Prove every compact set is closed and bounded.

**platinumpimp68plus1** · December 17th 2009, 07:26 PM

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!

**Jose27** · December 17th 2009, 08:09 PM

Originally Posted by platinumpimp68plus1

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!

Suppose (for a contradiction) that for all $y\in K$ there exists $\epsilon _y$ such that $B_{\epsilon _y} (y)$ contains only a finite number of terms of a given sequence $(x_n)\subset K$ then since $K$ is compact, the cover $\{ B_{\epsilon _y} (y) : y\in K \}$ has a finite subcover ie. there exists $y_0,...,y_m$ such that $K \subset \cup_{i=1}^{m} B_{\epsilon _{y_i}} (y_i)$ but this is clearly a contradiction since each of these balls contain only finitely many terms. So we conclude that for any given sequence $(x_n)\subset K$ there exists a $y\in K$ such that for all $\epsilon >0$ $B_{\epsilon } (y)$ contains infinitely many terms of said sequence ie. there exists a subsequence $(x_{n_k})$ converging to $y$ . This in turn proves that $K$ is closed trivially.

**platinumpimp68plus1** · December 17th 2009, 08:18 PM

Originally Posted by Jose27

Suppose (for a contradiction) that for all $y\in K$ there exists $\epsilon _y$ such that $B_{\epsilon _y} (y)$ contains only a finite number of terms of a given sequence $(x_n)\subset K$ then since $K$ is compact, the cover $\{ B_{\epsilon _y} (y) : y\in K \}$ has a finite subcover ie. there exists $y_0,...,y_m$ such that $K \subset \cup_{i=1}^{m} B_{\epsilon _{y_i}} (y_i)$ but this is clearly a contradiction since each of these balls contain only finitely many terms. So we conclude that for any given sequence $(x_n)\subset K$ there exists a $y\in K$ such that for all $\epsilon >0$ $B_{\epsilon } (y)$ contains infinitely many terms of said sequence ie. there exists a subsequence $(x_{n_k})$ converging to $y$ . This in turn proves that $K$ is closed trivially.

thanks for your response. that clears some of it up... but could you clarify what the contradiction is exactly? i seem to be missing it

**Jose27** · December 17th 2009, 08:28 PM

By how we defined them, each $B_{\epsilon _y} (y)$ 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 $K$ which contains the sequence.

**Plato** · December 18th 2009, 07:20 AM

Originally Posted by platinumpimp68plus1

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!

I assume that we are in a metric space.
Here are traditional proofs for both properties.
Suppose that $x$ is a limit point of $K$ but $x\notin K$ .
$\left( {\forall y \in K} \right)$ this is true $r_y = \frac{{d(x,y)}}{4} > 0$ .
The collection $\left\{ {B(y;r_y )} \right\}_{y \in K}$ covers $K$ .
So finite subcollection $K \subset \bigcup\limits_{j = 1}^n {B(y_j ;r_{y_j } )}$ also covers $K$ .

But note that $x \in \bigcap\limits_{j = 1}^n {B(x ;r_{y_j } )}$ . That is a open set that contains $x$ and no other point of $K$ . Contradiction.

For bounded, there is finite collection $\bigcup\limits_{j = 1}^n {B(y_j ;1)}$ covering $K$ .
Let $M = \max \left\{ {d(y_k ,y_j )} \right\} + 2$ . It is easy to show that $M$ is a bound for $K$ .

**HallsofIvy** · December 19th 2009, 03:46 AM

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 $B_p(n)$ 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, $\{B(q_1), B(q_2), \cdot\cdot\cdot, B(q_n)\}$ . Look at the corresponding collection of open sets $\{C(q_1), C(q_2), \cdot\cdot\cdot, C(q_n)\}$ . 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 $B(q_i)$ , it is not in the corresponding $C(q_i)$ and so not in there intersection.

That is, the intersection of the all the $C(q_i)$ 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.

**swimhotdragon** · July 13th 2010, 03:12 PM

At first, I thought Rudin made a typo on the index

, so I googled the prove, and found out the idea from this website. Thanks a lot everyone who post the proofs!

There are so many experts here!

**Bgrasty** · July 13th 2010, 08:55 PM

I love that one! It's so cool!

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Prove every compact set is closed and bounded.

Thanks! That is super clear...

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