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What is the sufficient condition on $X$ which makes the spce $C_p(X)$ is the first countable? Thanks!

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Can you answer the question for $X$ discrete? –  Martin Feb 22 '13 at 3:38
    
Does it depend on the cardinality of $X$? –  Paul Feb 22 '13 at 3:44
    
Sure. $C_p(X) = \mathbb{R}^X$ for $X$ discrete and this is first countable if and only if it is metrizable if and only if $\lvert X \rvert \leq \aleph_0$. –  Martin Feb 22 '13 at 3:47
    
Why could $C_p(X)=R^X$ be true? –  Paul Feb 22 '13 at 3:57
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@NateEldredge: see math.stackexchange.com/questions/298445/… –  Paul Feb 22 '13 at 7:24

1 Answer 1

up vote 1 down vote accepted

$C_p(X)$ is first countable iff $X$ is countable (so iff $C_p(X)$ is metrisable).

In fact just assuming that $C_p(X)$ has a countable $\pi$-base at some point is enough for this conclusion: in that case $C_p(X)$ has a countable $\pi$-base at the $0$-function, by homogeneity, so we have a countable family of open subsets $O_n$ of $C_p(X)$ such that every open subset that contains $0$ has one of the $O_n$ as a subset.

Suppose $X$ is not countable.

For every $n$ pick a basic open subset $V_n = B(f_n,F_n,\epsilon_n) = \left\{g \in C_p(X): \forall_{x \in F_n}\,|f_n(x) - g(x)| < \epsilon_n \right\} \subset O_n$, where $f_n \in O_n$, $F_n \subset X$ finite and $\epsilon_n > 0$. These $V_n$ also form a $\pi$-base, of course.

Let $F = \cup_n F_n$ and let $p \in X \setminus F$, which can be done as $X$ is not countable. Then let $O = B(0,\{p\},1)$, a (basic) open set that contains $0$ (the $0$-function), and so for some $n$ we must have $V_n \subset O$. But as $X$ is Tychonoff (standard assumption for all $C_p(X)$ spaces) we have a continuous function $f$ that agrees with the $f_n$ (from $V_n$'s definition) and is $1$ at $p$. This function $f$ is then in $V_n$ (and thus in $O$) but is not in $O$ by construction (not $|f(p)| < 1$, which is what it means to be in $O$).

This contradiction shows that $C_p(X)$ cannot have a countable $\pi$-base at any point for uncountable $X$. So in particular $C_p(X)$ first countable implies $X$ countable and thus $C_p(X)$ metrisable (as a subspace of the metrisable space $\mathbb{R}^\omega$).

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