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Problem : Suppose that $X$ is a finite set, and is Hausdorff as a topological space. Show that $X$ is discrete (as a topological space).

Thoughts: I'm not even quite sure what the question is asking. I know the definition of a discrete topology is that a set is open in $X$ if it is a subset of $\mathcal P(X)$, so then can't the discrete topology be applied to any set and so any $X$ is discrete? Any help appreciated.

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  • $\begingroup$ It is enough to prove that every singleton is open. Fix $a$ and for every $b\neq a$ find a neighborhood $U_b$ of $a$ not containing $b$. Then consider $\bigcap _{b\neq a}U_b$. $\endgroup$ – SMM Sep 22 '18 at 21:07
  • $\begingroup$ The discrete topology can be applied to any set, but that's not what's happening here. Here all we know is that the space is Hausdorff, and you are supposed to show that this means that the space is discrete. In general, Hausdorff is a weaker condition than discrete (e.g. $\Bbb R$ with standard topology is Hausdorff but not discrete). Your task is to show that when the space is finite, the notions of Hausdorff and discrete coincides. $\endgroup$ – Arthur Sep 22 '18 at 21:08
  • $\begingroup$ @Arthur I think I am confused what a discrete topology means. $\endgroup$ – IntegrateThis Sep 22 '18 at 21:09
  • $\begingroup$ It means that all subsets are open. $\endgroup$ – Arthur Sep 22 '18 at 21:09
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Since $X$ is Hausdorff, we know that singletons are closed. Therefore, all finite subsets of $X$ are closed (finite unions of closed sets are closed). Since all subsets of $X$ are finite, it follows that every subset of $X$ is closed. Using this it is not hard to see that every subset of $X$ is open.

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