Metric Spaces Are Compact Spaces If and Only If They're BW Spaces
Metric Spaces Are Compact Spaces If and Only If They're BW Spaces
Recall from the Compact Spaces as BW Spaces page that if $X$ is a compact space then $X$ is also a BW space.
Also recall from The Lebesgue Number Lemma page that if $(X, d)$ is a metric space that is also a BW space then for every open cover $\mathcal F$ of $X$ there exists an $\epsilon > 0$ such that for all $x \in X$ there exists a $U \in \mathcal F$ such that $B(x, \epsilon) \subseteq U$.
We will now use the Lebesgue Number Lemma to show that the converse of the first theorem mentioned above is true for metric spaces, that is, if $X$ is a metric space then $X$ is a compact space if and only if it is a BW space.
Theorem 1: Let $(X, d)$ be a metric space. Then $X$ is compact if and only if it is a BW space. |
- Proof: Let $(X, d)$ be a metric space.
- $\Rightarrow$ Suppose that $X$ is a compact space. From the first theorem mentioned at the top of this page we immediately have that $X$ is also a BW space.
- $\Leftarrow$ Suppose that $X$ is a BW space. Then every infinite subset of $X$ has an accumulation point. Let $\mathcal F$ be an open cover of $X$. Since $X$ is a metric space that is also a BW space, by the Lebesgue Number Lemma there exists a Lebesgue number $\epsilon > 0$ such that for all $x \in X$ there exists a $U \in \mathcal F$ such that $B(x, \epsilon) \subseteq U$.
- For each $x \in X$ there may be many such $U \in \mathcal F$ that satisfy the condition above. So, for each $x \in X$, select any $U_x \in \mathcal F$ such that $B(x, \epsilon) \subseteq U_x$.
- Step 1: Take any $x_1 \in X$. If $X \subseteq B(x_1, \epsilon) \subseteq U_{x_1}$ then $\mathcal F^* = \{ U_{x_1} \}$ is a finite subcover of $X$ and we're done. If not, proceed to step 2.
- Step 2: Take any $x_2 \in X$ such that $x_2 \not \in B(x_1, \epsilon)$. If $X \subseteq B(x_1, \epsilon) \cup B(x_2, \epsilon) \subseteq U_{x_1} \cup U_{x_2}$ then $\mathcal F^* = \{ U_{x_1}, U_{x_2} \}$ is a finite subcover of $\mathcal F^*$ and we're done. If not, proceed to step n.
- Step n: Take any $x_n \in X$ such that $\displaystyle{x_n \not \in \bigcup_{i=1}^{n-1} B(x_i, \epsilon)}$. If $\displaystyle{X \subseteq \bigcup_{i=1}^{n} B(x_i, \epsilon) \subseteq \bigcup_{i=1}^{n} U_{x_i}}$ then we're done, and if not, continue in this process.
- If at any point this process terminates then $\mathcal F^*$ is a finite subcover of $X$. We claim that this process cannot go on forever. Suppose instead that this process never terminates. Then we obtain an infinite set of points:
\begin{align} \quad A = \{ x_1, x_2, ... \} \end{align}
- Furthermore, from the choices of each $x_i$ we have that each $x_i$ is separated by an open ball of radius $\epsilon > 0$, i.e., for each $i, j \in \mathbb{N}$, $i \neq j$ we have that $d(x_i, x_j) \geq \epsilon > 0$. We claim that the set $A$ cannot have any accumulation points. Suppose otherwise, i.e., suppose that $A$ has an accumulation point $x \in X$. Then every open neighbourhood $U$ of $x$ contains infinitely many points of $A$. In particular, the open neighbourhood $B \left (x, \frac{\epsilon}{2} \right )$ contains infinitely many points of $A$. So there exists $x_i, x_j \in A$ such that $x_i, x_j \in B \left (x, \frac{\epsilon}{2} \right )$. But then $d(x_i, x_j) < \epsilon$ which is a contradiction.
- So $A$ has no accumulation points. But $(X, d)$ is a BW space, so this is a contradiction. Hence the assumption that the process above does not terminate was false. So the process above always terminates which implies that for every open cover $\mathcal F$ of $X$ there exists a finite subcover $\mathcal F^*$. Therefore, $X$ is compact. $\blacksquare$