0

My incorrect thought was to find uncountably many uniformly separated points. Now suppose countably many totally bounded sets $\{B_n\}$ cover these points, it is then apparent that each $B_n$ covers at most finitely many of them since each sequence in a totally bounded set has a Cauchy subsequence, and hence the contradiction.

Unfortunately there are a great many separable Banach spaces (eg. $\ell^2$), which invalidates my "proof". So I request an alternative way. And, as a side note, if it wouldn't complicate matters very much, please avoid using the fact that an infinite-dimensional Banach space admits an uncountable Hamel basis.

Best regards!

Community
  • 1
Vim
  • 13,640

1 Answers1

1

A metric space is compact if and only if it is complete and totally bounded, so a $\sigma$-totally bounded Banach space is $\sigma$-compact and hence finite-dimensional.

Community
  • 1
Brian M. Scott
  • 616,228
  • Thanks, this already solves my problem. Yet as a followup question, is there any easy way without appeal to BCT (in proving $\sigma$-compact $\Leftrightarrow$ finite-dimensional in particular)? – Vim Oct 10 '16 at 16:43
  • 1
    @Vim: You’re welcome. I don’t know, but I rather suspect that there is not. – Brian M. Scott Oct 10 '16 at 16:47