Does anyone have an elegant/easy to remember proof of this? I can prove bounded easily, and I have notes on why its closed, but they're kind of messy and confusing!

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- December 17th 2009, 08:26 PMplatinumpimp68plus1Prove every compact set is closed and bounded.
Does anyone have an elegant/easy to remember proof of this? I can prove bounded easily, and I have notes on why its closed, but they're kind of messy and confusing!

- December 17th 2009, 09:09 PMJose27
Suppose (for a contradiction) that for all there exists such that contains only a finite number of terms of a given sequence then since is compact, the cover has a finite subcover ie. there exists such that but this is clearly a contradiction since each of these balls contain only finitely many terms. So we conclude that for any given sequence there exists a such that for all contains infinitely many terms of said sequence ie. there exists a subsequence converging to . This in turn proves that is closed trivially.

- December 17th 2009, 09:18 PMplatinumpimp68plus1
- December 17th 2009, 09:28 PMJose27
By how we defined them, each contains only a finite number of terms of a sequence (a countable infinte set) so when we get our finite subcover we can get at most a finite number of terms of the sequence in the union of a finite number of balls, but these union contains which contains the sequence.

- December 18th 2009, 08:20 AMPlato
I assume that we are in a metric space.

Here are traditional proofs for both properties.

Suppose that is a limit point of but .

this is true .

The collection covers .

So finite subcollection also covers .

But note that . That is a open set that contains and no other point of . Contradiction.

For bounded, there is finite collection covering .

Let . It is easy to show that is a bound for . - December 19th 2009, 04:46 AMHallsofIvy
Here's what I think is the simplest proof that any compact set is bounded.

(Assuming that A is in a metric space, of course.)

Let p be any point in the compact set, A. Let be the open ball centered at p with radius n. Certainly every point in A has**some**distance from p and there exist an integer larger than that distance. That is, the set of all such open balls is an open cover of A. Since A is compact, there is a finite subcover of A, so there is a largest "N". Show that, if x and y are any points in A, d(x,y)< 2N.

To show that any compact set, A, is closed, show that its complement is open. (Again, in a metric space.)

Let p be a point in the complement of A. For any q in A, let B(q) be the open ball, of radius 1/2 the distance from p to q, centered on q. Let C(q) be the open ball, of radius 1/2 the distance from p to q, centered on p (note that C(q), though centered on p, is still "indexed" by q). The set of all open balls, B(q), for all q in A, is an open cover for A. Since A is compact, there exist a finite subcover, . Look at the corresponding collection of open sets . Since p is in each of them it is in there intersection. Further, since this is a**finite**collection, its intersection is an open set. Finally, since every member of A is in one of the sets , it is not in the corresponding and so not in there intersection.

That is, the intersection of the all the is an open set, containing p, which contains no member of A. That means that p is an interior point of the complement of A and, since p could be any member of the complement of A, the complement of A is open and A itself is closed. - July 13th 2010, 04:12 PMswimhotdragonThanks! That is super clear...
At first, I thought Rudin made a typo on the index(Crying), so I googled the prove, and found out the idea from this website. Thanks a lot everyone who post the proofs!(Rofl)

There are so many experts here!(Clapping) - July 13th 2010, 09:55 PMBgrasty
I love that one! It's so cool!