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I'm reading a proof about that disproves that the cofinite set

$$O_c= \{\emptyset,X\}\cup\{S\subset X|\quad X-S\quad \text{is finite }\}$$

is hausdorff. It goes as follows :

Let $U,V \subset X$ such that $U\cap V =\emptyset$, then this implies that $V\subset X-U$ and $X-U$ is finite which implies that $V$ is finite. But then $X-V$ is infinite, contradicting the fact that $V$ is open.

I don't see how this contradicts the fact that $V$ is open. Is the implication here that infinite sets cannot be open?

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The only open set $V$ such that $X\setminus V$ is infinite is the empty set (assuming $X$ infinite). This is because the only infinite closed set is $X$. The implication here is that closed sets are finite (except, possibly, the whole set). – egreg May 10 '14 at 19:38
To stress what egreg said: If $X$ is finite then the cofinite topology is the discrete topology an dhence Hausdorff. So the condition that $X$ is infinite is missing from th eproblem statement! – Hagen von Eitzen May 10 '14 at 19:49

3 Answers 3

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A subset of $X$ is open if either it is empty or its complement is finite. If a subset of $X$ has infinite complement, it is open if and only if it is empty. But the empty set is not a neighborhood of any point.

Let's see it in another way. In order for $X$ to be Hausdorff, you need to find, for $x,y\in X$ with $x\ne y$ an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that $U\cap V=\emptyset$. Note that $U$ and $V$ are not empty.

Now, suppose they exist. Then $F=X\setminus U$ and $G=X\setminus V$ are finite. Therefore $$ X=X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)=F\cup G $$ is finite.

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It contradicts $V$ being open, because if $V$ has an infinite complement, it cannot be in the finite complement topology, and thus cannot be open.

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No. Let $V$ be open, non-empty and finite. Because it is the cofinite topology, $X\setminus V$ is also finite. But then $X$ is finite, which is false.

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