Is there a countable basis for the finite complement topology on the natural numbers?

Question

I have the following question on a worksheet: Consider the finite complement topology τf on N. Does this topology have a countable base? If so give one such base and if not prove your claim.

Been trying to solve this for several hours but I have got nowhere. I think that there is no countable basis because it seems like to construct a basis you would need to take the entire topology, but I have no idea if that's right or how to show that.

Thanks

The collection of closed sets is uncountably infinite. I'm not sure how that helps though? — Dan Armour, Jan 31 '15 at 17:03
No. The closed sets are the finite subsets of $\Bbb N$. This collection is countable. So, the collection of open sets is countable as well. — David Mitra, Jan 31 '15 at 17:04
Sorry, I did search before posting but at the time I was assuming it had an uncountable number of sets so that post didn't seem relevant. — Dan Armour, Jan 31 '15 at 17:14

score 0 · Accepted Answer · answered Jan 31 '15 at 17:05

0

HINT: Calculate how many finite subsets $\Bbb N$ has, and conclude the number of cofinite sets.

answered Jan 31 '15 at 17:05

Asaf Karagila

393,674

Is there a countable basis for the finite complement topology on the natural numbers?

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