If $G=(V,E)$ is a simple, undirected graph, we say that $M\subseteq E$ is a *perfect matching* if the members of $M$ are pairwise disjoint and $\bigcup M = V$. Let $K\_\omega$ be the complete graph on $\omega$. 


Let ${\frak M}$ be a class of perfect matchings on $K\_\omega$ such that whenever $M\_1\neq M\_2\in{\frak M}$ we have $M\_1\cap M\_2 = \emptyset$.


What is the largest possible value that $|{\frak M}|$ can have?



Cardinality of a set of mutually disjoint perfect matchings of $K_\omega$
It's reasonably clear that there can be at most countably many disjoint perfect matchings, since each matching must contain an edge adjacent to each vertex and there are only countably many such edges.


Conversely it's not to hard to build a countable family of perfect matchings in the following manner:


Our aim will be to inductively construct sets $(M\_i(t) \colon i < \omega)$ such that for all $t < \omega$


* $M\_i(t)$ is a matching for all $i < \omega$;
* $M\_i(t) \supseteq M\_i(t-1)$ for all $i < \omega$;
* $M\_i(t) \cap M\_j(t) = \emptyset$ for all $i,j< \omega$;
* $M\_i(k)$ is finite and covers the vertices $\{1,\ldots,t\}$ for all $k \leq t$
* $M\_i(k) = \emptyset$ for all $k > t$.


Taking $M\_i := \bigcup\_{t < \omega} M\_i(t)$ gives a family of edge sets which are matchings by the first property, pairwise disjoint by the third, and perfect matchings by fourth property.


The inductive step is relatively easily; given $(M\_i(t) \colon i <\omega)$ we need to add at most one edge to each $M\_i(k)$ with $k \leq t$ so that $M\_i(k)$ covers the vertex $t+1$, and we need to find a matching $M\_i(t+1)$ which covers all the vertices $\{1,2,\ldots, t+1\}$.


However, since we've only used finitely many edges up to this point, we can just choose all of these edges greedily and continue.


