Update theorems/T000910.md

Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>

Author: JSMassmann

theorems/T000910.md

@@ -8,6 +8,10 @@ then:
   P000057: true
 ---
 
-Let $x \in X$ and $\mathcal{V}$ be a countable local $\pi$-base around $x$. Set $W = \bigcup\{X \setminus V: V \in \mathcal{V}\}$. For any $y \neq x$, $X \setminus \{y\}$ is a neighbourhood of $x$ and so there is $V \in \mathcal{V}$ with $V \subseteq X \setminus \{y\}$. Therefore $y \in X \setminus V \subseteq W$ so, since $y$ was arbitrary, $X \setminus \{x\} \subseteq W$. But $W$ is a countable union of finite sets, so $W$ is countable, so $X$ is countable.
+Let $x \in X$ and let $\mathcal{V}$ be a countable local $\pi$-base around $x$.
+The complement of every $V\in\mathcal V$ is finite; so $W := \bigcup\{X \setminus V: V \in \mathcal{V}\}$ is countable.
+For any point $y \neq x$, $X \setminus \{y\}$ is a neighbourhood of $x$ and so there is some $V \in \mathcal{V}$ with $V \subseteq X \setminus \{y\}$.
+Therefore $y \in X \setminus V \subseteq W$.
+It follows that $X \setminus \{x\} \subseteq W$ and $X$ is countable.
 
 Observe that the above argument only requires some point to have a countable local $\pi$-base, not every point.