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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!

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topsi
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  • Could you define $C_p(X)$? Could you also show what you've tried in this problem? – Dan Rust May 12 '14 at 11:38
  • $C_p(X)$ is the set of continuous real valued functions of $X$ with the topology of pointwise convergence. For example, basic open neighborhoods of $\overline 0$, the constant zero functions, are of the form ${ f_{K_\alpha,n}) }$ where $f_{K_\alpha,n})$ is the function that "takes" the finite set $K_\alpha$ to $(-\frac{1}{n},\frac{1}{n})$. I can prove the oposit direction since, if $X$ is countable, so is it's set of finite subsets. and every finite sebset of $X$ defines a coubtable set of open basis neighborhoods. The other direction is intuitively obvious but not it's formal proof. – topsi May 12 '14 at 11:55
  • No, I can prove that $X$ countable implies $C_p(X)$ first countable. But I can't prove $X$ not countable implies $C_p(X)$ not first countable. – topsi May 12 '14 at 12:09
  • Sorry just saw it corrected – topsi May 12 '14 at 12:12
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    For a function $f\colon X\to\Bbb R$ to be continuous, you need to specify a topology on $X$. Otherwise there's no accurate meaning to the term. – Asaf Karagila May 12 '14 at 12:17
  • I believe this is false if $|X|=|\mathbb{R}|$ and $X$ has the trivial topology - in this case $C_p(X)$ is the set of all constant functions $X\to\mathbb{R}$. In this case, I think $C_p(X)$ must be homeomorphic to $\mathbb{R}$ which is first countable. – Dan Rust May 12 '14 at 12:20
  • You are right. I forgot to mention that $X$ is Tychonoff. This is usually assumed when speaking of $C_p(X)$. Will be addd to my question – topsi May 12 '14 at 12:26

1 Answers1

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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 $X$.

Henno Brandsma
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