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Define $A$ as a nonempty set, $\mathcal{B}:=\{f: A \rightarrow \mathbb{R}: f(A) \text{is bounded} \} ,d_\infty:=\text{sup}\{|f(x)-g(x)|:x \in A\}$.

For which $A$ is $\overline {B_1(0)} \subset \mathcal{B}$ compact?

Notes: ${B_1(0)}$ is the unit sphere. I already proved that $(\mathcal{B},d_\infty)$ is a metric space.

My thoughts: I investigated a bit on this and found some proofs that the unit sphere in a banach space is compact when the banach space is finite dimensional. But I don't want to use this here, I don't even know if $(\mathcal{B},d_\infty)$ is a banach space. All proofs we did who depend on the dimension of a metric space is that the intersections of compact subsets is compact and that the finite union of compact subsets is compact. Maybe we can use this?

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    What do you $m$ean by "what ha$p$$p$ens"?2010-11-28

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The result you mention has a converse: the closed unit ball in a Banach space is compact if and only if the Banach space is finite-dimensional. That should suggest to you what is true here.

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    I am not suggesting that you use the general theorem. I am suggesting that the general theorem should guide you towards what is true in this case, which you should then prove as directly as possible.2010-11-28