1
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.