Here are two statements I want to show:
For cofinite topology space $(X, \mathcal{T})$:
(a) Show that X is first countable if and only if X has countably many points.
I have known how to prove " if X is first countable then X has countably many points". I use the contradiction to find if $X$ is uncountable, we could not find a countable local base.
But how about the inverse statement? If $X$ is countable, suppose $X=\{x_{i}\}_{i=1}^{\infty}$. For every $x_i$, I try to find their countable local basis. How to show that for every open set $U$, there exists a countable local basis $V_{i}$ such that $x\in V_i\subset U$? for some $i$.
(b) Show that X is second countable if and only if X has countably many points.
Best Answer
If $X$ is countable, it has only countably many finite sets (standard set theory, or use a mapping using prime numbers to show that all finite subsets of $\Bbb N$ are in bijection with $\Bbb N$, e.g.), so countably many closed sets, so countably many open sets.
The whole topology is then countable and is its own base, while (the also countable) subfamily, $\{O: O \in \mathcal{T}, x \in O\}$ is a local base at $x$, for any $x$.
As to the last question, $X$ uncountable implies $X$ not first countable and so $X$ not second countable, so it's the same argument.