1

Let $X$ be a compact Hausdorff space. Then show that $\mathcal{C}(X)$ is reflexive iff $X$ is finite, where $\mathcal{C}(X)$ is the set of all continuous from $X$ to the base field ($\mathbb{R}$ or $\mathbb{C}$ in this case).

Attempt: If $X$ is finite, then $\mathcal{C}(X)$ is clearly finite dimensional, and hence reflexive. How do I show the converse? (Some answers seem to require Urysohn's Lemma, anyway around this?)

DK26
  • 429
  • 2
    I suspect it will be hard to circumvent Urysohn's lemma, without essentially re-proving it. I don't know too many other tools for producing interesting functions in $C(X)$. Any special reason you want to avoid it? (If it's just that "it seems difficult", I would suggest instead getting more familiar with the lemma!) – Nate Eldredge Oct 01 '15 at 14:37
  • If you don't like Urysohn's lemma, you can first prove it for a metric space (where Urysohn's lemma is easy). Note that $X$ is metrizable iff $C(X)$ is separable. If $X$ is not metrizable, it may be hard to avoid Urysohn's lemma. – Prahlad Vaidyanathan Oct 01 '15 at 15:20
  • There is an answer here: http://math.stackexchange.com/questions/250325/space-of-bounded-functions-is-reflexive-if-the-domain-is-finite

    I liked the proof by Martin, however, I am unable to see how compactness is being used. Any pointers?

     – DK26 Oct 01 '15 at 16:43
  • @DK26 Compactness is used to ensure that sequence ${\delta_n}$ is in $\ell^\infty$. – user0 Jul 06 '24 at 16:56

0 Answers0