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In my topology book there is the following exercise.

Let $X$ be a infinite set with the cofinite topology. Show that the set $G=\{X\setminus M\mid M\text{ finite subset of } X\}$ is a filter and determine the set $A= \bigcap_{(X\setminus M) \in G} \overline{X\setminus M}$.

Showing that $G$ is a filter I have done easily, but I am unsure about the second part. I think $A=X$ because $x\in A\Leftrightarrow \forall\; U\in \mathcal{U}(x)\;\forall\; X\setminus M\in G\text{ we have } U\cup (X\setminus M)\neq \emptyset$, where $\mathcal{U}(x)$ is the set of all the neighbourhood of $x$. Then, this intersection should always be non empty. Am I right or do I miss something? Thank you for your help!

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If $A\subseteq X$ is infinite, what is $\overline{A}$? By definition, it is the smallest closed set containing $A$.

But what are the closed subsets of the cofinite topology? They are sets whose complement is open. So they are $X$ and finite sets. So if $A$ is infinite, there is only one closed set containing $A$.

From here, it should be apparent what this intersection ends up being.

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  • $\begingroup$ Let me see if I understand. $A$ is infinite and $A\subset \overline{A}$, $\overline{A}$ is a closed set so it is either $X$ or a finite set. A finite set cannot be because $A$ infinite is not contained in a finite set. Then $\overline{A}=X$. So the intersection I am looking for is $X$. Right? $\endgroup$ – Michela Jul 11 '17 at 15:13
  • $\begingroup$ Yeah, that's the idea. $\endgroup$ – Asaf Karagila Jul 11 '17 at 15:19
  • $\begingroup$ Ok, ,thanks a lot for making it clear $\endgroup$ – Michela Jul 11 '17 at 15:20

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