# Question about almost locally ccc and the Krom space

Definition 1. A family $$\mathcal{B}$$ of non-empty open sets in a topological space will be called $$\pi$$-base (or pseudo-base) if every non-empty open set contains at least one member of $$\mathcal{B}$$.

A $$\pi$$-base $$\mathcal{B}$$ is called countable-in-itself $$\pi$$-base (or locally countable) if each member of $$\mathcal{B}$$ contains only countably many elements of $$\mathcal{B}$$.

Definition 2. Let $$X$$ be a topological space. We will say that $$X$$ is almost locally ccc, provided every open set contains an open ccc subspace i.e., if the space has a $$\pi$$-base of open ccc subspaces.

Note that a ccc space is a almost locally ccc space.

The Krom ultrametric space

Notation: $$\omega$$ be the set of all non negative integers and for $$n \in \omega$$ let $${\omega}^{n} = \omega - \{0,1,2,...,n\}$$, in particular $$\mathbb{N}=\omega - \{0\}$$, also let $${\omega}^{n}_{+} = {\omega}^{n} \cup \{\omega\}$$.

For any sets $$S,T$$ and for $$n\in {\omega}^{0}$$ let $${}^{S}{T}{}$$ be the set of all functions from $$S$$ into $$T$$ and let $${}^{n}{T}{}$$ be the set of all functions from $$\{0,...,n-1 \}$$ into $$T$$.

For a set $$S$$ of sets and $$n\in{\omega}^{0}_{+}$$ let $$\downarrow {}^{n}{S}{} = \{\sigma\in{}^{n}{S} \hspace{0.1cm}: \hspace{0.1cm} \sigma(k+1)\subseteq \sigma(k)\hspace{0.2cm} \mbox{for all}\hspace{0.2cm} k\leq n\}$$

Definition 3. For any topological space $$X$$ and $$\pi$$-base $$\mathcal{B}$$ for $$X$$, the associated countable sequence space (or Krom space) $$\mathcal{K}(X)$$ is defined by $$\mathcal{K}(X) = \left\{\sigma\in\hspace{0.1cm}\downarrow\hspace{-0.1cm}{}^{\omega}{\mathcal{B}}{} : \bigcap_{n\in\omega}\sigma(n)\not=\emptyset \right\},$$ and the topology is that given by the Baire metric, for $$\sigma\not=\rho$$ the distance $$d(\sigma,\rho)=\frac{1}{n+1}$$ where $$n$$ is the least integer in $$\{m\in\omega : \sigma(m)\not=\rho(m)\}$$.

A base for $$\mathcal{K}(X)$$ is the family of all sets $$[f]$$, $$f\in \bigcup_{n\in\mathbb{N}}\downarrow\hspace{-0.1cm}{}^{n}{\mathcal{B}}{}$$ where, if $$n<\omega$$ and $$f\in \downarrow\hspace{-0.1cm}{}^{n}{\mathcal{B}}{}$$, then $$[f]=\{g\in \mathcal{K}(X) : g\upharpoonright_{n} =f \}$$

Some facts about the Krom space

• Let $$X, Y$$ be topological spaces, then $$X\times Y$$ is Baire iff $$X\times \mathcal{K}(Y)$$ is Baire iff $$\mathcal{K}(X)\times\mathcal{K}(Y)$$.
• Let $$X$$ be a topological space, then $$X$$ is a Baire space iff $$\mathcal{K}(X)$$ is a Baire space.
• The Banach-Mazur game (or Choquet game) played on $$X$$ and the Banach-Mazur game played on $$\mathcal{K}(X)$$ are equivalents.

My question is the following: If $$Y$$ is an almost locally ccc space then its associated Krom space $$\mathcal{K}(Y)$$ is also almost locally ccc ?

Remark : In the article More on products of Baire spaces of Rui Li and László Zsilinszky, they define the Krom space for a topological space $$(X, \tau)$$ as: $$\mathcal{K}(X) = \left\{\sigma\in\hspace{0.1cm}\downarrow\hspace{-0.1cm}{}^{\omega}{\tau}{} : \bigcap_{n\in\omega}\sigma(n)\not=\emptyset \right\},$$ where $$\downarrow\hspace{-0.1cm}{}^{\omega}{\tau}{}=\{f\in (\tau\setminus \{\emptyset\}) : f(k+1)\subseteq f(k)\}$$. A base for $$\mathcal{K}(X)$$ is the family of all sets $$[f]$$, $$f\in \bigcup_{n\in\mathbb{N}}\downarrow\hspace{-0.1cm}{}^{n}{\tau}{}$$ where, if $$n<\omega$$ and $$f\in \downarrow\hspace{-0.1cm}{}^{n}{\tau}{}$$, then $$[f]=\{g\in \mathcal{K}(X) : g\upharpoonright_{n}=f\}$$. The following theorem appears in the article:

Theorem 3.1 Let $$X, Y$$ be a Baire spaces, and $$Y$$ almost locally ccc. Then $$X\times Y$$ is a Baire space.

In the proof of the Theorem 3.1, a first statement is made which is $$\mathcal{K}(Y)$$ has a countable-in-itself $$\pi$$-base for this it is first shown that $$\mathcal{K}(Y)$$ is an almost locally ccc metric space.

It is in this part that I am confused, because $$\mathbb{R}$$ is an almost locally ccc ($$\mathbb{R}$$ is ccc) and its Krom space $$\mathcal{K}(\mathbb{R})$$ is not almost locally ccc, because consider the non-empty open set $$\mathcal{K}(\mathbb{R})$$ and let $$\mathcal{U}$$ be a non-empty open subset of $$\mathcal{K}(\mathbb{R})$$, then there exists $$f\in \downarrow\hspace{-0.1cm}{}^{n}{\tau}{}$$ such that $$[f]\subseteq \mathcal{U}$$, $$f(n-1)$$ is a non-empty open subset of $$\mathbb{R}$$, so there are continuum non-empty open subsets of $$\mathbb{R}$$ contained in $$f(n-1)$$, so enumerate $$\{A_{\lambda}\}_{\lambda\in\mathfrak{c}}$$. Then the family $$\{[f^{\smallfrown}A_{\lambda}]\}_{\lambda\in\mathfrak{c}}$$ is an uncountable pairwise disjoint family of non-empty open sets contained in $$\mathcal{U}$$, so $$\mathcal{U}$$ is a non-empty open set which is not ccc.

My question is whether the example above is correct and if not, is there another way to understand the proof of the Theorem 3.1?

Thanks a lot.

• Note that $[f ^\frown A]$ and $[f ^ \frown B]$ are disjoint if and only if $A$ and $B$ are disjoint. So your claim about the uncountable pairwise disjoint family seems incorrect, since the $A_\lambda$’s cannot be all disjoint (precisely because $\mathbb{R}$ is ccc!). – Ramiro de la Vega Nov 5 '19 at 23:12
• But I think that $\{[f^{\smallfrown}A_{\lambda}] \}_{\lambda<\mathfrak{c}}$ is a pairwise disjoint family of non-empty open sets, because otherwise there are $\lambda_{1}, \lambda_{2}\in\mathfrak{c}$ such that $[f^{\smallfrown}A_{\lambda_{1}}] \cap [f^{\smallfrown}A_{\lambda_{2}}]\not=\emptyset$, so there is $g\in \mathcal{K}(X)$ such that $g|_{n+1}=f^{\smallfrown}A_{\lambda_{1}}=f^{\smallfrown}A_{\lambda_{2}}$, contradiction. – Gabriel Medina Nov 5 '19 at 23:46
• Yes, I think you are right. I interpreted the definition of $[f]$ wrong, thinking that any sequence $g$ for which $g(k) \subseteq f(k)$ for $k \in dom(f)$ would be an element of $[f]$. Looking at the proof of Theorem 3.1 in the article that you mention, it seems to me that the authors are saying what I said before (that $[f ^\frown A] \cap [f^\frown B]= \emptyset$ implies $A \cap B = \emptyset$). Something is not quite right, maybe you could just e-mail one of the authors and ask. – Ramiro de la Vega Nov 6 '19 at 0:47
• Great, thank you very much, is strange because this argument is used more than once in the article. In addition, Theorem 3.1 is used to show that ccc spaces are productively Baire. – Gabriel Medina Nov 6 '19 at 0:54

Theorem. If the Continuum hypothesis holds, then there is a regular, Hausdorff space $$Y$$ such that:
1. $$Y$$ is a ccc space,
2. $$Y$$ is a Baire space, and
3. $$Y\times Y$$ is not a Baire space (in fact $$Y\times Y$$ is meager in itself).
• I studied the article and introduced the requirements (a little measure of Lebesgue in $\mathbb{R}$ and the density topology in $\mathbb{R}$), if anyone is interested, you can send me an email to andre_asmat@usp.br and I will gladly send you my summary. – Gabriel Medina Nov 28 '19 at 2:25