# Compact sets are closed

1. Mar 25, 2013

### bedi

Theorem: Let S be a compact subset of ℝ^n. Then S is closed.

Before looking at the book I wanted to come up with my own solution so here is what I've thought so far:

Fix a point x in S. Let Un V_n (union of V_n's...) be an open covering of S, where V_n=B(x;n). We know that there is a finite subcover of that open covering and it consists of neighborhoods. Now pick the neighborhood with maximum radius n' and consider the neighbourhood A with radius n'+1. Clearly A is open and S is a subset of A and additionally, A\S is nonempty and still open(?) (this part really needs verification). Hence, ℝ\(A\S)=ℝ\A disjoint union S is closed. Does that mean that S is closed? If not, why?

2. Mar 25, 2013

### jbunniii

You could conclude that $A \setminus S$ is open if you knew that $S$ is closed, but that is what you are trying to prove.

Hint: Try using a different covering of $S$, consisting of complements of closed neighborhoods of an appropriate point.

3. Mar 25, 2013

### Fredrik

Staff Emeritus
Edit: Ignore the first and third paragraph below. As jbunniii pointed out below, they're both wrong.

A\S is not open. Consider the intervals (0,1) and (0,2). Their difference is [1,2). This set is not open, because 1 is not an interior point.

I have seen at least three different definitions of the word neighborhood in different books, so that word shouldn't be used without an explanation. You seem to use it as a synonym for "open ball". I know that Rudin uses that definition.

S does not have to be a subset of A. Consider the compact set [0,5] and let the open cover consist of open intervals of length 2. The maximum radius is 1. So the radius of your A is 2.

Last edited: Mar 25, 2013
4. Mar 25, 2013

### jbunniii

Well, in this problem $A \setminus S$ is open, because $S$ is compact and therefore closed. But he can't use the fact that $S$ is closed because that is what he is trying to prove.
His open cover is an increasing sequence of nested sets, though, so $S$ is a subset of $A$ in this case.

5. Mar 25, 2013

### WannabeNewton

Do you know what sequential compactness is? In metric spaces it is often superior to the open cover definition. This proof is trivial with the former. Let $p\in \overline{S}$; there exists a sequence $(x_i)$ in $S$ that converges to $p$. Since $S$ is compact iff $S$ is sequentially compact for metric spaces, there exists a subsequence of $(x_i)$ that converges to some point in $S$. We are dealing with a Hausdorff space so if a sequence converges then every subsequence converges to the same point which implies $p\in S$ so $S$ is closed.

6. Mar 25, 2013

### Fredrik

Staff Emeritus
Agreed.

Yes, this is correct as well. I think I was somehow thinking that S is open, even though I know that it's not.

7. Mar 25, 2013

### bedi

Of course I did not assume that S is closed. So it is wrong in general that if we subtract (complement) any type of set (closed, open, clopen) from an open set, in this case A, the result is open, yes?

8. Mar 25, 2013

### Bacle2

Just to add that your result extends to Hausdorff spaces, i.e.,if U compact in X Hausdorff, then U is (are?) closed in X.

9. Mar 25, 2013

### jbunniii

It is wrong in general. See Fredrik's example: $(0,2) \setminus (0,1) = [1,2)$ is not open.

10. Mar 25, 2013

### Fredrik

Staff Emeritus
If E is open and F is closed, then $E-F=E\cap F^c$ is the intersection of two open sets, and is therefore open. But if F is open, then E-F is not necessarily open. The example I used earlier shows this. (0,2)-(0,1)=[1,2) is not open.

11. Mar 25, 2013

### WannabeNewton

Indeed there is nothing in the usual proof that would require more than a Hausdorff space. What is true for $\mathbb{R}^{n}$ is a much stronger statement which is that the compact subsets of $\mathbb{R}^{n}$ are exactly the closed and bounded ones (Heine - Borel). To the OP, if you are set on using the open cover definition then know that the proof that a compact subset of a Hausdorff space is closed is trivial still, if you know that disjoint compact subsets of a Hausdorff space can be separated by open sets.

12. Mar 25, 2013

### Fredrik

Staff Emeritus
"Trivial" may be too strong a word, because it suggests that it's very easy to think of the trick. It took me like five minutes even though I've solved this problem several times before. The first time I did this, it took a lot longer.

A hint for bedi: Try to prove that the complement of S is open, and make a clever choice of open cover of S.

Last edited: Mar 25, 2013
13. Mar 25, 2013

### WannabeNewton

True, it probably is too strong a word here. It certainly might be more rewarding than doing it brute force using open covers however. I still like the sequential compactness method the best becuase it's fast but if the OP is using Rudin, as I might suspect, then he does this stuff before talking about sequences so tis probably better left for 'nother day eh Fredrik?

14. Mar 25, 2013

### micromass

Staff Emeritus
I would say that it is trivial, but not completely obvious