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

• Fractal20
In summary, the convergent subsequence does not need to converge to something in the set for a compact set to be defined.
Fractal20

## Homework Statement

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!

## The Attempt at a Solution

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).

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.

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

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?

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.

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

Fractal20 said:
So then would my counter example show that {F} is not necessarily compact? I'm just wondering now if my teacher was mistaken.

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.

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

## 1. What is a compact function space?

A compact function space is a topological space in which every open cover has a finite subcover. In simpler terms, it is a space in which every sequence has a convergent subsequence.

## 2. How do you determine if a function space is compact?

To determine if a function space is compact, you can use the Heine-Borel theorem which states that a function space is compact if and only if it is both closed and bounded.

## 3. Can a compact function space contain an accumulation point?

No, a compact function space cannot contain an accumulation point. This is because a compact function space is a closed set, and accumulation points are limit points that are not necessarily in the set.

## 4. What is the significance of a compact function space not containing an accumulation point?

The fact that a compact function space cannot contain an accumulation point is important because it means that every sequence in the space must have a convergent subsequence. This property is useful in many mathematical proofs and applications.

## 5. Can a compact function space have an infinite number of accumulation points?

No, a compact function space cannot have an infinite number of accumulation points. This is because the definition of a compact function space requires that every sequence has a convergent subsequence, and an infinite number of accumulation points would mean that some sequences do not have a convergent subsequence.

• Calculus and Beyond Homework Help
Replies
14
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
942
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
26
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
10
Views
2K
• Calculus and Beyond Homework Help
Replies
26
Views
1K