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. $\blacksquare$