# Thread: Prove that the boundary of S is compact

1. ## 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.

2. ## Re: Prove that the boundary of S is compact

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?

3. ## Re: Prove that the boundary of S is compact

No I haven't proved that. That proof would be helpful.

4. ## Re: Prove that the boundary of S is compact

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.

5. ## 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.

6. ## Re: Prove that the boundary of S is compact

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 $\beta(S)$ for the boundary of $S$.

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

Note how I edited your reply. Say $U$ is a neighborhood of x.
Then there is a neighborhood of y, $V$, such that $x\not\in V\subset U.$.

Now apply the definition of a boundary point.

7. ## 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.

8. ## Re: Prove that the boundary of S is compact

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

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

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

Thus $\beta(S)$ is the intersection of two closed sets. SO?

9. ## 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?

10. ## Re: Prove that the boundary of S is compact

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.

11. ## 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?

12. ## Re: Prove that the boundary of S is compact

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 $\beta(S)$ is a closed set.
Any closed subset of a compact set in a Hausdorff space is compact.
Thus $\beta(S)\subseteq\overline{S}$ is compact.

13. ## 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.