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    MHF Contributor Drexel28's Avatar
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    Topology

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

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

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

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

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

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

    Tonio
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    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 $\displaystyle \{x_n\}^\infty_{n=1} $ be a dense countable set of $\displaystyle X$, and define $\displaystyle \forall\,n\in\mathbb{N}\,,\,\,f_n: X\rightarrow \mathbb{R}$ by $\displaystyle f_n(x):=|x-x_n|$ .

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

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

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

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

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

    Tonio
    You have yet to prove that $\displaystyle A$ is closed subalgebra.

    EDIT: Oops, I misread your post. Yes, that is good! I'll post my solution.
    Last edited by Drexel28; Mar 28th 2010 at 07:48 AM.
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    MHF Contributor Drexel28's Avatar
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    Proof:

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

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

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

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

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

    Thus,

    $\displaystyle \|\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}$$\displaystyle <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\vare psilon$.

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

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

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

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

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

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