# Compactness of the unit ball in infinite dimension

1. Feb 29, 2008

### quasar987

Is it a fact that in an infinite dimensional normed linear space, the closed unit ball is never compact?

If so, how does one go about seeing this?

2. Feb 29, 2008

### morphism

That's right. The compactness of the closed unit ball (in the norm topology) characterizes finite-dimensionality.

You can prove that the ball is not compact in an infinite dimensional space by invoking the Riesz lemma. (This is one of many ways.)

3. Feb 29, 2008

### quasar987

Any other way comes to mind? I wonder if I can prove this with what I know.

4. Feb 29, 2008

### morphism

What can you work with?

There's a really easy proof for Hilbert spaces (elements of an infinite o.n. set are a distance of sqrt(2) apart).

5. Feb 29, 2008

### quasar987

An idea: proving that the ball is dense in the whole space by means of showing the only functional null on the ball is the identically null functionnal.

(How can elements of an infinite o.n. set be a distance of sqrt(2) apart in the case where the set is a basis and hence dense???)

6. Feb 29, 2008

### quasar987

mmh my idea does not work directly... because I just double checked and this criterion for density applies to subspaces only.

And it doesn'T make sense that the ball be dense, since it's closed, it would be the whole space, which it obviously isnt. :p

7. Feb 29, 2008

### morphism

Why would this be a problem? If e and f are two distinct elements in any o.n. set, then ||e - f||^2 = <e-f, e-f> = ||e||^2 + ||f||^2 = 2.

By the way, the most elementary way to prove this is to use the Riesz lemma. In fact the Riesz lemma basically tells you how to do the Hilbert space trick when you don't have an inner product: it lets you find a vector that's 'nearly' orthogonal to any proper subspace.

8. Feb 29, 2008

### quasar987

Alright, I'll look into it. There's a nice wiki page about including a proof.

9. Mar 1, 2008

### jostpuur

There is a quite simple intuitive idea behind this, which also leads towards one possible proof. It goes like this:

Let E be a norm space, and B(0,1) the unit ball. Let us assume B(0,1) is compact. Since {B(x,1/2) | ||x|| < 1} is an open cover for the unit ball, there exists a finite number of points x_k so that

$$B(0,1) \subset \bigcup_{k=1}^N B(x_k, \frac{1}{2}).$$

Now there is two alternatives. Either the x_k span the space E, <x_1,...,x_N>=E, or then they don't. If they do, E is finite dimensional, and we are done. If they do not span, then we end up into a contradiction because the ball B(0,1) has points that have distance greater than 1/2 from the spanned subspace.

You can convince yourself of this by drawing a two dimensional picture. Pretend that x-axis describes the subspace <x_1,...,x_N>, and y-axis some non spanned direction. If you draw a ball x^2+y^2 < 1, you'll see it contains points that have distance greater than 1/2 from the x-axis.

Last edited: Mar 1, 2008
10. Mar 1, 2008

### mathwonk

if you read the book of dieudonne on foundations of modern analysis, you will find this and other beautiful results in the early chapters. in the 1960's this was taught in basic honors advanced calculus.

11. Mar 2, 2008

### quasar987

Well it's not in the chapter on normed space (ch. 5). Are you sure you've got the right book?

12. Mar 2, 2008

### mathwonk

look at thm 5.9.4 page 109.

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

### quasar987

I see! "A locally compact normed space is finite dimensional."

If the closed unit ball were compact, every closed ball would be too, and hence the space would be locally compact, and hence finite dimensional