Is the cofinite topology on an uncountable set first countable?

Let $X$ be any uncountable set with the cofinite topology. Is this space first countable?

I don't think so because it seems that there must be an uncountable number of neighborhoods for each $ x \in X$. But I am not sure if this is true.

Solution 1:

You are correct: it is not first countable. However, this is not because each point of $X$ has uncountably many nbhds: each point of $\Bbb R$ also has uncountably many nbhds, but $\Bbb R$, being a metric space, is certainly first countable.

To prove that $X$ is not first countable, you must show that some point of $X$ does not have a countable local base. All points of $X$ ‘look alike’ in the cofinite topology, so it doesn’t matter what point we pick, so let $x\in X$ be any point. Suppose that $\mathscr{B}=\{B_n:n\in\Bbb N\}$ is a countable local base of open sets at the point $x$, meaning that if $U$ is any open nbhd of $x$, then $x\in B_n\subseteq U$ for some $n\in\Bbb N$. For each $n\in\Bbb N$ let $F_n=X\setminus B_n$: $B_n$ is open, so by definition $F_n$ is finite. Let $F=\{x\}\cup\bigcup_{n\in\Bbb N}F_n$; $F$ is the union of countably many finite sets, so $F$ is countable. $X$ is uncountable, so there is some $y\in X\setminus F$. Let $U=X\setminus\{y\}$.

• Is $U$ an open nbhd of $x$?
• Is there any $n\in\Bbb N$ such that $x\in B_n\subseteq U$?