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How to show that space of all infinitely differentiable functions is separable? My idea is to construct approximation polynomial with rational coefficients. How to construct by using Weierstrass Approximation Theorem? Thanks!

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    $\begingroup$ What space? $C^\infty(\mathbb R, \mathbb R)$ ? What topology on it? $\endgroup$
    – GEdgar
    Sep 8, 2013 at 12:43
  • $\begingroup$ It is a C∞[a,b] space with normal distance function. Thanks! $\endgroup$ Sep 8, 2013 at 12:45
  • $\begingroup$ If you are compact you can use stone weierstraß, which says, that every continuous function can be approximated uniformly by polynomials. Hence you only need to check if you can approximate any polynomial with a polynomial with rational coefficients $\endgroup$ Sep 8, 2013 at 12:47
  • $\begingroup$ By normal distance function I presume the sup norm distance. Is that what you have in mind? $\endgroup$ Sep 8, 2013 at 13:09
  • $\begingroup$ @Dominic Michaelis What a nice idea, thanks! $\endgroup$ Sep 8, 2013 at 13:29

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The proof is outlined in Wikipedia, Stone-Weierstraß-Theorem for the $C([a,b],\mathbb{R})$ case. As every $C^\infty $ function is continuous this is a subspace, and as the polynomial with rational coefficients are $C^\infty$ you still can take the same countable set which is dense.

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So your space is $C^{\infty}([a,b])$. It is a subspace of $C([a,b])$ and a subspace of a separable metric space is separable. So we only need to prove that $C([a,b])$ is separable.

To do that let $f \in C([a,b])$. By Weirstrass, choose a polynomial $p = a_{0} + a_{1}x + \cdots a_{n}x^{n}$ such that $||f-p||_{\infty} < \epsilon$. Now choose a polynomial $q$ with rational coefficients such that $||p-q||_{\infty} < \epsilon$. Thus, $||f-q||_{\infty} < 2\epsilon$.

Now the set of polynomials with rational coefficients is countable (Why?) and we have just shown that it is dense in $C([a,b])$. Hence $C([a,b])$ is separable.

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  • $\begingroup$ The topology of $C^\infty([a,b])$ is strictly finer than the topology of $C([a,b])$. So $C^\infty([a,b])$ is not a subspace of $C([a,b])$ in the topological sense. $\endgroup$ Sep 8, 2013 at 13:08
  • $\begingroup$ @DanielFischer The OP is taking the sup norm topology (in his comments to the question). $\endgroup$ Sep 8, 2013 at 13:09

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