Uniform versus product topologies on $[0,1]^\mathbb{N}$, and their Borel $\sigma$-algebras

Question

Let $\tau_U$ and $\tau_P$ be the uniform (i.e. sup-metric) and product topologies on $[0,1]^\mathbb{N}$, respectively. Clearly, these topologies are not the same ($\tau_P$ is separable and $\tau_U$ is not, for instance). However, a typical basic open set in $\tau_U$,

$V_{x,\epsilon} = \{ y : \forall n\in\mathbb{N}\;\; |x_n - y_n| < \epsilon \}$

is easily seen to be a $G_\delta$ set in $\tau_P$.

Question 1: Is every open set in $\tau_U$ in the $\sigma$-algebra generated by $\tau_P$?

My guess is "no", although I haven't been able to come up with a counterexample so far. Assuming I'm right, though, I have the following weaker question;

Question 2: Does every open set in $\tau_U$ differ from an open set in $\tau_P$ by a meager set? I.e., does every open set in $\tau_U$ have the Baire Property?

Answer 1

The answer to question 1 is no. As shown in this answer, the $\sigma$-algebra generated by any countable family of sets (such as the Borel $\sigma$-algebra of a separable metrizable space, which $\tau_P$ is) has cardinality at most $\mathfrak{c}$.

On the other hand, $\tau_U$ contains $2^\mathfrak{c}$ distinct open sets. For any $\mathcal{A} \subset 2^\mathbb{N}$, let $$U_\mathcal{A} = \bigcup_{A \in \mathcal{A}} B(1_A, 1/2).$$ Clearly $U_\mathcal{A}$ is open. Since for any $A \subset \mathbb{N}$ we have $1_A \in U_\mathcal{A}$ iff $A \in \mathcal{A}$, we have $U_\mathcal{A} = U_{\mathcal{A}'}$ iff $\mathcal{A} = \mathcal{A}'$. Hence $\{U_\mathcal{A} : \mathcal{A} \in 2^\mathbb{N}\}$ are $2^{2^\mathbb{N}}$ distinct open sets in $\tau_U$.

Answer 2

Edit: After thinking about this a little longer, it seems the proof is much simpler.

Note that $[0,1]^\mathbb{N}$ contains the Cantor space $\{0,1\}^\mathbb{N}$. The Cantor space is closed in $[0,1]^\mathbb{N}$ in both $\tau_U$ and $\tau_P$, and the restriction of $\tau_P$ to $\{0,1\}^\mathbb{N}$ produces the usual topology on the Cantor space whereas the restriction of $\tau_U$ gives the discrete topology. Hence any subset of the Cantor space which doesn't have the Baire Property is already closed as a subset of $[0,1]^\mathbb{N}$ with $\tau_U$, and cannot have the Baire Property as a subset of $[0,1]^\mathbb{N}$ with $\tau_P$.

I'm leaving the previous argument below because I think it's interesting.

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Nate's suggestion in his comment above works, with a little tinkering, to show that the answer to question 2 is no.

First, notice that we may replace $[0,1]^\mathbb{N}$ with $\mathbb{T}^\mathbb{N}$, where $\mathbb{T}$ is the unit circle. (Consider the quotient map $[0,1]\to \mathbb{T}$; all of the relevant topological notions are preserved.)

Now fix a function $\phi : \mathbb{T}^\mathbb{N}\to \mathbb{T}$ which is multiplicative, continuous with respect to the uniform topology on $\mathbb{T}^\mathbb{N}$, and takes convergent sequences to their limits. (We won't need shift-invariance; any nonprincipal ultrafilter limit will do.) Divide $\mathbb{T}$ into three closed arcs $A_0$, $A_1$, and $A_2$ corresponding to the intervals $[0,1/3]$, $[1/3,2/3]$, and $[2/3,1]$, and let $E_k = \phi^{-1}(A_k)$. Note that each $E_k$ is uniformly closed, $\mathbb{T}^\mathbb{N} = E_0\cup E_1\cup E_2$, and

$$ E_0^2 = E_0\cup E_1,\quad E_1^2 = E_0\cup E_2,\quad E_2^2 = E_1\cup E_2 $$

One of the $E_k$'s must be nonmeager. Suppose for a contradiction that it also has the Baire Property; then it's comeager in some nonempty (product-)open set $U\subseteq\mathbb{T}^\mathbb{N}$. By Pettis's lemma (see Nate's reference above, or Theorem 9.9 in Kechris's Classical Descriptive Set Theory), $U^2\subseteq E_k^2$. Then the image of $U^2$ under $\phi$ is disjoint from the interior of some $A_j$. It's easy to see that this is a contradiction, since any (product-)open set in $\mathbb{T}^\mathbb{N}$ contains sequences which converge to any given limit in $\mathbb{T}$.