1
$\begingroup$

I'm slowly working my way through Munkres and need help with exercise 12 of section 48. The exercise reads:

Show that $\mathbb{R}^J$ is a Baire space in the box, product, and uniform topologies.

Here, Munkres' convention is that $J$ represents a potentially uncountable index set. In the uniform topology, Baireness of $\mathbb{R}^J$ follows immediately since for any metric space $X$, $X^J$ is complete under the uniform metric (and all complete metric spaces are Baire). However, the product topology is more difficult, and I haven't yet attempted the box toplogy. My proof so far for the product topology:

Let $\mathbb{R}^J$ be given the product topology and let $\{U_n\}$ be a countable collection of dense open subsets of $\mathbb{R}^J$. For any open subset $V$ of $\mathbb{R}^J$, there is some finite set $F \subset J$ such that $\pi_{\alpha_i}(V) \subset \mathbb{R}$ for every $\alpha_i \in F$. If $U_n$ is dense in $\mathbb{R}^J$, then $V$ must intersect $U_n$ at some point $x$, which is true if and only if $\pi_\alpha(V)$ intersects $\pi_\alpha(U_n)$ at $\pi_\alpha(x)$ for every $\alpha \in J$. Hence, $\pi_{\alpha_i}(U_n)$ is a dense open subset of $\mathbb{R}$ for every $\alpha_i \in F$.

By exercise $6(a)$ of section 43, a countable product of completely metrizable spaces is completely metrizable. Thus, $\mathbb{R}^F$ is completely metrizable and hence Baire. Further, because any finite product of open dense sets is itself dense, $\bigcap_n \pi_F(U_n)$ is dense in $\mathbb{R}^F$, where $$\pi_F(U_n) = \prod_{\alpha_i \in F} \pi_{\alpha_i}(U_n).$$

More specifically, there is at least one point $x \in \mathbb{R}^F$ such that $\pi_{\alpha_i}(x) \in \pi_{\alpha_i}(V) \cap \pi_{\alpha_i}(U_n)$ for every $n$ and every $\alpha_i \in F$.

It remains only to show that, for $\alpha \in J - F$, there is some $y_\alpha \in \bigcap_n \pi_\alpha(U_n)$ (because $\pi_\alpha(V) = \mathbb{R}$). Then, letting $z$ be such that $\pi_{\alpha_i}(z) = \pi_{\alpha_i}(x)$ and $\pi_{\alpha}(z) = y_\alpha$, we find that $z \in V \cap \bigcap_n U_n$. But how do I know that $\bigcap_n \pi_\alpha(U_n)$ is nonempty for any $\alpha \in J - F$? This answer suggests I can assume $U_n \supset U_{n+1}$ "without loss of generality", but I don't see how that is so.

I greatly appreciate any advice. I've been juggling products, projections, and intersections in my brain for so long I'm starting to forget my own name!

$\endgroup$
  • $\begingroup$ The product of countably many completely metrisable spaces is also completely metrisable. The standard product metric preserves completeness of metrics. $\endgroup$ – Henno Brandsma Jul 2 '17 at 21:36
  • $\begingroup$ It's not provable that a countable product of Baire spaces is Baire, though. That is the wrong way. $\endgroup$ – Henno Brandsma Jul 2 '17 at 21:37
  • $\begingroup$ The only thing that remains is $\mathbb{R}^J$ in the box topology. The uniform metric is complete metric and the countable product is completely metrisable, as said. $\endgroup$ – Henno Brandsma Jul 2 '17 at 21:40
  • $\begingroup$ I assume $J$ Is countable, or the uniform metric would not make sense? $\endgroup$ – Henno Brandsma Jul 3 '17 at 3:41
  • $\begingroup$ cf. mathoverflow.net/q/24424/2060 $\endgroup$ – Henno Brandsma Jul 3 '17 at 3:49
1
$\begingroup$

One cannot prove the product of Baire spaces is Baire; this might even fail for two spaces. (Oxtoby found/constructed some examples of this). But $\mathbb{R}$ is so nice that we can say more:

This paper by Frolík defines a notion of countably complete (def 2.3) : a space $X$ is called countably complete if there is a sequence of bases $\mathcal{B}_n$ for $X$ such that for every nested (i.e. decreasing) family of sets $A_{n_k}$ for some increasing sequence $n_k$ of integers and such that $A_{n_k} \in \mathcal{B}_{n_k}$, the intersection $\bigcap_{k} \overline{A_{n_k}}$ is non-empty.

Note that all such spaces are Baire: if $U_n$ are open and dense, we can make them decreasing open and dense by defining $V_n = \cap_{i=1}^n U_i$ and noting that all $V_n$ are also open and dense, $V_{k+1} \subseteq V_k$ for all $k$ and $D:= \bigcap_n V_n = \bigcap_n U_n$; so we continue with the $V_n$ instead. Now let $O$ be empty and non-empty Pick $B_n \in \mathcal{B}_n$ with $\overline{B_n} \subseteq B_{n-1} \cap V_n \cap O$ (work by recursion, with $B_0 = X$ as a start), which can be done as we have bases, and $V_n$ is open and dense. The $B_n$ are nested so the intersection of their closures is non-empty, and this lies in $O$ and in $\bigcap_n V_n = D$, so $D$ intersects every non-empty open set of $X$, hence is dense. So $X$ is Baire.

It's also clear that $\mathbb{R}$ has this property. Take $\mathcal{B}_n$ to be all open intervals of diameter $\le \frac{1}{n}$, and Cantor's theorem for complete metric spaces shows that $\mathbb{R}$ is countably complete for this choice.

Theorems 2.10 and 2.12 state that any (box) product of countably complete spaces is countably complete. The proof basically is: take all base sets formed from members of the $n$-th bases in the component spaces, and use this $n$-th base for the product.

So $\mathbb{R}^J$ is Baire in the product and the box topology, and in the uniform topology it's even completely metrisable, hence Baire.

$\endgroup$
  • $\begingroup$ Wow! First of all, I'm upset with myself for not coming up with the definition of the sets $V_n$. It's so obvious! (But everything is, in retrospect) Very straightforward arguments on almost every front, though I still don't quite grok Frolík's proof for a product of countably complete spaces. I'm sure, however, that I'll come to understand it if I set aside the time to really pore over it. Thank you so much for your answer! $\endgroup$ – Itserpol Jul 4 '17 at 11:41
  • $\begingroup$ I did give the essence of the proof. It's quite simple. $\endgroup$ – Henno Brandsma Jul 4 '17 at 20:51
  • $\begingroup$ True, it is a fairly simple proof, but unfortunately I found your proof outline to be a bit too brief. I sometimes need to belabor concepts with excess detail before I fully grasp them. No worries, though, as I've since come to understand Frolík's proof. Thanks again! $\endgroup$ – Itserpol Jul 5 '17 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.