Why is cofinite topology on a countable set a second countable space?

Asked May 06 '19 at 13:08

Active Apr 02 '23 at 21:29

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Consider a countable set $X$ with cofinite topology. Here the claim is that $X$ is second countable.

The simple argument is as follows :

$\color{red}{\text{Since $X$ is countable, the number of finite sets in $X$ are countable.}}$ Hence the number of open sets in $X$ are also countable. Since the topology itself is countable, the space is second countable.

Question: Why should it be true that the number of finite subsets of a countable set is countable?

edited Apr 02 '23 at 21:29

Steven Clontz

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asked May 06 '19 at 13:08

Bijesh K.S

2,604

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Enumerate the set $1,2,3,...$ and map each finite subset $A$ to $\sum_{n\in A}2^{n}$. This map is an injection into $\mathbb{N}$. – logarithm May 06 '19 at 13:11
@logarithm What set are you enumerating? And by enumerating it, you already assume that it is countable. – Mark Kamsma May 06 '19 at 13:17
@MarkKamsma The only set around that is given countable. – logarithm May 06 '19 at 13:18
He's enumerating your set $X$. And you have assumed that $X$ is countable. – Lee Mosher May 06 '19 at 13:18
@logarithm Ah yes, I read too quickly – Mark Kamsma May 06 '19 at 13:21

Why is cofinite topology on a countable set a second countable space?

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