Thread: 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.

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?

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

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.

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.

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 for the boundary of .

What is the exact meaning of . (i.e. what is a boundary point?)

Note how I edited your reply. Say is a neighborhood of x.
Then there is a neighborhood of y, , such that .

Now apply the definition of a boundary point.

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.

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: stands for the interior of .

By your definition .

But is open set. Therefore its complement is closed.

Thus is the intersection of two closed sets. SO?

I'm sorry about that, I will bear your advice in mind in future. Where have you used X is hausdorff?

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.

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?

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 is a closed set.
Any closed subset of a compact set in a Hausdorff space is compact.
Thus 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.