**Short and informal version**: Does the [stable marriage problem](https://en.wikipedia.org/wiki/Stable_marriage_problem) have a solution if there are $\kappa$ men and $\kappa$ women for any cardinal $\kappa \geq \aleph\_0$?


**Long and formal version**: Let $\kappa$ be an infinite cardinal. For any set $X$ we set $\text{Bij}(X)$ to be the set of bijections from $X$ to itself.


We are given two "preference functions" $$P\_M, P\_W: \kappa \to \text{Bij}(\kappa).$$ Let $\mu:M\to W$ be a bijection (also called "marriage"). We say that $(m\_0, w\_0)\in \kappa\times \kappa$ is a *blocking pair* for the marriage $\mu$ if


1. $\mu(m\_0) \neq w\_0$, and
2. $\big(P\_M(m\_0)\big)(w\_0) < \mu(m\_0)$ and vice versa, $\big(P\_W(w\_0)\big)(m\_0) < \mu^{-1}(w\_0)$.


A marriage is said to be *stable* if there are no blocking pairs.


Given $\kappa, P\_M, P\_W$ as above, is there always a stable marriage?


[This looks so ugly that I'm not even sure it's the correct formalisation, feel free to correct & downvote if I got it wrong.]



Stable marriages for infinite bipartite graphs
Your formal version does not look correct. For each boy $b$, there should be a total order $\leq\_b$ on the set of girls $G$ (this is the preference order for $b$) and for each girl $g$, there should be a total order $\leq^g$ on the set of boys (this is the preference order for $g$). In the special case that $G$ and $B$ are both countably infinite and each $\leq\_b$ and $\leq^g$ has order type $\omega$ (this is a natural generalization of the finite case), there is indeed a Stable Marriage Theorem. See Theorem 2.1 of this [paper](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.408.8184) of Cenzer and Remmel. The proof of correctness is to simply extend the Gale-Shapley algorithm to the infinite case. Theorem 2.5 of the same paper shows that as long as the preference orderings $\leq\_b$ and $\leq^g$ are all well-orderings, then there is a stable matching.


