# Proof that an uncountable set with the countable complement topology is not first countable

In an exercise set for my topology course, I was asked to prove that an uncountable set with the countable complement topology is first countable. I couldn't prove it, and I started to suspect that it was false. I wrote the following proof, and then checked to see if I was right. I did find that, as the title says, such a space would not be first countable (there was a mistake in the exercise set). However, all the proofs I found were different from mine, so I would appreciate it if someone checked it for me. Thanks in advance.

Proof:

Let $X$ be an uncountable set, and $\tau$ the countable complement topology. We choose any $x \in X$, and prove that there can't be a countable neighborhood basis for $x$. Suppose $V = \{V_i,\ i\in \mathbb{N} \}$ is a countable neighborhood basis. Then, we take $W= \bigcap_{i \in \mathbb{N}} V_i$. This set is a neighborhood of $x$ ($x$ belongs to $W$, and $W^\complement = \bigcup_{i \in \mathbb{N}}V_i^\complement$, which is countable, since it is a countable union of countable sets). We know that $W$ has an uncountable number of elements (its complement is countable), so we choose one of its elements $y\neq x$, and we have that $W\setminus\{y\}$ is a neighbourhood of $x$. But there is no $V_i$ in $V$ for which $V_i \subseteq W\setminus\{y\}$, since $y\in V_i$ for all $V_i$ (it belongs to the intersection of all of them), and $y\notin W\setminus\{y\}$. Then, $V$ is not a neighborhood basis for $x$.

• It's fine. I would rather say that $V$ is a countable family of neighbourhoods of $x$ than suppose it were a countable neighbourhood basis, but that's not an important difference. Now I wonder how the proofs that you found looked, because this one is what I consider the canonical proof. – Daniel Fischer Apr 7 '15 at 19:34
• Thank you! For example, this one. – Mauro Apr 7 '15 at 20:02
• Aha. That one first uses the assumption that $\mathcal{B}_x$ is a neighbourhood basis rather than the countability. When doing that, I'd prefer not to make a countability assumption and deduce that every neighbourhood basis must have cardinality at least $\operatorname{card} X$ (just in case: of course we're using choice; without choice, a countable union of countable sets might be uncountable), which gives more information.It's not so different, however, both proofs essentially state that if a point $x$ had a countable neighbourhood basis, its complement were countable. – Daniel Fischer Apr 7 '15 at 20:15
• There is a step I don't understand in that proof though. It says that $\bigcap \mathcal B_x = \left\{ {x}\right\}$, but I can't see why the only point in the intersection is $x$. – Mauro Apr 7 '15 at 21:48
• For every $y\in X\setminus \{x\}$, the set $X\setminus \{y\}$ is a neighbourhood of $x$. So there must be a $V\in \mathcal{B}_x$ with $V\subset X\setminus \{y\}$ and hence $y\notin \bigcap \mathcal{B}_x$. – Daniel Fischer Apr 8 '15 at 10:01