# Hausdorff Spaces Are BW Spaces If and Only If They're Countably Compact

Recall from the Bolzano Weierstrass Topological Spaces page that a topological space $X$ is said to be a Bolzano Weierstrass space (or BW space) if every infinite subset of $X$ has an accumulation.

Also recall from the Hausdorff Topological Spaces page that a topological space $X$ is said to be Hausdorff if for every pair of distinct points $x, y \in X$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$.

Also recall from the Lindelöf and Countably Compact Topological Spaces page that a topological space $X$ is said to be countably compact if every countably cover of $X$ has a finite subcover.

We will now look at a very nice theorem which characterizes BW spaces in terms of the Hausdorff and countably compact property.

Theorem 1: Let $X$ be a topological space. Then $X$ is a BW space if and only if $X$ is Hausdorff and countably compact. |

**Proof:**Let $X$ be a Hausdorff space.

- $\Rightarrow$ Suppose that $X$ is a BW space. Then every infinite subset of $X$ has an accumulation point.

- We first show that $X$ is countably compact. Let $\mathcal F$ be a countable open cover of $X$. Then $\mathcal F = \{ F_1, F_2, … \}$. For each $i \in \{ 2, 3, … \}$ we may assume that:

- In other words, we may assume that no element in the cover $F$ is contained in the union of the preceding elements in the cover.

- Now, suppose that $\mathcal F$ has no finite subcover that also covers $X$. Then consider the following set:

- Then the set $A$ is an infinite set by construction. Since $X$ is a BW space, this means that $A$ has an accumulation point, call it $x \in X$. Furthermore, there exists $F_m \in \mathcal F$ such that $x \in F_m$. But since $x$ is an accumulation point of $A$ and $X$ is a Hausdorff space then within any open neighbourhood of $x$ there exists infinitely many points of $A$ in that open neighbourhood. In other words, the open neighbourhood $F_m$ contains infinitely many points of $A$.

- But from how $A$ is constructed, this is a contradiction, since the open neighbourhood $F_m$ of $x$ can at most contain the points $\{ x_1, x_2, …, x_m \}$.

- Therefore the assumption that $X$ is not countably compact was false. So, for every countable open cover of $X$ there exists a finite open subcover.

- $\Leftarrow$ Suppose that $X$ is countably compact and let $A \subseteq X$ be an infinite subset of $X$. There are two cases to consider.

**Case 1:**Suppose that $A$ is a countably infinite subset of $X$. Then $A$ is of the following form:

- Assume that $A$ has no accumulation point. Recall that the closure of $A$ can be characterized as $\bar{A} = A \cup A'$. If $A$ contains no accumulation points then $A' = \emptyset$ which means that $\bar{A} = A$. This shows us that $A$ is therefore closed in $X$. Furthermore, $a_i \in A$ is an isolated point for each $i \in \{ 1, 2, … \}$ and so there exists open neighbourhoods $U_i$ of $a_i$ such that:

- Notice that then the collection of sets $\{ U_1, U_2, … \}$ is a countable open cover of $A$. Moreover, since $A$ is closed in $X$, $X \setminus A$ is an open cover of $A^c = X \setminus A$. Therefore we see that $\mathcal F = \{ U_1, U_2, … \} \cup \{ X \setminus A \}$ is a countable open cover of $X$. Since $X$ is countably compact, there exists a finite open subcover $\mathcal F^* \subset \mathcal F$ of $X$, say:

- But this implies that $\{ U_{i_1}, U_{i_2}, …, U_{i_n} \}$ is an open subcover of $A$. Then $\displaystyle{A \subseteq \bigcup_{j=1}^{n} U_{i_j}}$. But $\displaystyle{\bigcup_{j=1}^{n} U_{i_j}}$ contains only $n$ points from $A$ which implies $A$ is a finite set - a contradiction.

**Case 2:**If instead $A$ is an uncountably infinite set then $A$ contains a countably subset to which we can apply the same reasoning as in case 1.

- Hence the assumption that $A$ is not a BW space was false. So every countably compact space is a BW space. $\blacksquare$