# Is there a suitably generalized Baire property for topological spaces of arbitrary cardinalities?

Is there some suitable generalization to the notion of Baire property for topological spaces of arbitrary cardinalities which satisfies the following condition:

1. The meager sets are sets which are union of $\lambda$ nowhere dense sets where $\lambda < \kappa$ and $\kappa$ is the cardinality of the space.

2. If we consider a model of ZFC $\mathfrak{M}$ in which every (suitably definable) set of reals have the property of Baire, then every (suitably definable) set of a suitably defined toplogical space have the "generalized" Baire property.

• The usual Baire property involves {\it countable} union of nowhere dens sets. Maybe there is a generalization to $\lambda$ union of nowhere dense sets, for $\lambda < \kappa$ where $\kappa$ is the cardinality of the space. – user38200 Dec 10 '13 at 19:19
• I assume that we define the topological space in such a way that if it has cardinality $\kappa > 2^{\aleph_0}$ in $\mathfrak{M}$ and if it has cardinality $2^{\aleph_0}$ in the real world, the topology must coincide with the real topology. – user38200 Dec 10 '13 at 20:17
• You may already know this, but the most well-known generalisation of this is probably to Polish spaces, i.e. separable completely metrisable spaces. I don’t know if “every set of reals is Baire” implies “every subset of a Polish space is Baire”, but the standard models of the former are also models of the latter, if I remember right. However, this is a bit short of a full answer, since Polish spaces cannot have cardinality larger than the continuum. – Peter LeFanu Lumsdaine Dec 10 '13 at 20:47
• @EmilJeřábek: Yes that is exactly what I mean. For the models of ZFC, we can take the full Solovay model or a model of projective determinacy. Sorry I made some changes. – user38200 Dec 10 '13 at 21:09
• There is one generalization of the Baire property (that I know of) that seems to be in line with your first requirement due to Halko and Shelah. It can be found in shelah.logic.at/files/662.pdf. I believe that, modulo the notation, you want to mainly look at sections 3 and 4. – Shehzad Ahmed Dec 11 '13 at 5:46

I am not sure if this is the kind of answer you were looking for, but since no one has given an answer yet, I think it is a good idea to say what little I know about larger cardinal analogues of the Baire category theorem.

The only spaces that I can currently think of where one would want to consider larger cardinal generalizations of Baire spaces to larger cardinals are the $P_{\kappa}$-spaces, so I will say a few things about how Baire spaces relate to $P_{\kappa}$-spaces. In order to give motivation for this answer, I will first have to say a few things about $P_{\kappa}$-spaces in general before I talk about the Baire property.

$\textbf{Preliminaries concerning$P_{\kappa}$-spaces}$

Many of the basic results from general topology hold when one replaces each instance of "finite" with an instance of "less than $\kappa$" for some regular cardinal $\kappa$. In this context, instead of dealing with ordinary topological spaces where the intersection of finitely many open sets is open, one deals with spaces where the intersection of less than $\kappa$ many open sets is open.

If $\lambda$ is a cardinal, then a $P_{\lambda}$-space is a completely regular space where the intersection of less than $\lambda$ many open sets is open. If $\lambda$ is a singular cardinal, then every $P_{\lambda}$ space is automatically a $P_{\lambda^{+}}$-space. Therefore, without loss of generality, we may restrict our attention to $P_{\lambda}$-spaces where $\lambda$ is a regular cardinal.

A space $X$ is said to be $\lambda$-compact if the intersection of less than $\lambda$ many open sets is open. If $\lambda$ is a regular cardinal, then the product of finitely many $\lambda$-compact $P_{\lambda}$-spaces is $\lambda$-compact.

One can also generalize the notion of a metric space and a uniform space to $P_{\kappa}$-spaces. We therefore define a $P_{\kappa}$-uniform space to be a uniform space where the intersection of less than $\kappa$ many entourages is an entourage. If $\kappa$ is uncountable, then it is easy to show that every $P_{\kappa}$-uniform space is generated by equivalence relations. It is well known that a uniform space is induced by a metric if and only if it is generated by a countable sequence of entourages. The notion of a metric space is analogous to the notion of a uniform space generated by linearly ordered descending sequence of entourages of length $\kappa$.

$\textbf{$P_{\kappa}$-spaces and Baire spaces}$

If $\lambda$ is an infinite cardinal, then a $\lambda$-Baire space is a a topological space $X$ such that the intersection of less than $\lambda$ many dense open sets is dense. With this definition, every topological space is an $\aleph_{0}$-Baire space, and an $\aleph_{1}$-Baire space is simple a Baire space.

Fortunately, the Baire category theorem does hold for some $P_{\kappa}$-spaces where you might expect there to be some form of the Baire category theorem. Let $\kappa$ be a regular cardinal. Suppose that $I$ is a set and $X_{i}$ is a space for $i\in I$. Let $\prod_{i\in I}^{\kappa}X_{i}$ be the topology with underlying set $\prod_{i\in I}X_{i}$ generated by the basis consisting of products $\prod_{i\in I}U_{i}$ where each $U_{i}$ is open in $X_{i}$ and where $|\{i\in I|U_{i}\neq X_{i}\}|<\kappa$. If each $X_{i}$ is a $P_{\kappa}$-space, then $\prod_{i\in I}^{\kappa}X_{i}$ is also a $P_{\kappa}$-space.

$\mathbf{Proposition}$ If $I$ is a set, $\kappa$ is a regular cardinal, and $X_{i}$ is a discrete space for each $i\in I$, then $\prod_{i\in I}^{\kappa}X_{i}$ satisfies the $\kappa^{+}$-Baire property.

$\mathbf{Proof}$ The proof of this result is analogous to the proof of the ordinary Baire category theorem. Let $U_{\alpha}\subseteq\prod_{i\in I}^{\kappa}X_{i}$ be a dense open set for each $\alpha<\kappa$ and let $U\subseteq\prod_{i\in I}^{\kappa}X_{i}$ be a nonempty open set. We shall using transfinite induction construct sets $I_{\alpha}\subseteq I$ with $|I_{\alpha}|<\kappa$ and functions $f_{\alpha}\in\prod_{i\in I_{\alpha}}X_{i}$ such that

1. if $\alpha<\beta$, then $I_{\alpha}\subseteq I_{\beta}$ and $f_{\alpha}=f_{\beta}|_{I_{\alpha}}$

2. If $f\in\prod_{i\in I}^{\kappa}X_{i}$, and $f|_{I_{\alpha}}=f_{\alpha}$, then $f\in U\cap\bigcap_{\beta<\alpha}U_{\alpha}$.

Zero step- Since $U$ is a non-empty open set, there is some $I_{0}\subseteq I$ and some $f_{0}\in\prod_{i\in I_{0}}X_{i}$ where if $f\in\prod_{i\in I}X_{i}$ and $f|_{I_{0}}=f$.

Limit ordinal step- If $\lambda$ is a limit ordinal with $\lambda<\kappa$, then let $I_{\lambda}=\bigcup_{\alpha<\lambda}I_{\alpha}$ and let $f_{\lambda}=\bigcup_{\alpha<\lambda}f_{\alpha}$.

Successor ordinal step- Suppose that $I_{\alpha},f_{\alpha}$ have been defined already and $|I_{\alpha}|<\kappa$. Then $\{f\in\prod_{i\in I}X_{i}:f|_{I_{\alpha}}=f_{\alpha}\}$ is a non-empty open set, so $U_{\alpha}\cap\{f\in\prod_{i\in I}X_{i}:f|_{I_{\alpha}}=f_{\alpha}\}$ is a non-empty open set. Therefore, there is some set $I_{\alpha+1}$ with $I_{\alpha}\subseteq I_{\alpha+1}$ and some $f_{\alpha+1}\in\prod_{i\in I_{\alpha+1}}X_{i}$ where if $f\in\prod_{i\in I}X_{i}$ and $f|_{I_{\alpha+1}}=f_{\alpha+1}$, then $f\in U_{\alpha}$.

Let $J=\bigcup_{\alpha<\kappa}I_{\alpha}$ and let $f\in\prod_{i\in I}X_{i}$ be a function with $f|_{J}=\bigcup_{\alpha<\kappa}f_{\alpha}$. Then $f\in U\cap\bigcap_{\alpha<\kappa}U_{\alpha}$. Therefore, since $U\cap\bigcap_{\alpha<\kappa}U_{\alpha}$ is non-empty, the set $\bigcap_{\alpha<\kappa}U_{\alpha}$ is dense. $\mathbf{QED}$

Interestingly, for $P_{\lambda}$-spaces, the property of being a $\lambda$-Baire space is hereditary under taking dense subspaces.

$\mathbf{Proposition}$ Let $\lambda$ be a cardinal and suppose that $Y$ is a dense subspace of a space $X$.

1. If $Y$ is a $\lambda$-Baire space, then $X$ is also a $\lambda$-Baire space.

2. If $X$ is a $\lambda$-Baire $P_{\lambda}$-space, then $Y$ is also a $\lambda$-Baire space.

One can also construct $P_{\kappa}$-spaces with strong Baire properties using ultraproducts. Recall that an ultrafilter $U$ is $\lambda$-regular if there is some $E\subseteq U$ with $|E|=\lambda$ but where $\bigcap D=\emptyset$ for each infinite $D\subseteq E$. Furthermore, the countably incomplete $\lambda$-good ultrafilters are precisely the ultrafilters where the ultraproducts are always $\lambda$-saturated.

$\mathbf{Proposition}$ (Bankston)

1. The ultraproduct of regular spaces by a $\lambda$-regular ultrafilter is a $P_{\lambda^{+}}$-space.

2. Furthermore, the ultraproduct of topological spaces by a countably incomplete $\lambda$-good ultrafilter is $\lambda$-Baire.

$\textbf{Ideas of proof}$ A proof of these facts is given in the paper Topological reduced products via good ultrafilters by Paul Bankston. A proof of 1 uses basic facts about regular ultrafilters. A proof of 2 uses the fact that the ultraproduct of models by a $\lambda$-good ultrafilter is saturated.

$\textbf{The failure of the Baire property}$

One might expect for $\kappa$-compact $P_{\kappa}$-spaces to be Baire spaces, or complete uniform spaces generated by a linearly ordered set of entourages (i.e. generalized metric spaces) to be Baire spaces. Unfortunately, this is not the case. In fact, for most $P_{\kappa}$-spaces it is easy to find a closed subspace that is not even a Baire space. If $X$ is a $P_{\kappa}$-space, and there is a sequence of closed subsets $C_{n}$ where $C_{n}$ is nowhere dense in $C_{n+1}$ for all $n$, then $\bigcup_{n}C_{n}$ is a closed subspace of $X$ and each $C_{n}$ is nowhere dense in $\bigcup_{n}C_{n}$, so $\bigcup_{n}C_{n}$ is not a Baire space.

Let $X_{n}$ be a $P_{\kappa}$-space for all natural numbers $n>0$ and $x_{n}\in X_{n}$ be a non-isolated point for all $n$. Give $X=\prod_{n\in\mathbb{N}}X_{n}$ the box topology. Then $\prod_{n\in\mathbb{N}}X_{n}$ becomes a $P_{\kappa}$-space with this topology. For natural numbers $N$, let $C_{N}\subseteq\prod_{n\in\mathbb{N}}X_{n}$ consist of all sequences $(y_{n})_{n}$ such that $y_{n}=x_{n}$ whenever $n>N$. Then $C_{N}\simeq X_{1}\times...\times X_{N}$. Let $C=\bigcup_{N}C_{N}$. Then $C$ is not a Baire space. On the other hand, if each $X_{n}$ is $\kappa$-compact, then each finite product $X_{1}\times...\times X_{n}\simeq C_{n}$ is $\kappa$-compact, so $C$ is $\kappa$-compact. Furthermore, if each $X_{n}$ can be given a complete uniformity generated by a linearly ordered set of cofinality $\kappa$, then $C$ can also be given a complete uniformity generated by a linearly ordered set of cofinality $\kappa$ (i.e. $C$ is like a complete metric space), but $C$ is still not a Baire space.

$\textbf{An application of generalized Baire spaces}$

In this section, I will give motivation for the Baire property for $P_{\kappa}$-spaces in terms of point-free topology and Boolean algebras. The notions in this section come from my own personal research.

Recall that a frame is a complete lattice that satisfies the distributive law $x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i})$. Frames are the main objects of study in point-free topology since if $(X,\mathcal{T})$ is a topological space, then $\mathcal{T}$ is a frame.

If $L,M$ are frames, then a frame homomorphism from $L$ to $M$ is a function $f:L\rightarrow M$ such that $f(0)=0,f(1)=1,f(\bigvee R)=\bigvee f[R],f(x\wedge y)=f(x)\wedge f(y)$ for each $R\subseteq L$ and $x,y\in L$. Frame homomorphisms from $L$ to $M$ correspond roughly to continuous functions from $M$ to $L$.

A frame $L$ shall be called $\kappa$-distributive if whenever $|I|<\kappa$ and $R_{i}\subseteq L$ for $i\in I$, then $\bigwedge_{i\in I}\bigvee R_{i}=\bigvee\{\bigwedge_{i\in I}x_{i}|x_{i}\in R_{i}\,\textrm{for}\,i\in I\}.$ It can be shown that a $T_{1}$-space is $\kappa$-distributive if and only if the intersection of less than $\kappa$ many open sets is open. Therefore, the notion of $\kappa$-distributivity is a way to generalize the notion of a $P_{\kappa}$-space to point-free topology. It turns out that the surjective image of a $\kappa$-distributive frame under a frame homomorphism is not necessarily $\kappa$-distributive. On the other hand, one may characterize the completely regular $\kappa$-distributive frames $L$ such that the image of $L$ under a surjective frame homomorphism is $\kappa$-distributive, and this characterization requires the Baire property. We shall state the results here for topological spaces.

$\mathbf{Proposition}$ (Sikorski) Let $X$ be a $P_{\kappa}$-space. Then the regular open algebra $\mathrm{Ro}(X)$ is $\kappa$-distributive if and only if $X$ is a $\kappa$-Baire space.

$\mathbf{Theorem}.$ Let $\kappa$ be a regular cardinal. Let $(X,\mathcal{T})$ be a $P_{\kappa}$-space. Then the following are equivalent.

1. Every closed subspace of $X$ satisfies the $\kappa$-Baire property.

2. Every subspace of $X$ satisfies the $\kappa$-Baire property.

3. If $f:\mathcal{T}\rightarrow M$ is a surjective frame homomorphism, then $M$ is $\kappa$-distributive.

4. If $f:\mathcal{T}\rightarrow B$ is a surjective frame homomorphism and $B$ is a complete Boolean algebra, then $B$ is $\kappa$-distributive.