# Prove that the boundary of S is compact

• Dec 19th 2012, 03:09 AM
Plato13
Prove 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 AM
Plato
Re: Prove that the boundary of S is compact
Quote:

Originally Posted by Plato13
Let S be a compact subset of a hausdorff space X. Prove that the boundary of S is compact.

Have you proved that in a Hausdorff space boundaries of sets are closed?

What is known about closed subsets of compact sets?
• Dec 19th 2012, 03:59 AM
Plato13
Re: Prove that the boundary of S is compact
No I haven't proved that. That proof would be helpful.
• Dec 19th 2012, 04:15 AM
Plato
Re: Prove that the boundary of S is compact
Quote:

Originally Posted by Plato13
No I haven't proved that. That proof would be helpful.

Well then prove that the boundary is a closed set. Show it contains all its limit points.
• Dec 19th 2012, 04:59 AM
Plato13
Re: 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 AM
Plato
Re: Prove that the boundary of S is compact
Quote:

Originally Posted by Plato13
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 neighborhood of x contains an element y in B different to x.

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 AM
Plato13
Re: 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 AM
Plato
Re: Prove that the boundary of S is compact
Quote:

Originally Posted by Plato13
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.

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 AM
Plato13
Re: 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 AM
Plato
Re: Prove that the boundary of S is compact
Quote:

Originally Posted by Plato13
Where have you used X is hausdorff?

It was not use. Being Hausdorff is not necessary.

BTW: It is Hausdorff. That is a proper name.
• Dec 19th 2012, 08:01 AM
Plato13
Re: 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 AM
Plato
Re: Prove that the boundary of S is compact
Quote:

Originally Posted by Plato13
Are you sure you've proved the boundary is compact, and not closed? Something's wrong because it is a necessary hypothesis.

What was shown is that $\displaystyle \beta(S)$ is a closed set.
Any closed subset of a compact set in a Hausdorff space is compact.
Thus $\displaystyle \beta(S)\subseteq\overline{S}$ is compact.
• Dec 19th 2012, 08:38 AM
Plato13
Re: 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.