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.
 A: 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 $U=X\setminus\{y\}$.


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*Is $U$ an open nbhd of $x$?  

*Is there any $n\in\Bbb N$ such that $x\in B_n\subseteq U$?

A: (Just a rewrite of an already excellence answer above and its comment.)
Cofinite topology is not first countable. To prove it, we show that some point of $X$ does not have a countable local base.
Let $x \in X$, and suppose that $\mathscr{B_x}=\{B_n:n\in\Bbb N\}$ is a countable local base of open sets at the point $x$.
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 $b\in X\setminus F$ (and $b \neq x$).

Note that if $X$ is not a cofinite topology, then $F$ is not necessarily countable, then $b$ does not necessarily exist.

Let $O = X \setminus \{b\}$. $\{b\}$ is finite, so by definition $O$  is an open set, and it contains $x$ (because of how $b$ defined). For every $n \in \Bbb N$, we know that $b \notin F_n$, so $b \in B_n$, but $b \notin O$, so $b \in B_n \setminus O$, and therefore $B_n \notin O$.
In summary, there is an open set $O$ containing $x$ but not containing $B_n$. This is a contradiction.

We may wonder though what happens if we define $O = X \setminus \{b\}$ in the usual topology. If the local base is an open ball $B(x, \epsilon)$, there would always be $\epsilon < d(x, b)$.


