TITLE: Does a topology on a countable set always have a countable base?
QUESTION [23 upvotes]: I'm quite new to topology, and a homework question (which I solved without knowing the answer to this question) got me thinking: 
If $X$ is a countable set, and $\tau$ is a topology on it, does it necessarily have a countable basis? 
Since $\tau \subset 2^{X}$, it might have an uncountable number of sets in it, but a base can be made of a very small subset of $\tau$, so pure intuition says that the answer is yes, but I couldn't prove it, and now I'm not sure that it's true at all.
Thanks!

REPLY [6 votes]: The Hewitt-Marczewski-Pondiczery theorem implies that $[0,1]^{\mathbb{R}}$ in the product topology has a countable dense subset $D$. This $D$, as a space in its own right (in the subspace topology) has a local base of minimal size $\mathfrak{c}$ at every point, i.e. its character at every point is $\mathfrak{c}$.  I show this and more in this answer, where you can find other countable examples based on ultrafilters as well.