# $C(X)$ is separable when $X$ is compact

Let $X$ be a compact space and let $\Bbb U =\{(U,V); U,V \mbox{ are open subsets of }X \mbox{ and }\mathrm{cl} U \subset V\}$. for $u=(U,V)$ in $\Bbb U$ , let $F_u:X\to [0,1]$ be a continuous function such that $f_u=1$ on $\mathrm{cl} U$ and $f_u=0$ on $X \setminus V$. Show

a- the linear span of $\{f_u;u\in \Bbb U\}$ is dense in $C(X)$.

b- If X is a metric space, then $C(X)$ is separable.

c-If X is a $\sigma-$ compact metrizable locally compact space, then $C_0(X)$ is separable.

My attempt: a- put $M=\{f_u; u\in \Bbb U\}$. suppose $\mu \in M^{\perp}$. thus for every open subset $U$, $\mu(U)=0$. Also $||\mu||=|\mu|(X) =0$ which shows that $M^{\perp}=0$.

b- For every $n$, put $B_n=\{B(x,\frac{1}{n}) ; x\in X\}$. X is compact so there is a finite set $F_n\subset X$ such that $\{B(x,\frac{1}{n}) ; x\in F_n\}$ is an open finite cover for X. put $F=\cup F_n$. I can show F is dense in X.

put $u_x= (B(x,\frac{1}{n}), B(x,\frac{1}{n-1}))$ for every $x\in F.$ I want to show $M=\{f_{u_x}, x\in F\}$ is dense in $C(X)$. But I can not.

c- $X=\cup X_n$ when every $X_n$ is compact.suppose $A_n$ is a countable dense set for each $X_n$. put $A=\cup A_n$. clearly A is dense in X. Can I claim $C_0(X)=\cup C(X_n)$? so in this case $C_0(X)$ is separable.

I do not know my proof in part (a) is correct or not. Also I have problem in parts b,c.

Please help me. Thanks in advance.

• Your argument for part $a$ does not make sense unless you tell us what you mean by $\perp$. – Mariano Suárez-Álvarez Jul 23 '14 at 5:21
• @AdamHughes, your hint does not mean anything in this context, as the domain of the functions is not a subset of a field... – Mariano Suárez-Álvarez Jul 23 '14 at 5:21
• Yeah, I just realized that after rereading. I saw $[0,1]$ and my brain made that into $X$. – Adam Hughes Jul 23 '14 at 5:22
• I mean $M^\perp=\{x^*\in X^* ; (m,x^*)=0 ~for ~ every ~ m\in M\}$ – niki Jul 23 '14 at 5:23
• Please help me to find the answer. – niki Jul 26 '14 at 11:21

## 1 Answer

Hint.

It is enough to show that one can approximate non-negatve functions by rational linear combinations of the bump functions$F_u$ you constructed, for every function is a difference of non-negative functions.

So let $f:X\to\mathbb R$ be continuous and non-negative, and suppose that $f$ is not identically zero. let $\mathcal G$ be the set of all finite linear combinations $g$ with rational coefficients of the functions $F_u$ such that $0\leq g\leq h$, and let $d=\inf\{\lVert f-g\rVert_\infty:g\in\mathcal G\}$. If we show that $d=0$, we will be done.

Can you do that? (One has to show that $\mathcal G$ is not empty to get things started, of course)

• (-1) The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns. – Norbert Aug 5 '14 at 9:47
• @Norbert, while that may be true, why should I be the one required to provide that canonical answer? I am pretty sure that that one of the silliest reasons for a downvote I've seen in a while... – Mariano Suárez-Álvarez Aug 5 '14 at 17:59
• If you post an answer to the question during the bounty you are confrming that you agree with requirements of this bounty. – Norbert Aug 5 '14 at 18:01
• That is simply an absurd claim. I could not care less for the bounty, and I could not agree less with any «requirements» whatsoever. The person offering the bounty is entirely free to judge that my answer does not satisfy him. – Mariano Suárez-Álvarez Aug 5 '14 at 18:02
• Firstly, hint is not a detailed canonical answer, which was required. There is nothing to argue here. Secondly I have a right to upvote and downvote and I use this "power". I even left a commment, so you don't guess why and who was that. Please don't bother me anymore. – Norbert Aug 5 '14 at 18:16