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quick question. There is a question in the book saying "Every finite subset of a Hausdorff space is closed." The proof uses the fact that for each point in $X\setminus\{p\}$ there is a neighborhood contained in $X\setminus\{p\}$ the Hausdorff condition and hence $X\setminus\{p\}$ is open so $\{p\}$ is closed the set is the finite union finite sets.

The very next problem states: "The only Hausdorff topology on a finite set is the discrete topology." Well, in the discrete topology every $\{p\}$ is an open set, not closed as the proof above states. What is going on here?

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  • $\begingroup$ Open sets can be closed. $\endgroup$ – Stefan Hamcke Sep 30 '13 at 18:45
  • $\begingroup$ Take an arbitrary subset. The complement is finite and thus closed, so...? $\endgroup$ – Tobias Kildetoft Sep 30 '13 at 18:46
  • $\begingroup$ It's also the complement of an open set, hence closed. In the discrete topology, all subsets are both, open and closed. $\endgroup$ – Daniel Fischer Sep 30 '13 at 18:46
  • $\begingroup$ got it, thanks. $\endgroup$ – masszz Sep 30 '13 at 18:47
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Both statements are true, and the issue here is that openness/closedness aren't mutually exclusive properties for a subset (as long as the space isn't connected, that is).

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While others have addressed the issue of sets being both open and closed, it seems that there might be another point of misunderstanding for the OP, which is that being a closed or open set is relative to the ambient topological space.

If you have a Hausdorff space $X$ and $F$ is a finite subset of $X$, then the argument you give shows that $F$ is closed in $X$. Now, $F$ itself carries the subspace (or induced) topology from $X$, with which it is a topological space in its own right. As with any topological space, $F$ is open in itself. That does not mean that it is open in $X$. These are different concepts. Since the Hausdorff condition passes to subspaces, $F$ is Hausdorff and finite, and this does indeed imply that it is discrete, so each of its subsets is open in itself. There is no reason that this has to imply that each subset of $F$ is open in $X$.

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