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Let $X$ be an infinite metric space , let $C_c(X,\mathbb R)$ be the set of all compactly supported real valued continuous functions on $X$ (a function $f:X \to \mathbb R$ is called compactly supported iff $\exists $ a compact $K \subseteq X$ such that $f|_{X \setminus K}=0$ ) . Then $C_c(X,\mathbb R)$ is a metric space ( actually more , a normed linear space ) under the sup metric . Then my question is ; is the set $\{f \in C_c(X,\mathbb R) : ||f||_{\infty}\le 1\}$ not compact ? Or equivalently asking , is $C_c(X,\mathbb R)$ infinite dimensional vector space over $\mathbb R$ ? If not always true then can some sufficient condition on $X$ make $C_c(X,\mathbb R)$ infinite dimensional ? ( I only know that it is infinite dimensional if $X$ itself is compact )

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It is always true that if $X$ is an infinite metric space, then $C_c(X,\mathbb{R})$ is infinite-dimensional, because the dual space of $C_c(X,\mathbb{R})$ contains infinitely many linearly dependent functionals, namely, the Dirac-measures concentrated at single points of $X$, and so the dual space is infinite-dimensional, hence so is the space itself.

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    $\begingroup$ If no point of $X$ has a compact nbhd then the only member of $C_c(X,\mathbb R)$ is the constant function $0.$ For if $p\in X$ and $f:X\to \mathbb R$ with $f(p)\ne 0$ then $f$ is non-zero on a nbhd of $p,$ and hence $f$ does not have compact support. $\endgroup$ May 11 '17 at 7:15
  • $\begingroup$ What's an example of a metric space where no point has a compact nbd? $\endgroup$ May 11 '17 at 9:18
  • $\begingroup$ If $X$ itself is any infinite-dimensional normed linear space, for example. $\endgroup$ May 11 '17 at 9:23
  • $\begingroup$ So you claim that $C_c(X,\mathbb{R})$ is trivial for every infinite dimensional normed linear space? You claim that there are no continuous functions with compact support on $l_2$? $\endgroup$ May 11 '17 at 12:48
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    $\begingroup$ If $f$ has compact support then there exists compact $K$ such that $\{x: f(x)\ne 0\}\subset K.$ If $f:l_2\to \mathbb R$ is continuous and $f(x)=1$ for some $x$ then $S=f^{-1}(1/2,3/2)$ is a non-empty open subset of $l_2.$ But if $K$ is compact and $S\subset K$ then there is an open ball $B(x,r)\subset K$ with $r>0$. Let $\{e_n\}_n$ be a Hilbert-space basis for $l_2$. Then $\{x+re_n/2\}_n$ is an infinite closed discrete set lying inside $K,$ so $K$ would not be compact. $\endgroup$ May 11 '17 at 13:16
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Let $X,$ with metric $d,$ be locally compact. Then for every $p\in X$ there exists $r>0$ such that $Cl(B_d(x,r))$ is compact.

Let $P:\mathbb N\to X$ be $1$-to-$1$ such that no sub-sequence of $(P(n))_n$ converges to a member of $P(\mathbb N).$ (See footnote.)

For $n\in \mathbb N$ let $a_n>0$ such that $P(m)\not \in B_d(P(n),a_n)$ for any $m\ne n.$ Let $b_n\in (0,a_n)$ such that $Cl(B_d(P(n),b_n))$ is compact.Let $c_n=b_n/2.$

Then $Cl(B_d(P(n),c_n) \subset B_d(P(n),b_n).$ So the sets $$D_n=Cl(B_d(P(n),c_n)), \quad E_n=Cl(X \backslash B_d(P(n),b_n))$$ are disjoint.

Since a metric space is a normal space there exists a continuous $f_n:X\to [0,1]$ with $f_n(D_n)=\{1\}$ and $f(E_n)=\{0\}.$ Now $f_n^{-1}(0,1]$ is a subset of the compact set $Cl(B_d(P(n),b_n))$, so $f_n$ has compact support.

Now $\{f_n: n\in \mathbb N\}$ is an infinite linearly independent subset of $C_c(X,\mathbb R)$ because $f_n(P(m))=0$ when $m\ne n,$ and $f_n(P(n))=1.$

Footnote: If $X$ has a non-isolated point $q,$ let $(P(n))_n$ converge to $q$ with $0<d(q,P(n+1))<d(q,P(n)).$ If every point of $X$ is an isolated point let $P:\mathbb N\to X$ be any $1$-to-$1$ function.

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