# Let $(X,\large\tau)$ be a normal topology, then show that the weak topology induced by the cont. real-valued functions on $X$ is $\large\tau$

Let $(X,\large\tau)$ be a normal topological space and $\cal F$ the collection of continuous real-valued functions on $X$. Show that $\large\tau$ is the weak topology induced by $\cal F$

My cohorts and I have been agonizing over this proof for days now without success. My interpretation of this statement is that a normal topology $(X,\large\tau)$ is 'nice' enough that you can look at its associated set of continuous real-valued functions, collect together all their pre-images of open sets in $\mathbb{R}$, and this collection will form a sub-basis for $\large\tau$; please correct me if this interpretation is wrong.

We were able to show fairly easily that the weak topology induced by $\cal F$, call it $\large\sigma$, is weaker/coarser than $\large\tau$, but we haven't had any luck proving that $\large\sigma=\large\tau$. We have Urysohn's Lemma and Tietze's Extension Theorem at our disposal, also some consequences of Urysohn's Lemma which imply the existence of certain continuous functions which seem promising; Section 12.1 of Royden's Real Analysis. Anyways they seem useful but we've been unsuccessful in their application.

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Are you sure that this is right? I've read in Willard's General Topology that a space is completely regular iff it has the weak topology with respect to all continuous real-valued functions. But there are normal spaces which are not completely regular. –  Stefan Hamcke Apr 14 '13 at 10:35

Hint: So you need only show that every $\tau$-open set if $\sigma$-open. Note that if $U \subseteq X$ is $\tau$-open to show that it is $\sigma$-open you need to show the following
For each $x \in U$ there are $\tau$-continuous functions $f_1 , \ldots , f_n : X \to \mathbb{R}$ and open sets $V_1 , \ldots , V_n \subseteq \mathbb{R}$ such that $$x \in f_i^{-1} [ V_1 ] \cap \cdots \cap f_n^{-1} [ V_n ] \subseteq U.$$
The fact that $( X , \tau )$ is normal (and all singletons are closed) should be helpful (and you should be able to take $n = 1$).