Topology

• Mar 27th 2010, 07:23 PM
Drexel28
Topology
Problem: Let $X\subseteq \mathbb{R}$ be compact. Prove or disprove that $\mathcal{C}\left[X,\mathbb{R}\right]$ is separable in general.
• Mar 28th 2010, 03:25 AM
tonio
Quote:

Originally Posted by Drexel28
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
• Mar 28th 2010, 08:25 AM
Drexel28
Quote:

Originally Posted by tonio
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.
• Mar 28th 2010, 09:10 AM
Drexel28
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, $\Omega=\left\{p\mid X:p\in\Omega^*\right\}$ 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.