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Let $X$ be a regular ($T_{1}$) space such that for each non-empty open subset $U$ of $X$ the complement $X \setminus U$ is a finite set. Why this implies $X$ is finite?

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    @Theo: I see. While I am certain that both versions of my answers are correct and useful to user10, I would rather wait for him to give a final clear for the interpretation so I could rest in peace.2011-08-14

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If $X=\{x\}$ then of course it is finite, otherwise there are $x_1,x_2\in X$ which are two distinct points.

Separate $x_1$ from $x_2$ by open sets $U_1$ and $U_2$ which are disjoint. Since $X\setminus U_1$ is finite, we have that $U_2$ is finite, $X\setminus U_2$ is also finite. Therefore $X=U_2\cup X\setminus U_2$ is the disjoint union of two finite sets and therefore finite.

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    @user10: $T_1$ in the sense that distinct points can be separated, which also implies singletons are closed?2011-08-17