# A cocountable topology is not first countable

I'm trying to understand this example:

Let $X$ be any uncountable set and declare $U\subset X$ to be open if $X\setminus U$ is countable. Then $X$ is not first countable.

Here is my reasoning:

Suppose for a contradiction every $x\in X$ has a countable neighborhood basis. So let $x\in X$ and let $\{N_i\mid i\in\mathbb{N}\}$ be a countable neighborhood basis at $x$. Each $N_i$ contains an open neighborhood $U$, and since $U$ is open, $X\setminus U$ is countable, so $U$ must be uncountable. But $U\subseteq N_i$, so $N_i$ is uncountable.

Hence every $N_i$ is uncountable. But how does that give a contradiction?

I'm also struggling to understand the next statement: if $A$ is a proper uncountable subset of $X$. then $\overline{A}\neq A$. Clearly if $A$ is uncountable, then $A$ is not open, so $X\setminus A$ is non-empty and not closed, but then what?

I understood that $X$ is not first countable and $A\neq\overline{A}$. Now the claim is that $\overline{A}\neq\{x\in X\mid$ there is a sequence $(x_n)_{n\in\mathbb{N}}$ in $A$ which converges to $x\}$.

It suffices to show that $S:=\{x\in X\mid$ there is a sequence $(x_n)_{n\in\mathbb{N}}$ in $A$ which converges to $x\}=A$.

Clearly $A\subseteq S$, by setting $x_n=x$ for each $n\in\mathbb{N}$. But How do I prove $S\subseteq A$?

• No. Every nonempty open set is uncountable. – William Elliot Sep 7 '17 at 23:03