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.

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?

share|improve this question
    
@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
add comment

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
geez! can't believe didn't see that, time to sleep, thank you! –  user10 Aug 14 '11 at 9:10
    
@Asaf: Sorry, I deleted my confused comment. I'm going to take a break. –  Jonas Meyer Aug 14 '11 at 9:30
    
@Jonas: It's fine, I was also confused about your comment. I'm not quite sure anymore if user10 means regular in the sense of separating points from closed sets, or just separating points, etc etc. –  Asaf Karagila Aug 14 '11 at 9:32
    
@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
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.