# Does having a countable subbasis also satisfy the second axiom of countability? [duplicate]

My guess is "yes".

Unfortunately, the course I've taken this semester on "point-set topology" did not cover the concept of subbasis, however I have a few words on it in mind, subbasis unlike basis uses "intersection" instead of "union" of its elements to build the topology.

I need this proposition (If it's true, otherwise can you give me a hint on proving the following ) to prove this; $\mathbb{Z}$ with the finite-complement topology is second-countable. My insight into proving this proposition is this: $S=\{\{x\}^c| x \textrm{ is a point in the space}\}$ and $|S|=\aleph_0$, thus since any open set can be created by such subsets, then the space is second-countable.

If $\mathcal{S}$ is a subbasis for a topological space $\mathcal{X}=\langle X,\tau\rangle$, then you can construct a basis for $\mathcal{X}$ by taking the closure of $\mathcal{S}$ under pairwise intersections.
What I mean by this is that if we set $\mathcal{S}_0=\mathcal{S}$ and put $$\mathcal{S}_{k+1}=\mathcal{S}_k\cup\big\{S\cap T \big\vert S,T\in\mathcal{S}_k\big\}$$ then the basis generated by $\mathcal{S}$ is just $\mathcal{B}=\bigcup_{k=0}^{\infty}\mathcal{S}_{k}$.
1. I'm telling the truth and $\mathcal{B}$ really is a basis for $\mathcal{X}$, and
2. $\mathcal{B}$ is countable.
(You can simplify the notation slightly if you append $\{X\}$ to $\mathcal{S}$ at the start).