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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove that a compact set $K$ in a Hausdorff topological space $\mathbb{X}$ is closed? I seek a proof that is as self contained as possible.

Thank you.

share|cite|improve this question
up vote 18 down vote accepted

Fix $x\in\mathbb{X}\setminus K$. Since $\mathbb{X}$ is Hausdorff, for each $y\in K$ there are disjoint open sets $U_y$ and $V_y$ such that $x\in U_y$ and $y\in V_y$. $\{V_y:y\in K\}$ is an open cover of $K$, so it has a finite subcover, say $\{V_y:y\in F\}$, where $F$ is some finite subset of $K$. Let $$U=\bigcap_{x\in F}U_x\;;$$ clearly $U$ is an open nbhd of $x$ disjoint from $K$. Since $x$ was an arbitrary point of $\mathbb{X}\setminus K$, $K$ must be closed.

share|cite|improve this answer
I thought it would be very laborious. Thanks, Brian. – MathOverview Nov 18 '11 at 13:05
But "$ \bigcap_{y\in F} U_y$" would not be out instead of "\bigcap_{x\in F} U_x$"? – MathOverview Nov 18 '11 at 13:12
@user19266: They say exactly the same thing: $x$ (in mine) and $y$ (in yours) are dummy variables. You could just as well say $\bigcap_{\xi\in F}U_\xi$ if you really wanted to. – Brian M. Scott Nov 18 '11 at 13:30

A "sequential" proof: Let $x_\alpha \in K$ be a net with limit $x \in \mathbb{X}$. By compactness of $K$, there exists a subnet $x_{\alpha_{\beta}}$ which converges in $K$. Let $y \in K$ denote its limit. Since it's a subnet of $x_\alpha$, it follows that also $x_\alpha \to y$. Since $\mathbb{X}$ is Hausdorff, nets have unique limits, so $y=x$ and in particular $x \in K$.

share|cite|improve this answer
This proof is interesting Mark. I like it. – MathOverview Nov 18 '11 at 13:23

Your Answer


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.