2
$\begingroup$

Q1: Is there a separable normed space that is not sigma-compact, but is a countable union of totally bounded closed subsets?

A test case is the space $C^1(I)$ with the $C^0$ norm where $I=[0,1]$. The space is the countable union of closed subsets $\{f: ||f||_{C^1}\le n\}$ which are totally bounded by Ascoli's theorem but I have trouble seeing if the space is sigma-compact.

Here is the difficulty. The usual proof of non-sigma-compactness invokes the following fact: a sigma-compact space is a countable union of nowhere dense subsets if and only if it is nowhere locally compact. For example, if a normed space $X$ contains a closed infinite dimensional Banach subspace $Y$, then $X$ is not sigma-compact because if it were, then $Y$ would be sigma-compact, nowhere locally compact, and complete contradicting the Baire category theorem. For normed spaces (and probably in general) this italised statement holds with "compact" replaced by "closed totally bounded". Thus in the setting of Q1 there seems to be lack of tools to prove non-sigma-compactness.

Q2: Is there another method to prove non-sigma-compactness for separable normed spaces?

Improve this question
$\endgroup$
4
  • $\begingroup$ Just a small question: Is there an easy argument why $C^1[I]$ with the $C^0$-norm does not contain an infinite dimensional Banach space? I assume this somehow comes out of Stone-Weierstrass, correct? $\endgroup$
    – Tom
    Jun 9, 2015 at 9:32
  • $\begingroup$ @Tom: I do not know whether what you say is true. $\endgroup$
    – Igor Belegradek
    Jun 9, 2015 at 11:15
  • 1
    $\begingroup$ Here is an idea (which might not yet really work) to find $c_0$ as a subspace of $C^1(I)$: Choose $\phi_n \in C^1(I)$ with disjoint supports (contained in $[1/n - 1/n^2, 1/n+1/n^2]$), $\phi_n(1/n)=1$ and $|\phi_n(x)|\le 1$ and define $T$ on $c_0$ by $T(\alpha)=\sum\limits_{n=2}^\infty \alpha_n \phi_n$. This will be isometric from $c_0$ to $C(I)$ and the only problem is, that $T(\alpha)$ is not differentiable at $0$ (only continuity follows from $\alpha_n\to 0$). $\endgroup$
    – Jochen Wengenroth
    Jun 9, 2015 at 15:44
  • $\begingroup$ That is a nice idea, but as you already said, This function will in general not be differentiable. You can for example construct such a function such that the maxima of the bumb functions follow the graph of a sqare root function. Unfortunately I do not see a possibility to repair this argument... $\endgroup$
    – Tom
    Jun 11, 2015 at 8:10

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.