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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.


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up vote 17 down vote accepted

this can be found in "counterexamples in topology" as referenced on the wikipedia page (a book i sometimes wish i had)

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Good question! I suspect the answer is no. Here is my suggestion for a counterexample. Let $U$ be a nonprincipal ultrafilter on $\mathbb N$. Then I claim $U\cup\{\emptyset\}$ is a topology on $\mathbb N$, which follows from the axioms of an ultrafilter. Now I suspect that this topology does not have a countable basis, mainly because there are $2^{2^{|\mathbb N|}}$ many ultrafilters, implying you need to make uncountably many choices to specify it. If you could get away with specifying a countable basis, that would only be countably many choices.

Edit: Yoyo just posted a definitive answer.

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Indeed, for any free ultrafilter this space has uncountable base. This is well-known. – Henno Brandsma May 25 '11 at 6:00

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 $\mathfrak{c}$.

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