I need to show that if $X$ is an uncountable Tychonoff space, then $C(X)$ is not metrizable. All I've been able to show so far is that that $F(X)$, the space of all functions with pointwise topology, is homeomorphic to $\mathbb{R}^X$ (the product) which is not metrizable, but I can't seem to get much further. Thanks.

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    $\begingroup$ The following result is known: $C_p(X)$ is first countable if and only if $X$ is countable. See McCoy, Ntantu: Topological Properties of Spaces of Continuous Functions, Theorem 4.4.2; or Tkachuk: A $C_p$-Theory Problem Book, Problem 169, p.142. This question might also be interesting for you. $\endgroup$ – Martin Sleziak Aug 7 '12 at 7:39

Hint: If the pointwise convergence topology was metrizable, it would be first countable, so $0$ would have a countable neighbourhood base. Thus there would be a sequence of open neighbourhoods $U_i$ of $0$ such that every neighbourhood of $0$ contains some $U_i$.
By definition of the topology of pointwise convergence, for each $i$ there exist some finite set $J_i$ and some $\epsilon_i > 0$ such that $f \in U_i$ if $|f(x)| < \epsilon_i$ for all $x \in J_i$. Now define another neighbourhood $U$ of $0$, and use the fact that $X$ is completely regular to show that for any $i$ there is some $f \in C(X)$ such that $f \in U_i$ but $f \notin U$.

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    $\begingroup$ Thank you very much, I managed to do the rest. $\endgroup$ – James Aug 7 '12 at 10:05
  • $\begingroup$ Why by definition of the topology of pointwise convergence do we have the for each $i$ there exists this finite set $J_i$ and $\epsilon_i$ such that $f \in U_i$ if $|f(x)| < \epsilon_i$ ? I am not clear as to why there is such a finite set. $\endgroup$ – oliverjones Dec 28 '17 at 1:03

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