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Can you help me please with this question?

Let $X$ be a non-empty set with the cofinite topology.

Is $\left ( X,\tau_{\operatorname{cofinite}} \right ) $ a metrizable space?

Thanks a lot!

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2 Answers 2

up vote 7 down vote accepted
  • If $X$ is finite, the cofinite topology is the discrete one, which is metrizable, for example using the distance $d$ defined by $d(x,y):=\begin{cases}0&\mbox{ if }x=y,\\ 1&\mbox{ otherwise} .\end{cases}$
  • If $X$ is infinite, it's not a Hausdorff space. Indeed, let $x,y\in X$, and assume that $U$ and $V$ are two disjoint open subsets of $X$ containing respectively $x$ and $y$, then $U=X\setminus F_1$ and $V=X\setminus F_2$ where $F_1$ and $F_2$ are finite. Hence $\emptyset =X\cap F_1^c\cap F_2^c=X\setminus (F_1\cup F_2)^c$, which is a contradiction since $X$ is infinite. But each singleton is closed in such a space.
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Thank you very much! –  Mushka May 15 '12 at 20:01
You're welcome. –  Davide Giraudo May 15 '12 at 20:03

Amusingly enough, if $X$ is infinite and given the cofinite topology, then a sequence $\langle x_n:n\in\mathbb{N}\rangle$ of points of $X$ having all but finitely-many terms distinct will converge to every point in $X$! (It actually turns out that the only sequences of points of $X$ that converge everywhere are those with all but finitely-many terms distinct.)

Thus, when $X$ is infinite, the space $(X,\tau_\mathrm{cofinite})$ doesn't even have unique sequence convergence, so certainly isn't Hausdorff (there exist non-Hausdorff spaces with unique sequence convergence, but convergent sequences in Hausdorff spaces always have unique limits). It is $T_1$, though, as the closed sets are precisely the finite point sets.

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