Problem: Let be compact. Prove or disprove that is separable in general.
I shouldn't be dealing with analysis/topology, but what to do: I loved a lot the great Stone-Weierstrass Theorem (this is a hint) when I studied it and
I couldn't help it. Let's see if don't mess the thing up too much.
Spoiler:
Let be a dense countable set of , and define by .
Let be the algebra generated by the functions . Clearly and also:
i) If then clearly for some
ii) If then for some
Then, by the S-W theorem is dense in and we're done.
Note: I'm taking as metric spaces with the standard euclidean metric and with the supremum metric. Then the same proof as above works if in general is compact, second countable and Hausdorff (because then is metrizable and we define the given metric).
I shouldn't be dealing with analysis/topology, but what to do: I loved a lot the great Stone-Weierstrass Theorem (this is a hint) when I studied it and
I couldn't help it. Let's see if don't mess the thing up too much.
Spoiler:
Let be a dense countable set of , and define by .
Let be the algebra generated by the functions . Clearly and also:
i) If then clearly for some
ii) If then for some
Then, by the S-W theorem is dense in and we're done.
Note: I'm taking as metric spaces with the standard euclidean metric and with the supremum metric. Then the same proof as above works if in general is compact, second countable and Hausdorff (because then is metrizable and we define the given metric).
Tonio
You have yet to prove that is closed subalgebra.
EDIT: Oops, I misread your post. Yes, that is good! I'll post my solution.
Last edited by Drexel28; March 28th 2010 at 08:48 AM.
I didn't use the whole Stone-Weierstrass theorem, only the Weierstrass.
Lamma: Let be the set of all polynomials in with rational coefficients. Then, is dense in Proof: Let be arbitrary. Choose such that .
Now, the mapping given by is continuous and so it assumes a maximum .
Choose such that .
Then, . It follows that .
Thus,
.
It follows that is dense in .
It follows that since given by is a surjection that is countable.
Now, for every there exists, by Tietze's extension theorem, an extension such that .
Also, for every there exists some such that . So, let and . Clearly and thus countable. Clearly then, \in\Omega^*\right\}" alt="\Omega=\left\{p\mid X\in\Omega^*\right\}" /> is also countable.
Also, let and be arbitrary. Letting and the subsequent be as before. Then, and .
It follows that is a countable dense subset of . The conclusion follows.