How to prove that $X$ is not countable implies ,$C_p(X)$ not first countable? How can I prove, that, Given a Tychonoff topological space $X$, If $X$ is not countable, then, $C_p(X)$ is not first countable?
Thank you!
 A: Let $X$ be uncountable and let $u$ be the all-zero function on $X$, so $u \in C_p(X)$, then we will show that $u$ does not have a countable local base in $C_p(X)$. (We could have used any function, but for definiteness let's take $u$.) So suppose $U_n, n \in \omega$ is a countable local base at $0$. 
As the sets $[K, \frac{1}{n}] = \{f \in C_p(X): f[K] \subset (-\frac{1}{n}, \frac{1}{n}) \}$ form a standard local base at $u$, where $K$ ranges over the finite subsets of $X$, and $n$ over the positive integers, for each $n$ we pick some $[K_n, \frac{1}{m(n)}]$, with $K_n \subset X$ finite, $m(n) \in \mathbb{N}^+$ such that $[K_n, \frac{1}{m(n)}] \subset U_n$.
Consider the set $K = \cup_n K_n \subset X$, which is at most countable. So as $X$ is uncountable, we find some $p \in X \setminus K$, and define $O \subset C_p(X)$ by $O = [\{p\}, 1] = \{f \in C_p(X): f(p) \in (-1,1) \}$, and note that $u \in O$. 
By assumption, for some $k \in \omega$, $U_k \subset O$ and so also $[K_k, \frac{1}{m(k)}] \subset O$. But if $f$ is any function that is $0$ on all of $K_k$ and $1$ on $p$, which exists by Tychonoff-ness of $X$, then $f \in [K_k, \frac{1}{m(k)}]$ but $f \notin O$, which contradicts our subset relation. 
This contradiction shows that $u$ does not have a local countable base. 
It's clear we could adapt this to any $f \in C_p(X)$ instead of $u$, and if the local base is of size $< |X|$. We can see that $\chi(C_p(X)) = w(C_p(X)) = |X|$ for any Tychonoff space, with hardly more work. 
If $X$ is countable, $C_p(X) \subset \mathbb{R}^X$, which is a separable metrisable space,
and hence so is $C_p(X)$. This shows the strong contrast between $C_p(X)$ for countable $X$ and uncountable $X$.  
