Showing that a closed unit ball is not compact

Question

Let $B[0, 1]$ be the space of all bounded functions on $[0, 1]$. Show that the closed unit ball $D = \{f \in B[0, 1] : \|f\| \leq 1 \}$ is not compact in $(B[0, 1], \|.\|_{\infty})$.

can we reason as follows. If a closed unit ball which is bounded of a Normed Linear Space is compact then the space should be finite dimesional. But $X=B[0,1]$ is not finite dimensional. Hence it is not compact? Will this work?

That's correct, but there's more direct ways to show it without that fact. You can try to violate sequential compactness, for example. — perplexed, Sep 17 '19 at 06:31

Kavi Rama Murthy · Answer 1 · 2019-09-17T06:42:26.900

3

If $(E_n)$ is any disjoint sequence of subsets of $[0,1]$ then $(I_{E_n})$ has no convergent subsequence. [This avoids using the theorem on finite dimensionality under compactness of the unit ball].

edited Sep 17 '19 at 06:42

answered Sep 17 '19 at 06:36

Kavi Rama Murthy

311,013

You mean the indicator functions? And distinct subsets will do. – orangeskid Sep 17 '19 at 06:40
@OrestBucicovschi You are right. – Kavi Rama Murthy Sep 17 '19 at 06:43

Showing that a closed unit ball is not compact

1 Answers1