What is the Local Compactness of [0, 1]ω in the Uniform Topology?

[radou]

Homework Statement

This one has been bothering me for a while.

One needs to show that [0, 1]ω is not compact in the uniform topology.

The Attempt at a Solution

As a reminder, the uniform topology on Rω is induced by the uniform metric, which is defined with d(x, y) = sup{min{|xi - yi|, 1} : i = 1, 2, ...}.

For any ε > 1, and for any x in Rω, the open ball B(x, ε) equals Rω.

Now, clearly, for any x, y in [0, 1]ω, d(x, y) = sup{|xi - yi|}.

The main problem seems to be that I can't figure out what kind of sets (except finite sets) are compact in Rω (if at all)?

 

[rasmhop]

Let $x_k = (0,\ldots,0,1,0,\ldots)$ be the sequence with 1 in the kth place.

Depending on your definition of compactness or the theorems you know this sequence x1,x2,... can be used to show that $[0,1]^\omega$ is not compact.

For instance you can show that (x1,x2,...) does not have a convergent subsequence (and thereby that $[0,1]^\omega$ is not sequentially compact).

 

[radou]

OK, but how does this help me?

I'm aware that compactness implies local compactness, i.e. the contrapositive is that non local compactness implies non compactness.

But I don't see how this helps here.

I'm aware too that, since this space is metrizable, compactness, sequential compactness and limit point compactness are equivalent. Hence, our space cannot be compact (by proving what you suggested).

But I don't know of an implication which says that a non compact space is non local compact.

 

[Office_Shredder]

In the title you say you need local compactness, but the OP question just asks about compact. Regardless, you can modify rasmhop's sequence to show that the space is not locally compact at (0,0,0,...)

 

[rasmhop]

Oh sorry. At some point I convinced myself that we were just dealing with compactness, not local compactness.

Anyway the general idea should still work. If $X=[0,1]^\omega$ is locally compact, then the point b = (0,0,0,...) has a compact neighborhood. In particular we can find an $\epsilon > 0$ such that $U = B(b,\epsilon) \cap X$ is contained in a compact set $K \subseteq X$.

Now instead form the sequence
[tex]x_k = (0,0,\ldots,0,2\epsilon/3,0,\ldots)[/itex]
where x_k has $2\epsilon/3$ in the kth place and 0 in all other places.

Then $(x_k)$ is a sequence in K, so it has a convergent subsequence, but this is a contradiction since no subsequence of it is Cauchy.

EDIT: Posted before I saw office_shredder's reply which to some extent renders this redundant, but I'll leave it since it has some additional details.

 

[radou]

No, I'm sorry - I didn't notice I mistyped - I need to show it's not locally compact.

OK, thanks to both for your help. So, we assumed it to be locally compact, and found a sequence which doesn't converge in a compact, i.e. sequentially compact subspace, hence a contradiction.