Compact Spaces as BW Spaces

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

On the Hausdorff Spaces Are BW Spaces If and Only If They're Countably Compact we have seen that if $X$ is a Hausdorff space that $X$ is a BW space if and only if $X$ is countably compact.

Of course, not all topological spaces are Hausdorff, so, we will now look at a new theorem that does not impose such a requirement. We will see that every compact space is a BW space.

Theorem 1: Let $X$ be a topological space. If $X$ is compact then $X$ is a BW space.

• Proof: Let $X$ be a compact space. Then every open cover of $X$ has a finite subcover.

• Suppose instead that $X$ is not a BW space. Then there exists an infinite subset $A$ of $X$ that does not have an accumulation point. In other words, $A' = \emptyset$.

• Since the closure of $A$ is equal to $A$ union its set of accumulation points, we see that $\bar{A} = A \cup A' = A \cup \emptyset = A$. So $\bar{A} = A$ which implies that $A$ is a closed set.

• Since $X$ is compact and $A$ is a closed subset of $X$ we know by the theorem presented on the Closed Sets in Compact Topological Spaces page that then $A$ is compact in $X$. Furthermore, since $A$ has no accumulation points, for each $a \in A$ there exists open neighbourhoods $U_a$ of $a$ such that:

(1)

\begin{align} \quad A \cap U_a = \{ a \} \end{align}

• So then $\{ U_a : a \in A \}$ is a open cover of $A$ that covers $A$. Since $A$ is compact in $X$ there exists a finite subcover, say $\{ U_{a_1}, U_{a_2}, ..., U_{a_n} \} \subset \{ U_a : a \in A \}$ with:

(2)

\begin{align} \quad A \subseteq \bigcup_{i=1}^{n} U_{a_i} \end{align}

• However $\displaystyle{\bigcup_{i=1}^{n} U_{a_i}}$ contains only $n$ points from $A$ which implies that $A$ is a finite set which is a contradiction.

• Therefore the assumption that $X$ was not a BW space was false. So if $X$ is a compact space then $X$ is also a BW space. $\blacksquare$