# Each open set is cofinite in a regular space implies the space is finite

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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@Asaf: I think $T_3$ and $T_1$ is meant. At some point in the past user10 made it clear that the main source of the questions is Dugundji's book, where regular means $T_3$, i.e., separating points from closed sets (without $T_1$). Btw. While regular can mean just about anything under the sun, I never heard it used for $T_1$. Also, if the cofinite topology (or a weaker one) is regular and $T_1$ then it is certainly Hausdorff, hence the space must be finite. I see no other interpretation than yours. – t.b. Aug 14 '11 at 10:22
@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. – Asaf Karagila Aug 14 '11 at 10:25

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.
@Asaf Karagila: regular and $T_{1}$, regular in the sense you can separate a point from a closed set. Thanks. – user10 Aug 17 '11 at 11:51
@user10: $T_1$ in the sense that distinct points can be separated, which also implies singletons are closed? – Asaf Karagila Aug 17 '11 at 15:47