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In my notes we prove Stone-Weierstrass which tells us that if we have a subalgebra $A$ of $C(X)$ such that it separates points and contains the constants then its closure (w.r.t. $\|\cdot\|_\infty$) is $C(X)$.

A few chapters later there is a lemma that if $X$ is a compact metric space then $C(X)$ is separable. The proof constructs a subalgebra that separates points by taking a dense countable subset of $X$, $\{x_n\}$, and defining $f_n (x) = d(x,x_n)$.

Question: could we treat this as a corollary of Stone-Weierstrass and say that polynomials with rational coefficients are a subalgebra containing $1$ and separating points? Thank you.

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    I'm not sure what your question is. Yes, separability is a consequence of the algebra you constructed and the application of Stone-Weierstrass. And yes, polynomials are dense in, say, $C(K)$ if $K$ is a compact subset of the reals or the complex numbers, because the polynomials fulfil the prerequisites of Stone-Weierstrass. But what has your metric space to do with polynomials?2012-08-04
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    @Thomas So basically, having a metric space enables us to prove separability of $C(X)$ without having Stone-Weierstrass?2012-08-04
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    @MattN. No. Once you have the algebra you have shown that the assumption of Stone Weierstrass are fulfilled and then S-W implies separabilty. Maybe I'm missing something here.2012-08-04
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    @Martin Sleziak How do you define polynomials on arbitrary metric space?2012-08-04
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    BTW this is Matt's question, which I have posted instead of him (because of problems with 30 days/months quota). If you want to see more details, you can have a look at [meta](http://meta.math.stackexchange.com/questions/4806/removal-of-upper-limit-of-questions-you-can-ask).2012-08-04
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    This raises a lot of additional questions. E.g. when you get an upvote/downvote you choose an arbitrary question from Matt and pass it on? ;-).2012-08-04
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    @Thomas That's why I didn't upvote : ) I suppose we can have one of the mods pass over the questions (or not, I don't know if it's that important)2012-08-04
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    I provided an answer to test your reactions on answers to this. Could not resist ;-)2012-08-04
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    @Norbert Could we define a polynomial on $X$ as given a function $f: X \to \mathbb R$ then $p(x) = \sum a_k f(x)$, for $a_k \in \mathbb R$?2012-08-04
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    @MattN, we can define anything we want, but will this newborn object good for our needs? So we come another more difficult question.2012-08-04
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    @MattN. Sure you can define anything as long it is well defined. With this particular defintion of polynomial, however, I assume that a lot of people will complain. But these are then _not_ the polynomials for which you have shown that S-W can be applied to them. But just by defining it you did not proof it fulfils the requirements of S-W.2012-08-04
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    @Thomas Oh, right, they are not. I guess I cannot get around constructing a different algebra for this.2012-08-04
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    @Norbert Right, thank you, it's dawning on me that what I wanted to do is actually more work rather than less.2012-08-04
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    @Thomas It was a joint venture : )2012-08-04

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In the hope I did understand the question correctly by now: It is a corollary of Stone-Weierstrass. Add the constant functions to the algebra you constructed and you are done.

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    Now after reading Norbert's comment I'm not so sure anymore.2012-08-04
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    Sorry, the question is, why do we have to construct an algebra other than polynomials. And by now I suspect the answer could be because we don't have a definition of polynomial on a general metric space.2012-08-04
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    @MattN. Yes, that's correct. That's what I wanted to express with the last sentence of my first comment. (_And_ the algebra is rather easy to construct, so it is indeed a corollary of S-W).2012-08-04
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    Oh ok, now I see what you meant there : )2012-08-04
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    (I was hoping to get around having to construct an algebra other than polynomials so that I could save some work.)2012-08-04
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    @MattN. For a compact metric space you can take Lipshcitz functions in place of polynomials. That is your appropiate algebra would be $ C^{0,1}(X) $ which would be dense by Stone Weierstrass theorem. All you have to show is that $ C^{0,1}(X) $ is seperable.2013-03-15