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.

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


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
    $\begingroup$ (-1) The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns. $\endgroup$ – Norbert Aug 5 '14 at 9:47
  • 1
    $\begingroup$ @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... $\endgroup$ – Mariano Suárez-Álvarez Aug 5 '14 at 17:59
  • 1
    $\begingroup$ If you post an answer to the question during the bounty you are confrming that you agree with requirements of this bounty. $\endgroup$ – Norbert Aug 5 '14 at 18:01
  • 1
    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Aug 5 '14 at 18:02
  • 1
    $\begingroup$ 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. $\endgroup$ – Norbert Aug 5 '14 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.