Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

2 Answers

up vote 5 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.
share|improve this answer
    
Thank you very much! –  Mushka May 15 '12 at 20:01
    
You're welcome. –  Davide Giraudo May 15 '12 at 20:03
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.