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?

  • $\begingroup$ Well, $\mathcal B$ is a topological vector space (the topology is even given by a metric, as you have proved), so that unit ball is compact iff $\mathcal B$ is locally compact iff $\mathcal B$ is finite dimensional. So your problem reduces to classifying those $A$ for which $\mathcal B$ is finite dimensional. You don't need $\mathcal B$ to be a Banach space for this, see the first chapter of Rudin's Functional Analysis. $\endgroup$ Nov 28 '10 at 13:16
  • $\begingroup$ What is A? A metric space? A topological space? A set? $\endgroup$ Nov 28 '10 at 13:46
  • $\begingroup$ I adjusted it to be more clear. $\endgroup$
    – Listing
    Nov 28 '10 at 13:55
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    $\begingroup$ Suggestion: Since $A$ is merely a set and has no other structure, the answer can only depend on the cardinality of $A$. Try seeing what happens for some sets of various cardinalities, and then see if you can formulate a general statement and prove it. $\endgroup$ Nov 28 '10 at 16:54
  • $\begingroup$ What do you mean by "what happens"? $\endgroup$
    – Listing
    Nov 28 '10 at 18:55

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.

  • $\begingroup$ Thanks, but is there a proof without using general theorems like the one you mentioned? $\endgroup$
    – Listing
    Nov 28 '10 at 13:56
  • $\begingroup$ 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. $\endgroup$ Nov 28 '10 at 13:58

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