KindiakMath

Compact Products on Steroids

Let’s talk about the Tychonoff theorem, or as I like to call it, compact products on steroids.

Theorem 1 (Tychonoff Theorem). Let \{ K_\alpha \} be an arbitrary collection of compact spaces. Then \prod_\alpha K_\alpha is compact when equipped with the product topology.

This theorem will be useful in helping us obtain the “largest” compactification, but before we get there, we’ll take inspiration from this paper to prove the Tychonoff theorem.

Oftentimes, tackling difficult problems requires us to shift our perspective by just a little bit, so that this sub-problem requires a little less effort to conquer.

Definition 1. A collection \mathcal C of subsets of K is said to possess the finite intersection property if any finite sub-collection \{C_1,\dots, C_n\} \subseteq \mathcal C yields a nonempty intersection: \bigcap_{k=1}^n C_k \neq \emptyset.

Lemma 1. A topological space K is compact if and only if for any collection \mathcal C of closed subsets of K, if \mathcal C possesses the finite intersection property, then \bigcap_{C \in \mathcal C} C \neq \emptyset.

Proof. We prove this lemma in two directions.

For the direction (\Rightarrow), assume that K is compact and let \mathcal C be any collection of closed subsets of K that possesses the finite intersection property. Suppose for a contradiction that \bigcap_{C \in \mathcal C} C = \emptyset. Then \{ K \backslash C : C \in \mathcal C\} forms an open cover for K. Since K is compact, we can extract a finite sub-cover \{ K \backslash C_1, \dots, K \backslash C_n\}, so that \bigcap_{i=1}^n C_i = \emptyset, a contradiction to the finite intersection property.

For the direction (\Leftarrow), let \mathcal U be any open cover of K. Suppose for a contradiction that for any finite sub-collection \{U_1,\dots,U_n\}, \bigcup_{i=1}^n U_i \neq K. Then \bigcap_{i=1}^n K \backslash U_i \neq \emptyset. By hypothesis, \bigcap_{U \in \mathcal U} K \backslash U \neq \emptyset, implying that \bigcup_{U \in \mathcal U} U \neq K, which contradicts \mathcal U being an open cover of K.

Lemma 2. For any topological space K and \mathcal C \subseteq 2^K that possesses the finite intersection property, there exists a collection \mathcal C_{\max} of subsets of K that has the finite intersection property and contains \mathcal C.

Proof. We first note that the power set 2^K of K denotes all subsets of K and that \mathcal C \subseteq 2^K so that \mathcal C \in 2^{2^K}. Hence, 2^{2^K} denotes all sub-collections of subsets of K.

Let \mathcal F \subseteq 2^{2^K} denote the sub-collection of sub-collection of subsets of K that possess the finite intersection property and contains \mathcal C. More precisely, we say that \mathcal A \in \mathcal F if and only if \mathcal C \subseteq \mathcal A and \mathcal A is a sub-collection of subsets of K that possesses the finite intersection property.

Define the partial order \leq on \mathcal F via \mathcal A \leq \mathcal B \! \! \iff \! \! \mathcal A \subseteq \mathcal B. Fix any chain \mathcal G \subseteq \mathcal F, the elements of which are subsets of K. Then the sub-collection \mathcal H:= \bigcup_{\mathcal B \in \mathcal G} \mathcal B is an upper bound of \mathcal G, in the sense that for any \mathcal B \in \mathcal G, \mathcal B \leq \mathcal H. To check that \mathcal H \in \mathcal F, we need to prove that it possesses the finite intersection property.

Let H_1,\dots, H_n be elements in \mathcal H. By construction, each H_i is an element of some sub-collection \mathcal B_i \in \mathcal G. Suppose \mathcal B_i \subseteq \mathcal B_n for any i without loss of generality. This implies that H_i \in \mathcal B_n for any i. Since \mathcal B_n \in \mathcal G \subseteq \mathcal F, it possesses the finite intersection property. Hence, \bigcap_{i=1}^n H_i \neq \emptyset, as required.

Since this upper bound holds given any chain \mathcal G in \mathcal F, by Zorn’s lemma, \mathcal F has a maximal element; call this element \mathcal C_{\max}.

Lemma 3. Define \mathcal C_{\max} according to Lemma 2.

  • A finite intersection of elements in \mathcal C_{\max} is an element of \mathcal C_{\max}.
  • L \subseteq K belongs to \mathcal C_{\max} if and only if L intersects every element of \mathcal C_{\max}.

Proof. For the first claim, fix C_1,\dots, C_n \in \mathcal C_{\max}. Define C := \bigcap_{i=1}^n C_i. We leave it as an exercise to verify that \mathcal C_{\max} \cup \{C\} possesses the finite intersection property, so that the maximality of \mathcal C_{\max} yields

C \in \mathcal C_{\max} \cup \{C\} \subseteq \mathcal C_{\max}.

For the second claim, we prove in two directions. For the direction (\Rightarrow), fix L \in \mathcal C_{\max}. For any M \in \mathcal C_{\max}, since \{L, M\} is finite, L \cap M \neq \emptyset, as required.

For the direction (\Leftarrow), fix L \subseteq K. Suppose L \cap M \neq \emptyset for any M \in \mathcal C_{\max}. By a similar argument as before, the collection \mathcal C_{\max} \cup \{L\} possesses the finite intersection property, so that L \in \mathcal C_{\max}, as required.

These tools equip us to properly prove the Tychonoff theorem (Theorem 1).

Proof of Theorem 1. Let \{ K_\alpha \} be an arbitrary collection of compact spaces and define K := \prod_{\alpha} K_\alpha. Let \mathcal C be any collection of closed subsets of K that possess the finite intersection property. By Lemma 1, we need to prove that \bigcap_{C \in \mathcal C} C \neq \emptyset. By Lemma 2, there exists a maximal collection \mathcal C_{\max} of subsets of K that contain the finite intersection property and contains \mathcal C. Since \bigcap_{C \in \mathcal C} C \supseteq \bigcap_{T \in \mathcal C_{\max}} \bar T, it suffices to prove that the latter is nonempty.

For each \alpha, let \pi_{\alpha} : K \to K_\alpha denote the projection map. Consider the set \mathcal P_\alpha := \{ \pi_\alpha(T): T \in \mathcal C_{\max}\}. Since \mathcal C_{\max} has the finite intersection property, so does \mathcal P_\alpha. Hence, the set \{\overline{\pi_\alpha(T)} : T \in \mathcal C_{\max}\} is a collection of closed subsets of K_\alpha that have the finite intersection property. Since K_\alpha is compact, by Lemma 1, \bigcap_{T \in \mathcal C_{\max}} \overline{\pi_\alpha(T)} \neq \emptyset. Choose x_\alpha \in \bigcap_{T \in \mathcal C_{\max}} \overline{\pi_\alpha(T)}.

Define \mathbf x \in K by \mathbf x(\alpha) := x_\alpha. We claim that \mathbf x \in \bigcap_{T \in \mathcal C_{\max}} \bar T. This means we need to prove that \mathbf x \in \bar T for any T \in \mathcal C_{\max}. In particular, for any T \in \mathcal C_{\max}, we need to prove that for any basic neighbourhood B of x, B \cap T \neq \emptyset. By Lemma 3, this criterion is equivalent to the proposition B \in \mathcal C_{\max}.

To that end, fix B = \prod_{k=1}^n U_k \times \prod_{\alpha \neq 1,\dots, k} K_\alpha containing x. Recall that

\displaystyle B = \prod_{k=1}^n U_k \times \prod_{\alpha \neq 1,\dots, k} K_\alpha = \bigcap_{k=1}^n \pi_k^{-1}(U_k),

so by Lemma 3, it suffices to verify that \pi_k^{-1}(U_k) \in \mathcal C_{\max} for each k.

Since x_k\in \bigcap_{T \in \mathcal C_{\max}} \overline{\pi_k(T)} and U_k contains x_k, for any T \in \mathcal C_{\max},

\displaystyle U_k \cap \pi_k(T) \neq \emptyset \quad \Rightarrow \quad \pi_k^{-1}(U_k) \cap T \neq \emptyset.

By Lemma 3, \pi_k^{-1}(U_k) \in \mathcal C_{\max}, as required.

Credits to James Munkres, the author of the famous Topology textbook, who outlined this proof.

Now that we have established the Tychonoff theorem, we are ready to construct the largest compactification of a (suitably generically well-behaved topological space).

—Joel Kindiak, 6 May 25, 1746H

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