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Math 344-1,2: Kursobjekt

Aaron Greicius

Section 1.15 Compactness in \(\R^n\)

Before investigating the notion of compactness in the setting of \(\R^n\text{,}\) we first drop a real whopper of a theorem. Tychonoff’s theorem states simply that arbitrary products of compact topological spaces are compact with respect to the product topology. The proof of this most general result requires the axiom of choice; in fact, as we will show, it is equivalent to the axiom of choice. We will provide a full proof of Tychonoff’s theorem later using the concept of nets. In the meantime, we will officially state the result, make use of the result in all its generality henceforth, and provide a proof for the case where we have a finite product.

Proof.

For now we only prove the result for finite products. It suffices, by induction, to show that the product of two compact spaces is compact, and for this case we will use Lemma 1.15.2, which is quite useful in its own right.
Let \(X\) and \(Y\) be compact topological spaces, and suppose we have an open covering \(\{U_i\}_{i\in}\) of \(X\times Y\text{.}\) For any \(x\in X\) the subspace \(\{x\}\times Y\) is homeomorphic to \(Y\) and hence compact. Since \(\{U_i\}_{i\in I}\) is a covering of \(\{x\}\times Y\text{,}\) there is a finite set \(J_x\subseteq I\) such that
\begin{equation*} \{x\}\times Y\subseteq \bigcup_{j\in J_x}U_j\text{.} \end{equation*}
Since \(Y\) is compact, it follows by the tube lemma that there is an open set \(U_x\subseteq X\) such that
\begin{equation*} U_x\times Y\subseteq \bigcup_{j\in J_x}U_j\text{.} \end{equation*}
Since \(X=\bigcup_{x\in X}U_x\) and \(X\) is compact, there are elements \(\{x_1,x_2,\dots, x_n\}\) such that
\begin{equation} X=\bigcup_{i\in I}U_{x_i}\text{.}\tag{1.15.1} \end{equation}
We claim that
\begin{equation*} X\times Y=\bigcup_{i=1}^n \left (\bigcup_{j\in J_{x_i}} U_j \right)\text{,} \end{equation*}
in which case \(\{U_i\}_{i\in \bigcup_{i=1}^n J_{x_i}}\) is a finite subcovering of \(X\times Y\text{,}\) as desired.
Let’s see why (1.15.1) holds: given any \((x,y)\in X\times Y\text{,}\) we have \(x\in U_{x_i}\) for some \(i\text{,}\) in which case
\begin{equation*} (x,y)\in U_i\times Y\subseteq \bigcup_{j\in J_{x_i}}U_j\subseteq \bigcup_{i=1}^n \left (\bigcup_{j\in J_{x_i}} U_j \right), \end{equation*}
as desired. The reverse inclusion is automatic.

Proof.

If \(U\) is an open set satisfying \(\{x\}\times Y\subseteq U\text{,}\) then for all \(y\in Y\text{,}\) there exists open sets \(U_{y}\subseteq X\) and \(V_y\subseteq Y\) such that \(x\in U_y\) and \(y\in V_y\) and \(U_y\times V_y\subseteq U\text{.}\) Since \(\{V_y\}_{y\in Y}\) is an open covering of \(Y\) and \(Y\) is compact, we have
\begin{equation*} Y=\bigcup_{i=1}^n V_{y_i} \end{equation*}
for some elements \(y_1,y_2,\dots, y_n\text{.}\) Let \(W=\bigcap_{i=1}^n U_{y_i}\text{.}\) Clearly we have \(\{x\}\times Y\subseteq W\times Y\text{.}\) Furthermore, given any \((x,y)\in W\) we have \(y\in V_{y_i}\) for some \(1\leq i\leq n\text{.}\) Since \(x\in W\subseteq U_{y_i}\text{,}\) we conclude that
\begin{equation*} (x,y)\in U_{y_i}\times V_{y_i}\subseteq U\text{.} \end{equation*}
This proves \(W\times Y\subseteq U\text{,}\) as desired.

Example 1.15.3. Boxes and balls in \(\R^n\).

Show that the following subsets of \(\R^n\) are compact.
  1. A closed solid box \(B=\prod_{i=1}^n[a_i,b_i]\text{.}\)
  2. A closed ball \(\overline{B}_\epsilon(\boldx_0)=\{\boldx\in \R^n\colon \norm{\boldx-\boldx_0}\leq \epsilon\}\text{.}\)
Solution.
The box is compact by the (finite version) of Tychonoff’s theorem, since any closed interval is compact.
The closed ball \(\overline{B}_\epsilon(\boldx_0)\) is homeomorphic to \(\overline{B}_\epsilon(\boldzero)\text{,}\) which in turn is homeomorphic to the box \(B=[-\epsilon, \epsilon]^n\) via the map
\begin{equation*} f(\boldx)=\begin{cases} \frac{\norm{\boldx}}{\max\{\abs{x_i}\colon 1\leq i\leq n\}} \amp \text{if } \boldx=(x_1,x_2,\dots, x_n)\ne \boldzero \\ \boldzero \amp \text{if } \boldx=\boldzero \end{cases}\text{.} \end{equation*}

Definition 1.15.4. Bounded, diameter, and distance to sets.

Let \((X,d)\) be a metric space and let \(A\) be a subspace of \(X\text{.}\)
We say \(A\) is bounded if there is a positive number \(M\) such that \(d(x,y)\leq M\) for all \(x,y\in A\text{.}\) If \(A\) is nonempty, we define the diameter of \(A\text{,}\) denoted \(\operatorname{diam} A\text{,}\) as \(\operatorname{diam} A=\sup\{d(x,y)\colon x, y\in A\}\in \R\cup \{\infty\}\text{.}\)
The distance \(d(x,A)\) from \(x\) to \(A\) is defined as \(d(x,A)=\inf\{d(x,y)\colon y\in A\}\text{.}\)

Proof.

It is clear that the empty set is compact, closed, and bounded. Henceforth, we assume \(A\) is nonempty.
Assume \(A\) is compact. Since \(\R^n\) is Hausdorff, we know from Theorem 1.14.7 that \(A\) is closed. We now show that \(A\) is bounded. Pick any \(\boldx\in A\text{.}\) We have \(A\subseteq \bigcup_{n\in \Z_+}B_n(\boldx)\text{.}\) Since \(A\) is compact, we have \(A\subseteq \bigcup_{n=1}^N B_n(\boldx)=B_{N}(\boldx)\) for some \(N\text{.}\) It follows that \(d(\boldy,\boldy')\leq 2N\) for all \(\boldy, \boldy'\in A\text{,}\) in which case \(\operatorname{diam}(A)\leq 2N< \infty\text{.}\)
Now assume \(A\) is closed. If furthermore \(\operatorname{diam}A=N < \infty\text{,}\) then \(A\) is a closed subset

Definition 1.15.7. Lebesgue number.

Let \((X,d)\) be a metric space. A Lebesgue number for an open covering \(\mathcal{U}=\{U_i\}_{i\in }\) is a positive real number \(\delta\) satisfying the following condition: if \(\operatorname{diam} A< \delta\text{,}\) then \(A\subseteq U_i\) for some \(i\in I\text{.}\)
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