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.



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


