Is $C[0,1]$ normal with the topology of pointwise convergence?

I was asked a question by someone the other day regarding the topology of pointwise convergence, and I can't seem to get anywhere with it. I was wondering if anyone could be of any assistance...

The question was: is $C[0,1]$, the set of continuous real-valued functions from $[0,1]$ to $\mathbb{R}$, with the topology of pointwise convergence (defined by the sub-basis $A_{a,x,b} :=\{f\in C[0,1] : a< f(x) < b\}$ for $x \in [0,1]$ and $a<b \in \mathbb{R}$) a normal topological space?

Wikipedia suggests that $C(\mathbb{R})$ is not normal, so I've tried to show this as a starting point, and then work out whether the same argument holds or fails when we restrict to $[0,1]$: but I can't really get anywhere with this either.

My gut instinct seems to tell me that $C[0,1]$ is not normal, and I have some rather vague hand-wavey intuition for why but I can't really pin anything down at all and have given myself a bit of a headache.

Could anyone point me in the right direction with a good hint or reference please? The question is quite interesting so I'd prefer to try and finish it myself than be given a full solution.

Thanks!

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I removed the metric-spaces tag because $C([0,1])$ with the product topology is not metrizable (if it were, this question would be trivial). –  Nate Eldredge Jun 9 '12 at 15:45
According to Problem 215 from Tkachuk: A $C_p$-Theory Problem Book for a metrizable space $X$ we havep $C_p(X$) is Lindelof $\Leftrightarrow$ $C_p(X)$ is normal $\Leftrightarrow$ $X$ is second countable. Solution is given on p.170, but it relies heavily on some problems done eralier in that book. –  Martin Sleziak Jun 9 '12 at 16:19

Let $\mathscr{B}=\{B_n:n\in\omega\}$ be a countable base for $[0,1]$. For $m,n\in\omega$ let $$F(m,n)=\{f\in C_p[0,1]:f[B_m]\subseteq B_n\}\;.$$ Let $\mathscr{F}_0=\{F(m,n):m,n\in\omega\}$, and let $\mathscr{F}=\{\bigcap\mathscr{A}:\mathscr{A}\subseteq\mathscr{F}_0\text{ and }|\mathscr{A}|<\omega\}$; note that $\mathscr{F}$ is countable.

Suppose that $U$ is an open subset of $C_p[0,1]$, and $f\in U$. Then for some finite $\{x_1,\dots,x_n\}\subseteq[0,1]$ and $B_{k_1},\dots,B_{k_n}\in\mathscr{B}$ we have

$$f\in\Big\{g\in C_p[0,1]:g(x_i)\in B_{k_i}\text{ for }i=1,\dots,n\Big\}\subseteq U\;.$$

Since $f$ is continuous, for each $i=1,\dots,n$ there is a $B_{j_i}\in\mathscr{B}$ such that $x_i\in B_{j_i}$ and $f[B_{j_i}]\subseteq B_{k_i}$, i.e., $f\in F(j_i,k_i)$; clearly

$$f\in\bigcap_{i=1}^nF(j_i,k_i)\subseteq\Big\{g\in C_p[0,1]:g(x_i)\in B_{k_i}\text{ for }i=1,\dots,n\Big\}\subseteq U\;,$$ where $$\bigcap_{i=1}^nF(j_i,k_i)\in\mathscr{F}\;,$$ so $\mathscr{F}$ is a countable network for $C_p[0,1]$.

This is enough to ensure that $C_p[0,1]$ is Lindelöf. To see this, let $\mathscr{U}$ be an open cover of $C_p[0,1]$. For each $f\in C_p[0,1]$ there are a $U_f\in\mathscr{U}$ and an $F_f\in\mathscr{F}$ such that $f\in F_f\subseteq U_f$. Let $$\mathscr{F}_\mathscr{U}=\Big\{F_f:f\in C_p[0,1]\Big\}\;,$$ and for each $F\in\mathscr{F}_\mathscr{U}$ choose $U_F\in\mathscr{U}$ such that $F\subseteq U_F$; $\mathscr{F}_\mathscr{U}$ is countable, so $\{U_F:F\in\mathscr{F}_\mathscr{U}\}$ is a countable subcover of $\mathscr{U}$.

Finally, $C_p[0,1]$ is certainly $T_3$, and a Lindelöf $T_3$-space is normal, so $C_p[0,1]$ is normal.

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Dont'cha just love the brute simplicity of point set topology? Most of the proofs look a lot harder then they actually are. : ) –  Mathemagician1234 Jun 10 '12 at 1:49
Very nice proof. I'll add links to posts about spaces with countable networks and normality of Lindelof spaces at Dan Ma's Topology Blog. –  Martin Sleziak Jun 10 '12 at 4:07
@Martin: Thanks! –  Brian M. Scott Jun 10 '12 at 4:18
Sorry for the delayed response. That's fantastic - goes to show how wrong gut instincts can be! Thanks. –  Tom H Jun 15 '12 at 19:18