# Homework Help: Can a compact function spaces no contain an accumulation point?

1. Feb 5, 2013

### Fractal20

1. The problem statement, all variables and given/known data
So I had a homework problem which was to show that a certain function space was compact via Ascoli-Arzela Theorem. I was okay with doing this, accept, it appears the corresponding uniformly convergent sequence I found in any infinite set need not converge to something in the set. I brought this up with my professor and was told that outside of finite Euclidean space a compact set does not need to be closed.

So I tried to dig up different definitions of compact because I this seemed to be the opposite of what I remember and everything seemed to say that the convergent subsequence needs to converge to something in the set for it to be compact.

So my real question is, is this true? A compact function space need not be closed? If it helps the original question is:

Let M be a bounded subset of C[a,b] show the set {R(x) =∫ax f(t) dt | f is in M} is compact.

So I am fine with showing that this fits the criteria for Ascoli-Azela but again it seems like I can construct convergent sequences that don't converge to something in the set. For example, suppose M = {fn| fn = 1/n for n=1,2,3...}. Then the corresponding set {R} is equal to the series I will call Rn = (1/n)(x-a). This converges uniformly to the function f = 0, but this is not in {R}.

Again, my question is if compact spaces need not be closed. If so, when is this true? Thanks!

2. Relevant equations

3. The attempt at a solution

2. Feb 5, 2013

### CompuChip

Usually it does, for example on Rn compact is equivalent to closed and bounded. More generally, a compact subset of a Hausdorff is closed.

This covers (no pun intended) most "normal" vector spaces that you will encounter. However, function spaces - and what open and closed mean - are a little less intuitive. The fact that you can construct a sequence that is not closed under convergence means that it is not Banach, which is a necessary condition for many theorems (though I don't recall one which says that in Banach spaces compact sets are closed).

3. Feb 5, 2013

### micromass

In a metric space, a compact set is always closed. In particular, your space C[a,b] is a metric space and thus all compact sets are closed.

Spaces where compact sets are not closed arise in the more general theory of topological space. Specifically, it are non-Hausdorff spaces which behave badly. But your space is Hausdorff.

I guess that your version of Ascoli-Arzela says that your set should be uniformly bounded and equicontinuous. But that is not enough to ensure compactness. So the set you found is not compact. Ascoli-Arzela says which spaces are pre-compact. That is: which sets are such that its closure is compact!!

If you want Ascoli-Arzela to ensure the space is actually compact, then you should demand equicontinuity, uniformly bounded and closed. The set you mention is not closed.

So the thing has nothing to do with compact sets not being closed, since every compact set in C[a,b] is always closed.

4. Feb 5, 2013

### Fractal20

Thank you! That confirms my suspicions and clears a lot of things up.

5. Feb 6, 2013

### Fractal20

Actually one more question. In the problem, a metric on F is never given. Without a metric, can we even discuss whether it is compact?

6. Feb 6, 2013

### CompuChip

In general, compactness is defined in terms of open covers, for which all you need is a topology.

Often, for "standard" example like Rn and C([a,b]) the "usual" metric (Euclidean metric and $d(f, g)^2 = \int_a^b |f(x) - g(x)|^2 \, dx$, respectively) are silently assumed.

7. Feb 6, 2013

### Fractal20

So then would my counter example show that {F} is not necessarily compact? I'm just wondering now if my teacher was mistaken.

8. Feb 6, 2013

### micromass

Ascoli-Arzela usually works with C[a,b] and the norm $\|f\|_\infty=sup_{x\in [a,b]} |f(x)|$. A metric should have been given, but I guess that this is the one they look at.

In my opinion, your teacher was mistaken.

9. Feb 6, 2013

### CompuChip

Thanks micromass, reading your answers I realized how far away this has sunk in my brain :)