Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – Gunnar Þór Magnússon Nov 28 '10 at 13:16
What is A? A metric space? A topological space? A set? – Qiaochu Yuan Nov 28 '10 at 13:46
I adjusted it to be more clear. – Listing Nov 28 '10 at 13:55
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. – Nate Eldredge Nov 28 '10 at 16:54
What do you mean by "what happens"? – Listing Nov 28 '10 at 18:55
up vote 1 down vote accepted

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.

share|cite|improve this answer
Thanks, but is there a proof without using general theorems like the one you mentioned? – Listing Nov 28 '10 at 13:56
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. – Qiaochu Yuan Nov 28 '10 at 13:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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