Is any topology on a countable set first countable?
I think the answer is yes, but I just want a clarification of my reasoning:
Let $X$ be any countable set. First, I would like to recall the fact that any countable set with discrete topology is indeed second countable because it has a countable base $\{\{x\}\mid x\in X\}$.
We know that if a topological space is second countable, then it remains second countable under a coarser topology (i.e., if $\tau_1$ and $\tau_2$ are two topologies on $X$ such that $\tau_1\subseteq\tau_2$ then $(X,\tau_1)$ is second countable whenever $(X,\tau_2)$ is second countable).
Now any topology $\tau$ on $X$ is indeed coarser than the discrete topology, and hence $X$ with any topology is second countable. But any second countable topological space is first countable, and hence $X$ with any topology is first countable.
Is this reasoning correct?
Countable+~First countable– Steven Clontz Mar 29 '25 at 01:58