, and by (b) it belongs to $\mathcal{M}$. So $\mathcal{M} = \{A : p \in A\}$.
Find out for yourself: build a collection on five points, one set at a time, and see where you're forced to stop.
Your turn on paper. Every wrong choice below either loses the finite intersection property or could still grow.
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Infinite sets have maximal collections that aren't "everything containing $p$", and those are the ones that do the work. Here's the theorem.
An arbitrary product of compact spaces is compact in the product topology.
Proof. Let $X = \prod_{\alpha \in J} X_\alpha$ with each $X_\alpha$ compact, and let $\mathcal{A}$ be a collection of subsets of $X$ with the finite intersection property. We show $\bigcap_{A \in \mathcal{A}} \overline{A} \neq \emptyset$. For a collection of closed sets this is the closed-set form of compactness, so it proves $X$ is compact.
By the first lemma, choose a maximal collection $\mathcal{M} \supseteq \mathcal{A}$ with the finite intersection property. It is enough to find a point $x$ in $\overline{D}$ for every $D \in \mathcal{M}$.
Choosing $x$. Fix $\alpha \in J$. The collection $\{\pi_\alpha(D) : D \in \mathcal{M}\}$ has the finite intersection property in $X_\alpha$, since $$\pi_\alpha(D_1) \cap \cdots \cap \pi_\alpha(D_n) \supseteq \pi_\alpha(D_1 \cap \cdots \cap D_n) \neq \emptyset.$$ So do the closed sets $\overline{\pi_\alpha(D)}$, and since $X_\alpha$ is compact, they have a point in common. Choose one, $x_\alpha$, and let $x = (x_\alpha)_{\alpha \in J}$.
$x$ works. Let $\pi_\beta^{-1}(U_\beta)$ be a subbasis element containing $x$, with $U_\beta$ open in $X_\beta$ and $x_\beta \in U_\beta$. For each $D \in \mathcal{M}$, $x_\beta \in \overline{\pi_\beta(D)}$, so $U_\beta$ meets $\pi_\beta(D)$: some $y \in D$ has $y_\beta \in U_\beta$, and then $y \in \pi_\beta^{-1}(U_\beta) \cap D$. So $\pi_\beta^{-1}(U_\beta)$ meets every element of $\mathcal{M}$, and by part (b) of the second lemma, it belongs to $\mathcal{M}$.
Every basis element containing $x$ is a finite intersection of subbasis elements containing $x$, so by part (a) it belongs to $\mathcal{M}$. Since $\mathcal{M}$ has the finite intersection property, it meets every $D \in \mathcal{M}$. Every basis element around $x$ meets $D$, so $x \in \overline{D}$. $\blacksquare$
Two things to notice. "All but finitely many" is in the last paragraph: a basis element of the product topology is a finite intersection of subbasis elements, which is exactly what part (a) can absorb. And choice is used twice, once for $\mathcal{M}$ and once for picking $x_\alpha$ in every factor. Here's the shadow argument in two coordinates.
The engine under both proofs is the closed-set form of compactness. In a compact space, closed sets with the finite intersection property can't escape each other. In a space that isn't compact, they can. Try to make them escape.
Tychonoff's converse is easy, with one catch about empty factors.
Suppose $\prod_{\alpha \in J} X_\alpha$ is nonempty and compact. Show that each $X_\alpha$ is compact. Why is "nonempty" needed?
Proof. Since the product is nonempty, each projection $\pi_\beta$ is surjective: given $b \in X_\beta$, take any point of the product and change its $\beta$ coordinate to $b$. Projections are continuous, and a continuous image of a compact space is compact, so $X_\beta = \pi_\beta\big(\prod X_\alpha\big)$ is compact.
Without "nonempty" it fails: $\mathbb{R} \times \emptyset = \emptyset$ is compact, but $\mathbb{R}$ isn't. $\blacksquare$
Now decide which products are compact. Watch the topology and watch every factor.
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The first payoff is the cube that Urysohn's metrization theorem lives in.
Is $[0, 1]^\omega$ compact in the product topology? Is it compact in the uniform topology?
Product topology: yes, by Tychonoff's theorem, since $[0, 1]$ is compact. (It's a copy of the Hilbert cube.)
Uniform topology: no. Let $e_k$ have $1$ in coordinate $k$ and $0$ elsewhere. For $j \neq k$, $\overline{\rho}(e_j, e_k) = 1$, so no subsequence of $(e_k)$ is Cauchy and none converges. A compact metric space is sequentially compact, so $[0, 1]^\omega$ with $\overline{\rho}$ is not compact.
So among the three topologies on the same set, box, uniform, and product, only the coarsest is compact. More open sets means more open covers to defeat.
The cube $[0, 1]^J$ for a huge index set $J$ is where Stone–Čech Compactification builds $\beta X$: imbed $X$ in the cube and take the closure. Tychonoff is the reason the closure is compact.
The Cantor set, which looks like it was built by hand, turns out to be a product in disguise.
Let $C \subseteq [0, 1]$ be the Cantor set, the numbers with a ternary expansion using only the digits $0$ and $2$. Show that $g\colon \{0, 1\}^\omega \to C$, $g(a) = \sum_{n \ge 1} \frac{2a_n}{3^n}$, is a homeomorphism, where $\{0, 1\}$ is discrete and $\{0, 1\}^\omega$ has the product topology.
Proof. Onto: every point of $C$ has the form $\sum 2a_n/3^n$ with $a_n \in \{0, 1\}$, by the description of $C$.
One-to-one: if $a \neq b$, let $n$ be the first index where they differ, say $a_n = 1$ and $b_n = 0$. Then $$g(a) - g(b) \ge \frac{2}{3^n} - \sum_{k > n} \frac{2}{3^k} = \frac{2}{3^n} - \frac{1}{3^n} > 0.$$
Continuous: given $\varepsilon > 0$, choose $N$ with $3^{-N} < \varepsilon$. The points $b$ that agree with $a$ in the first $N$ coordinates form a basic open set around $a$, and for them $|g(a) - g(b)| \le \sum_{k > N} \frac{2}{3^k} = 3^{-N} < \varepsilon$.
By Tychonoff's theorem $\{0, 1\}^\omega$ is compact, and $C$ is Hausdorff as a subspace of $\mathbb{R}$. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism (Compact Spaces). $\blacksquare$
The last payoff is a warning. Compact used to mean "every sequence has a convergent subsequence", at least in metric spaces. In a big product that's false, and nets were never optional.
Let $X = [0, 1]^{[0, 1]}$, the set of all functions $g\colon [0, 1] \to [0, 1]$ with the product topology, so convergence is pointwise. For $n \ge 1$, let $f_n(x)$ be the $n$th binary digit of $x \in [0, 1)$, using the expansion that doesn't end in all $1$s, and let $f_n(1) = 0$. Show that $X$ is compact but $(f_n)$ has no convergent subsequence.
$X$ is compact by Tychonoff's theorem, and each $f_n$ takes values in $\{0, 1\}$, so $f_n \in X$.
Take any subsequence $n_1 < n_2 < n_3 < \cdots$. Let $x$ be the number whose binary digit in place $n_k$ is $1$ for even $k$, and whose other digits are all $0$: $$x = \sum_{k \text{ even}} 2^{-n_k}.$$ Its expansion has infinitely many $0$s, so it's the expansion the definition uses, and $x < 1$. Then $f_{n_k}(x) = 0, 1, 0, 1, \ldots$ for $k = 1, 2, 3, 4, \ldots$, which doesn't converge. A sequence converges in $X$ only if it converges at every coordinate, so $(f_{n_k})$ doesn't converge.
Every subsequence gets beaten by its own point $x$. Watch one get beaten.
Now you pick the subsequence, as cleverly as you like, and try to escape.
Both proofs used choice. You can't avoid it: Kelley showed in 1950 that Tychonoff's theorem gives the axiom of choice back. The axiom says a product of nonempty sets is nonempty, and the trick is to turn each set into a compact space with one extra point.
Suppose every product of compact spaces is compact. Then for every family $\{X_\alpha\}_{\alpha \in J}$ of nonempty sets, the product $\prod_{\alpha \in J} X_\alpha$ is nonempty.
Proof. Pick an object $\infty$ that lies in none of the $X_\alpha$ (for instance, $\infty = \bigcup_\alpha X_\alpha$ works). Let $Y_\alpha = X_\alpha \cup \{\infty\}$ with the topology $\{\emptyset, \{\infty\}, X_\alpha, Y_\alpha\}$. This is a topology: the union of $\{\infty\}$ and $X_\alpha$ is $Y_\alpha$, and their intersection is $\emptyset$. It has only four open sets, so $Y_\alpha$ is compact.
By hypothesis $Y = \prod_\alpha Y_\alpha$ is compact. (It's nonempty without any choice: the point with every coordinate $\infty$ is in it.) Let $C_\alpha = \pi_\alpha^{-1}(X_\alpha)$. Each $X_\alpha$ is closed in $Y_\alpha$, since its complement $\{\infty\}$ is open, so each $C_\alpha$ is closed in $Y$.
The $C_\alpha$ have the finite intersection property. Given $\alpha_1, \ldots, \alpha_n$, choose $x_i \in X_{\alpha_i}$ (finitely many choices need no axiom), and let $y$ have coordinate $x_i$ at $\alpha_i$ and $\infty$ everywhere else. Then $y \in C_{\alpha_1} \cap \cdots \cap C_{\alpha_n}$.
By compactness, $\bigcap_\alpha C_\alpha$ contains a point $z$. Every coordinate $z_\alpha$ lies in $X_\alpha$, so $z \in \prod_\alpha X_\alpha$. $\blacksquare$
Together with the proofs above: given the other axioms of set theory, Tychonoff's theorem and the axiom of choice are equivalent. It's worth looking at one of Kelley's spaces up close.
Write out Kelley's space $Y$ for $X = \{a, b\}$. Is it compact? Is it Hausdorff?
$Y = \{a, b, \infty\}$ with open sets $\emptyset$, $\{\infty\}$, $\{a, b\}$, and $Y$. It's compact, since there are only finitely many open sets. It's not Hausdorff: the only open set containing $a$ besides $Y$ is $\{a, b\}$, which also contains $b$, so $a$ and $b$ can't be separated.
That failure of Hausdorff is the whole story of the weaker versions.
Two versions need less than full choice. Finite products need no choice at all (the tube lemma). Products of compact Hausdorff spaces need only the ultrafilter lemma, which is strictly weaker than the axiom of choice: with unique limits, the proof never has to choose which limit to use, and the universal net can come from an ultrafilter. Kelley's spaces Y_α are not Hausdorff, and that's exactly where full choice sneaks back in.
Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim about anything on this page. Decide, then read why.
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Next, the proofs. Each one stops partway, and you pick the line that comes next. Every wrong line is false or doesn't follow.
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Last, the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. They're all multiple choice.
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Last, the Spicy Bonus: a third proof of Tychonoff. Zorn's lemma shows you only have to check covers by subbasis elements, and for a product those are easy.
(a) Alexander's subbase theorem. Let $\mathcal{S}$ be a subbasis for the topology of $X$. Prove that if every cover of $X$ by elements of $\mathcal{S}$ has a finite subcover, then $X$ is compact. (Hint: use Zorn's lemma to get an open cover with no finite subcover that is maximal among such covers.)
(b) Use (a) to prove Tychonoff's theorem.
Proof. (a) Suppose $X$ is not compact. Let $P$ be the set of open covers of $X$ with no finite subcover, ordered by $\subseteq$; it is nonempty. If $\mathcal{K}$ is a nonempty chain in $P$, its union $\mathcal{U}$ is an open cover, and it has no finite subcover: finitely many of its sets lie in finitely many members of $\mathcal{K}$, hence all in the largest of them, which has no finite subcover. (The empty chain is bounded by any element of $P$.) By Zorn's lemma there is a maximal $\mathcal{C} \in P$.
Claim: $\mathcal{C} \cap \mathcal{S}$ covers $X$. Let $x \in X$. Some $U \in \mathcal{C}$ contains $x$, and since $\mathcal{S}$ is a subbasis there are $S_1, \ldots, S_n \in \mathcal{S}$ with $x \in S_1 \cap \cdots \cap S_n \subseteq U$. Suppose no $S_i$ is in $\mathcal{C}$. By maximality, each $\mathcal{C} \cup \{S_i\}$ is an open cover with a finite subcover, which must use $S_i$: so $X = S_i \cup W_i$, where $W_i$ is a finite union of elements of $\mathcal{C}$. Then $$U \cup W_1 \cup \cdots \cup W_n = X,$$ because a point outside every $W_i$ lies in every $S_i$, hence in $U$. That is a finite subcover from $\mathcal{C}$, a contradiction. So some $S_i \in \mathcal{C} \cap \mathcal{S}$ contains $x$.
Now $\mathcal{C} \cap \mathcal{S}$ is a cover by subbasis elements, so by hypothesis it has a finite subcover. That is a finite subcover from $\mathcal{C}$, a contradiction. So $X$ is compact.
(b) Let $X = \prod_{\alpha \in J} X_\alpha$ with each $X_\alpha$ compact. If some $X_\alpha$ is empty, $X$ is empty and compact; assume not. The sets $\pi_\alpha^{-1}(U)$, $U$ open in $X_\alpha$, form a subbasis. Let $\mathcal{C}$ be a cover of $X$ by such sets, and for each $\alpha$ let $\mathcal{U}_\alpha = \{U : \pi_\alpha^{-1}(U) \in \mathcal{C}\}$. If some $\mathcal{U}_\alpha$ covers $X_\alpha$, compactness gives $U_1, \ldots, U_n \in \mathcal{U}_\alpha$ covering $X_\alpha$, and then $\pi_\alpha^{-1}(U_1), \ldots, \pi_\alpha^{-1}(U_n)$ is a finite subcover of $\mathcal{C}$. Otherwise, for every $\alpha$ choose $x_\alpha \in X_\alpha$ outside every set of $\mathcal{U}_\alpha$. The point $x = (x_\alpha)$ lies in no $\pi_\alpha^{-1}(U) \in \mathcal{C}$, so $\mathcal{C}$ doesn't cover $X$, a contradiction. Either way $\mathcal{C}$ has a finite subcover, and by (a) $X$ is compact. $\blacksquare$