# closed and bounded but not compact set of real-valued bounded functions

I'm trying out a problem I was given and this is the statement:

Prove, or disprove, that every bounded and closed subset of the set of real-valued and bounded functions on [0,1] equipped with the sup norm is compact.

I have a sneaking suspicion that this statement is false but I am unable to find a suitable counterexample. I have proven that the set of real-valued and bounded functions equipped with the sup norm is complete and I have bounded but do not have totally bounded so I believe that I must find a subset that is not totally bounded for my counterexample. My efforts thus far have not been fruitful. Does anyone have any idea how to approach this?

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Consider the functions $f_n(x) = 1_{\{{1 \over n}\}}(x)$ and $f_0(x) = 1_{ \{ 0 \} } (x)$. – copper.hat Jan 3 '14 at 20:29
I don't quite understand that notation. Could you please explain it? – usr2014 Jan 3 '14 at 20:33
By $1_A$ I mean the indicator of the set $A$, that is, $1_A(x) = \begin{cases} 0 & x \notin A \\ 1 & x \in A \end{cases}$. – copper.hat Jan 3 '14 at 20:34

This is not true. For if it were, the bounded functions on an interval would be a finite-dimesional vector space. Clearly it is not.

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Let $f_0(x) = 1_{ \{ 0 \} } (x)$, $f_n(x) = 1_{ \{ {1 \over n} \} } (x)$ for $n =1,2,..$.

Let $F = \{ f_n \}_{n=0}^\infty$.

Note that the $f_n$ are bounded and $\|f_n-f_m\| = 1$ for all $n \neq m$, hence $F$ is closed. Since $F$ cannot contain any convergent subsequence we see that $F$ is not compact.

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The unit ball is closed and bounded. Now if the unit ball were compact, then every closed and bounded subset was compact, because every such set is a closed subset of some ball. So if there is a counter-example, then the unit ball is one.

Of course, this does not help much, but it shows that you just have to find a bounded sequence of function that does not have a convergent subsequence. copper.hat has already given one.

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