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.
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)