Let S be a compact subset of a hausdorff space X. Prove that the boundary of S is compact.

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- Dec 19th 2012, 03:09 AMPlato13Prove that the boundary of S is compact
Let S be a compact subset of a hausdorff space X. Prove that the boundary of S is compact.

- Dec 19th 2012, 03:48 AMPlatoRe: Prove that the boundary of S is compact
- Dec 19th 2012, 03:59 AMPlato13Re: Prove that the boundary of S is compact
No I haven't proved that. That proof would be helpful.

- Dec 19th 2012, 04:15 AMPlatoRe: Prove that the boundary of S is compact
- Dec 19th 2012, 04:59 AMPlato13Re: Prove that the boundary of S is compact
Let S be a subset of a hausdorff space X and let B be it's boundary. Let x be a limit point of B. Then every neighbourhood of x contains an element y in B different to x. So there exists disjoint open sets U,V such that x is in U, y is in V.

Now i'm stuck. - Dec 19th 2012, 06:27 AMPlatoRe: Prove that the boundary of S is compact
Let's use $\displaystyle \beta(S)$ for the boundary of $\displaystyle S$.

What is the exact meaning of $\displaystyle t\in\beta(S)$. (i.e. what is a boundary point?)

Note how I edited your reply. Say $\displaystyle U$ is a neighborhood of x.

Then there is a neighborhood of y, $\displaystyle V$, such that $\displaystyle x\not\in V\subset U.$.

Now apply the definition of a boundary point. - Dec 19th 2012, 07:08 AMPlato13Re: Prove that the boundary of S is compact
Why do you say V is a subset of U? Also I don't know much about x being a boundary point except the boundary is closure/interior.

- Dec 19th 2012, 07:28 AMPlatoRe: Prove that the boundary of S is compact

That is your whole problem in a nutshell.

Let this be a lesson to you. Always post what you know about the question. I had no way of knowing that you were working with what I consider a non-standard definition.

Notation: $\displaystyle \mathcal{I}(S)$ stands for the interior of $\displaystyle S$.

By your definition $\displaystyle \beta(S)=\overline{S}\setminus\mathcal{I}(S)$.

But $\displaystyle \mathcal{I}(S)$ is open set. Therefore its complement is closed.

Thus $\displaystyle \beta(S)$ is the intersection of two closed sets. SO? - Dec 19th 2012, 07:33 AMPlato13Re: Prove that the boundary of S is compact
I'm sorry about that, I will bear your advice in mind in future. Where have you used X is hausdorff?

- Dec 19th 2012, 07:49 AMPlatoRe: Prove that the boundary of S is compact
- Dec 19th 2012, 08:01 AMPlato13Re: Prove that the boundary of S is compact
Are you sure you've proved the boundary is compact, and not closed? Something's wrong because it is a necessary hypothesis

BTW: are you highlighting the capital or that it should be droff and not dorff? - Dec 19th 2012, 08:24 AMPlatoRe: Prove that the boundary of S is compact
- Dec 19th 2012, 08:38 AMPlato13Re: Prove that the boundary of S is compact
Thanks for your help. In future please don't respond to my questions as other (better) helpers will see you're dealing with it and I will be stuck with your minimalist answers. I know your ethos is let the student do it themselves but sometimes it is more instructive for me to see the proof.