Borel $\sigma$-algebra of separable $T_{3 \frac 1 2 }$ space generated by bounded continuous functions A topological space $\Omega$ satisfies the $T_{3\frac 1 2}$ separation axiom if for every $A \subset \Omega$ closed, and every $x \in \Omega \setminus A$, there is a continuous function $f : \Omega \to [0,1]$ for which $f(x) = 0$ and $f(y) = 1$ for every $y \in A$. I want to prove the following:

If $\Omega$ is a separable $T_{3 \frac 1 2 }$ space, then $\mathcal B(\Omega) = \sigma(C_b(\Omega))$; that is, the Borel $\sigma$-algebra is generated by the bounded continuous functions.

Obviously we have $\sigma(C_b(\Omega)) \subset \mathcal B(\Omega)$, since each $f \in C_b(\Omega)$ is Borel measurable. Conversely, it's enough to show that $\sigma(C_b(\Omega))$ contains all of the closed sets; in particular, I want to show that for every $A \subset \Omega$ closed, there is $f \in C_b(\Omega)$ for which $f^{-1}(\{1\}) = A$.
My original strategy was to construct a sequence of nested open sets $(B_n)$ for which we have $B_1 \supset \overline B_2 \supset B_2 \supset \overline B_3 \supset B_3 \supset \cdots$, and $A = \mathop{\bigcap}_{n \geq 1} B_n$; then let $f_n : \Omega \to [0,1]$ be such that $f_n(y) = 1$ for all $y \in \overline{B_n}$, and let $f = \sum_{n=1}^\infty 2^{-n} f_n$, which may be shown to be continuous (since the sequence of partial sums is continuous and converges uniformly to $f$). But as was recently pointed out to me here, such a sequence of open sets may not exist.
Any suggestions?
 A: It isn't true.  Counterexamples include spaces like $\Omega = \{0,1\}^{[0,1]}$, a product of uncountably many copies of $\{0,1\}$ with the product topology.  It is compact Hausdorff, hence $T_{3 \frac{1}{2}}$, and it is well known to be separable.  But your $\sigma(C_b(\Omega))$ is the $\sigma$-algebra of Baire sets which here is strictly smaller than the Borel $\sigma$-algebra.
Let's represent elements of $\Omega$ as functions $\omega : [0,1] \to \{0,1\}$.  The fact is that any Baire set in $\Omega$ can only depend on countably many coordinates, and in particular cannot be a singleton.  To make this precise, for any countable $A \subset [0,1]$ let $\pi_A : \Omega \to \{0,1\}^A$ be the restriction map $\pi_A(\omega) = \omega|_A$.  Then let $\mathcal{F}$ be the collection of subsets $E$ of $\Omega$ that are of the form $\pi_A^{-1}(F)$ for some countable $A \subset [0,1]$ and some $F \subset \{0,1\}^A$.  Clearly $\Omega, \emptyset \in \mathcal{F}$, and $\mathcal{F}$ is closed under complements because $(\pi_A^{-1}(F))^c  = \pi_A^{-1}(F^c)$.  It is also closed under countable unions: suppose $E_1, E_2, \dots \in \mathcal{F}$, so that each $E_n = \pi_{A_n}^{-1}(F_n)$ for some countable $A_n \subset [0,1]$ and some $F_n \subset \{0,1\}^{A_n}$.  Let $A = \bigcup_n A_n$ which is again countable, and let $\pi_n : \{0,1\}^A \to \{0,1\}^{A_n}$ be restriction, so that $\pi_{A_n} = \pi_n \circ \pi_A$.  Then
$$\bigcup_n E_n = \bigcup_n \pi_A^{-1}(\pi_n^{-1}(F_n)) = \pi_A^{-1}\left(\bigcup_n \pi_n^{-1}(F_n)\right).$$
Hence $\mathcal{F}$ is a $\sigma$-algebra.
Consider the class $C_{\mathcal{F}}(\Omega)$ of continuous functions $f : \Omega \to \mathbb{R}$ which are $\mathcal{F}$-measurable.  This is clearly an algebra containing the constants.  It also separates points: for any $x \in [0,1]$ the function $\pi_x(\omega) = \omega(x)$ is continuous (by definition of the product topology) and $\mathcal{F}$-measurable, but if $\pi_x(\omega) = \pi_x(\omega')$ for all $x \in [0,1]$ then $\omega = \omega'$.  Therefore, by Stone-Weierstrass, $C_\mathcal{F}(\Omega)$ is uniformly dense in $C(\Omega)$.  That means every $f \in C(\Omega)$ is a pointwise limit of $\mathcal{F}$-measurable functions and is thus $\mathcal{F}$-measurable, so in fact $C(\Omega) = C_\mathcal{F}(\Omega)$.
Hence, $\sigma(C(\Omega)) \subset \mathcal{F}$.
On the other hand, for any fixed $\omega$, we have $\{\omega\} \notin \mathcal{F}$.  For suppose $E \in \mathcal{F}$ with $\omega \in E$.  Then $E = \pi_A^{-1}(F)$ for some countable $A$ and some $F \in \{0,1\}^A$.  Since $A$ is countable, there exists $y \in [0,1] \setminus A$, so let
$$\omega'(x) = \begin{cases}
\omega(x), & x \ne y \\
1 - \omega(x), & x = y
\end{cases}$$
Since $y \notin A$, we have $\omega|_A = \omega'|_A$ and thus $\omega' \in E$.  Therefore $E$ contains at least two points and cannot be the singleton $\{\omega\}$.
This is discussed a bit more in 6.10(i) of Bogachev's Measure Theory.  He gives Engelking 2.7.12(c) as a reference for the theorem about uncountable products.
