# Compactness and space dimension question

1. Nov 13, 2008

### Epsilon36819

I am having trouble accepting two well known results of analysis as non contradictory.

First, given a vector space equipped with norm ||.||, the unit ball is compact iff the space is finite dimensional.

Second, the Arzela-Ascoli theorem asserts that given a compact set X, a set S contained in the space of all continuous functions f(X) equipped with the supremum norm is compact iff it is uniformly closed, uniformly bounded and equicontinuous.

My problem is the following: let S be the set of all equicontinuous functions on X with ||f|| <= 1, where ||.|| is the supremum norm. Is the space of all equicontinuous functions not an infinite dimensional subspace? Yet S is compact.

Last edited: Nov 13, 2008
2. Nov 13, 2008

### Office_Shredder

Staff Emeritus
The limit of an equicontinuous uniformly converging sequence is continuous, but not necessarily equicontinuous if I recall correctly. Hence you don't have closure, and thus the space isn't compact.

I've never really done anything with equicontinuous functions, so I could be wrong

3. Nov 13, 2008

### Epsilon36819

Can't we use an epsilon/3 argument, just as in the proof that uniform convergence of continuous functions => continuity of the limit? In this case, the given delta valid for fn will be the one used for f, so f will be part of the equicontinuous family.

Now that I think of it, though, I don't think that the set of equicontinuous functions is closed under addition and scalar multiplication, hence it is not a subspace.

Nonetheless, it's still foggy to me why the first thm applies only to vector spaces, and not sets.

4. Nov 14, 2008

### Office_Shredder

Staff Emeritus
Well, it needs to be a vector space for norm, unit ball and finite dimensional to be defined.

5. Nov 14, 2008

### Epsilon36819

It can be a subset of a vector space.

6. Nov 14, 2008

### morphism

It doesn't make sense to speak of an equicontinuous function. Equicontinuity is a property of sets of functions.

So in your original post, the following bit is nonsensical:

7. Nov 14, 2008

### Office_Shredder

Staff Emeritus
In such a case there's no reason to believe it contains any unit vectors. And the set still has no dimension on it unless it's a vector space itself

8. Nov 14, 2008

### Hurkyl

Staff Emeritus
One common thing to do in mathematics is to generalize a theorem by carefully studying its proof, and removing unused hypotheses, or weakening overly strong hypotheses.

So, you could try to do that here -- study the proof to see exactly what properties of "normed vector spaces", "unit balls", and "dimension' are being used here, and which aren't.