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  1. #1
    MHF Contributor Drexel28's Avatar
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    Topology

    Problem: Let X\subseteq \mathbb{R} be compact. Prove or disprove that \mathcal{C}\left[X,\mathbb{R}\right] is separable in general.
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    Quote Originally Posted by Drexel28 View Post
    Problem: Let X\subseteq \mathbb{R} be compact. Prove or disprove that \mathcal{C}\left[X,\mathbb{R}\right] 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 \{x_n\}^\infty_{n=1} be a dense countable set of X, and define \forall\,n\in\mathbb{N}\,,\,\,f_n: X\rightarrow \mathbb{R} by f_n(x):=|x-x_n| .

    Let A be the algebra generated by the functions f_n . Clearly A\subset C(X,\mathbb{R})\,\,\,and\,\,\,|A|=\aleph_0 and also:

    i) If x\in X then clearly f_n(x)\neq 0 for some n\in \mathbb{N}

    ii) If x,y\in X\,,\,\,x\neq y then f_n(x)\neq f_n(y) for some n\in\mathbb{N}

    Then, by the S-W theorem A is dense in C(X,\mathbb{R}) and we're done.

    Note: I'm taking X,\,\mathbb{R} as metric spaces with the standard euclidean metric and C(X,\mathbb{R}) with the supremum metric.
    Then the same proof as above works if X in general is compact, second countable and Hausdorff (because then X is metrizable and we define f_n(x)=\delta(x,x_n)\,,\,\,\delta= the given metric).

    Tonio
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    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 \{x_n\}^\infty_{n=1} be a dense countable set of X, and define \forall\,n\in\mathbb{N}\,,\,\,f_n: X\rightarrow \mathbb{R} by f_n(x):=|x-x_n| .

    Let A be the algebra generated by the functions f_n . Clearly A\subset C(X,\mathbb{R})\,\,\,and\,\,\,|A|=\aleph_0 and also:

    i) If x\in X then clearly f_n(x)\neq 0 for some n\in \mathbb{N}

    ii) If x,y\in X\,,\,\,x\neq y then f_n(x)\neq f_n(y) for some n\in\mathbb{N}

    Then, by the S-W theorem A is dense in C(X,\mathbb{R}) and we're done.

    Note: I'm taking X,\,\mathbb{R} as metric spaces with the standard euclidean metric and C(X,\mathbb{R}) with the supremum metric.
    Then the same proof as above works if X in general is compact, second countable and Hausdorff (because then X is metrizable and we define f_n(x)=\delta(x,x_n)\,,\,\,\delta= the given metric).

    Tonio
    You have yet to prove that A 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 07:48 AM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Proof:

    I didn't use the whole Stone-Weierstrass theorem, only the Weierstrass.

    Lamma: Let \mathbb{Q}[a,b] be the set of all polynomials in [a,b] with rational coefficients. Then, \mathbb{Q}[a,b] is dense in \mathcal{C}[a,b]
    Proof: Let \varphi\in\mathcal{C}[a,b] be arbitrary. Choose a_0+\cdots+a_n x^n\in\mathbb{P}[a,b] such that \|\varphi-p\|_{\infty}<\frac{\varepsilon}{2}.

    Now, the mapping \eta:[a,b]\to\mathbb{R} given by x\mapsto\sum_{j=0}^{n}|x|^j is continuous and so it assumes a maximum M>1.

    Choose q_0,\cdots,q_n\in\mathbb{Q} such that |a_0-q_0|,\cdots|a_n-q_n|<\frac{\varepsilon}{2M}.

    Then, \left|a_0+\cdots+a_n x^n-\left(q_0+\cdots+q_n x^n\right)\right|=\left|(a_0-q_0)+\cdots+(a_n-q_n)x^n\right| \leqslant |a_0-q_0|+\cdots+|a_0-q_0||x|^n <\frac{\varepsilon}{2M}\left(1+\cdots+|x|^n\right)  \leqslant\frac{\varepsilon}{2}. It follows that \|a_0+\cdots+q_n x^n-\left(q_0+\cdots+q_n x^n\right)\|_{\infty}\leqslant\frac{\varepsilon}{2  }.

    Thus,

    \|\varphi-(q_0+\cdots+q_n x^n)\|_{\infty}\leqslant \|\varphi-(a_0+\cdots+q_n x^n)\|_{\infty}+\|a_0+\cdots+a_n x^n-(q_0+\cdots+q_n x^n)\|_{\infty} <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\vare  psilon.

    It follows that \mathcal{Q}[a,b] is dense in \mathcal{C}[a,b]. \blacksquare

    It follows that since \psi:\bigcup_{n=1}^{\infty}\mathbb{Q}^n\to\mathcal  {Q}[a,b] given by (q_0,\cdots,q_{n-1})\mapsto q_0+\cdots+q_{n-1}x^{n-1} is a surjection that \mathcal{Q}[a,b] is countable.

    Now, for every \varphi\in\mathcal{C}[X,\mathbb{R}] there exists, by Tietze's extension theorem, an extension \varphi^*\in\mathcal{C}[a,b] such that \varphi^*\mid X=\varphi.

    Also, for every \varepsilon>0 there exists some q_\varepsilon\in\mathcal{Q}[a,b] such that \|\varphi^*-q_\varepsilon\|_{\infty}<\varepsilon. So, let \Omega_\varphi=\left\{q_\varepsilon:\varepsilon>0\  right\} and \Omega^*=\bigcup_{\varphi\in\mathcal{C}[X,\mathbb{R}]}\Omega_\varphi. Clearly \Omega^*\subseteq\mathcal{Q}[a,b] and thus countable. Clearly then, \in\Omega^*\right\}" alt="\Omega=\left\{p\mid X\in\Omega^*\right\}" /> is also countable.

    Also, let \varepsilon>0 and \varphi\in\mathcal{C}[X,\mathbb{R}] be arbitrary. Letting \varphi^* and the subsequent p_\varepsilon be as before. Then, p_{\varepsilon}\mid X=q\in\Omega and \sup_{x\in X}|\varphi(x)-q(x)|=\sup_{x\in X}|\varphi^*(x)-p_{\varepsilon}(x)|\leqslant\sup_{x\in [a,b]}|\varphi^*(x)-p_{\varepsilon}(x)|<\varepsilon.

    It follows that \Omega is a countable dense subset of \mathcal{C}[X,\mathbb{R}]. The conclusion follows.
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