Compact Products

Previously, we strove to characterize compact sets in $\mathbb R^n$ as closed and bounded sets. This is known as the Heine-Borel theorem, and we state it below.

Theorem 1 (Heine-Borel Theorem). Let $K \subseteq \mathbb R^n$ be any subset. Then $K$ is compact if and only if it is closed and bounded.

Partial Proof. We prove the direction $(\Leftarrow)$ and leave the direction $(\Rightarrow)$ as an exercise. Since $K$ is bounded, there exists $M > 0$ such that for any $x \in K$ , $d(x, 0) \leq M$ . Therefore, $K \subseteq [-M, M]^n$ . Since $K$ is closed and $[-M, M]^n$ is compact, $K$ is compact, as required.

Now, why is the proof partial? It’s not because we only proved $(\Leftarrow)$ . It’s because we didn’t justify a crucial assertion: how do we know that $[-M, M]^n$ is compact? To be sure, we have actually proven Theorem 1 for $n = 1$ . The challenge is to verify the general case. For simplicity, we will prove the compactness of $[0, 1]^n$ —its results can readily generalize to other closed and bounded rectangles.

Lemma 1. $K := [0, 1]^n$ is compact.

Proof. Let $\{ \mathbf x_k \}$ be any $K$ -sequence. For each $k$ , write $\mathbf x_k = (x_{1,k}, \dots, x_{n,k})$ . We observe that $\{ x_{1,k} \}$ is a $[0,1]$ -sequence. Since $[0, 1]$ is compact, and hence sequentially compact, we can pass to a subsequence and assume that $\{ x_{1,k} \}$ converges to some $x_1$ . We can repeat for each $i$ so that $x_{i, k} \to x_i$ . Thus, $\{ \mathbf x_k \}$ contains a subsequence that converges to $(x_1, \dots, x_n) \in K$ . Hence, $K$ is compact.

While the arguments of Lemma 1 almost apply immediately to Theorem 1, we can formalize this connection through a homeomorphism.

Definition 1. Let $K, L$ be topological spaces. A bijection $f : K \to L$ is called a homeomorphism if both $f$ and $f^{-1}$ is continuous. If such a homeomorphism exists, we say that $K$ and $L$ are homeomorphic, and denote $K \cong L$ .

Example 1. The map $f : [0, 1] \to [a, b]$ defined by $f(t) = (1-t)a + tb$ is a homeomorphism.

Proof. We leave it as an exercise to verify that $f$ is bijective. Furthermore, the observation $f^{-1}((s, t)) = (f^{-1}(s), f^{-1}(t))$ yields the continuity of $f$ and $f^{-1}$ respectively.

Complete Proof of Theorem 1. By Lemma 1, $[0, 1]^n$ is compact. Define the function $\mathbf f : [0, 1]^n \to [a, b]^n$ by $\mathbf f = (f, \dots, f)$ with inverse $\mathbf f^{-1} = (f^{-1}, \dots, f^{-1})$ . Since each coordinate is continuous, so are both maps, so that $\mathbf f$ is a homeomorphism. Therefore, $[a, b]^n = \mathbf f([0, 1]^n)$ is compact as well. In particular, $[-M, M]^n$ is compact, completing the proof of Theorem 1.

This proof does technically complete our study of compact sets in $\mathbb R^n$ , but not in general. Is Lemma 1 just a coincidence, or a corollary of a much deeper result? The nontrivial result we have proved before is that $[0, 1]$ is compact. If we can prove that a finite product of compact sets is compact, then $[0, 1]^n$ is compact and Lemma 1 becomes a corollary.

We aim to prove this compactness result below.

Theorem 2. Let $K_1, \dots, K_n$ be compact spaces. Then $\prod_{i = 1}^n K_i$ is compact.

Proof. It suffices to prove the result for $n = 2$ , then extend the result inductively to the general case. Thus, suppose $K_1, K_2$ are compact spaces. Consider an open cover $\{ V_\alpha \}$ of $K_1 \times K_2$ .

Now, fix $x \in K_1$ . We remark that the map $K_2 \to \{ x \} \times K_2$ defined by $t \mapsto (x, t)$ is a homeomorphism, so that $\{ x \} \times K_2$ is a compact space. Hence, there exists a finite subcover $\mathcal V_x := \{ V_i(x) \}_{i = 1}^{N_x}$ that covers $\{ x \} \times K_2$ . This means that $\mathcal V := \bigcup_{x \in K_1} \mathcal V_x$ covers $K_1 \times K_2$ .

Now, $\mathcal V$ is by no means finite, but each $\mathcal V_x$ is. We still need to extract a finite sub-cover, but $\mathcal V_x = \{ V_i(x) \}_{i = 1}^{N_x}$ gives us a hint to resolve this issue. For each $i$ , find a basis element $B_i \times B_i' \subseteq V_i(x)$ where $x \in B_i$ . Define the neighborhood $U_x := \bigcap_{i = 1}^n B_i$ of $x$ .

By construction, $\mathcal V_x$ contains $U_x \times K_2$ , which in turn contains $\{x\} \times K_2$ . Hence, we can create an open cover $\{ U_x : x \in K_1 \}$ for $K_1$ . Since $K_1$ is compact, we can extract a finite subcover $\{ U_{x_1}, \dots, U_{x_n} \}$ for $K_1$ .

Some bookkeeping then yields the finite sub-cover $\bigcup_{i = 1}^n \mathcal V_{x_i}$ of $K_1 \times K_2$ , and we are done.

Corollary 1. The set $[0, 1]^n$ is compact.

Now, let’s switch up the question a little bit: is it true that $[0, 1]^\omega \equiv \mathcal F(\mathbb N, [0, 1])$ is compact? The answer is given by Tychonoff’s theorem, which we will return to explore after analyzing compact spaces in more detail.

—Joel Kindiak, 10 Apr 25, 1934H

KindiakMath

Compact Products

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Compact Products

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