If a topological space X is not first countable, is it second countable?

Question

I know that topological space is first countable if each point has a countable neighborhood basis. A neighborhood basis at a point. Consider the following topological space, $X=R$ with $\mathcal T=\{A\subseteq R:0 \not\in A\} \cup \{A\subseteq R:0\in A\text{ and } R \setminus A\text{ is finite}\}$. I know it is not first countable. Is this because at the local base $x=0$, $R\setminus A$ is finite, so the elements of A are infinite and uncountable? If a topological space such as the one above is not first countable, will it have to be second countable? or not second countable for the same reason as it is not first countable?

Answer 1

This is actually quite a reasonable question if you are learning about first/second-countability after learning about first/second category.

As Moishe Kohan mentioned in the comments, while the definition of second category is "not first category", this is not true for first/second countability. Instead, "first countable" means every point has a local countable basis, while "second countable" means the entire space has a countable basis.

This is why I encourage authors to use "meager" and "nonmeager" rather than "first category" and "second category".

Likewise, I prefer the terminology "countable local weight" and "countable weight" in place of "first countable" and "second countable", but that is far less conventional as of today.