Are countable graphs with infinite minimal degree $1$-factorizable? We say that a simple, undirected graph $G=(V,E)$ is $1$-factorizable if there is a partition of $E$ such that every member of the partition is a perfect matching of $G$. It is easy to see that any $1$-factorizable graph is regular (every vertex has the same degree).
If $G =(\omega, E)$ is a graph on the vertex set $\omega$ such that every vertex has degree $\aleph_0$, does this imply that $G$ is $1$-factorizable?
(It is a fact that such a graph has a perfect matching.)
 A: Yes, any $\aleph_0$-regular graph $G=(V,E)$ is $1$-factorizable. (By "graph" I mean "simple graph" as in the question. Actually a loopless multigraph is OK provided no two vertices are joined by an infinite number of edges.)
Let $C$ be a set of colors, $|C|=\aleph_0$. We will color the edges of $G$ with colors from $C$ so that each vertex is incident with just one edge of each color.
Since each connected component of $G$ is countable, we may assume that $|V|=|E|=\aleph_0$. The coloring will be done in $\omega$ steps, at most one edge being colored at each step, so that at each step only a finite number of edges will have been colored. Fix enumerations $E=\{e_i:i\in\omega\}$ and  $V\times C=\{(v_i,c_i):i\in\omega\}$.
At step $2i$ consider the edge $e_i$. If it has already been colored, do nothing; otherwise, give it a color which has not been given to any edge adjacent to $e_i$.
At step $2i+1$ consider the pair $(v_i,c_i)$. If the vertex $v_i$ is already incident with an edge of color $c_i$, do nothing; otherwise find an edge incident with $v_i$ which has not yet been colored and is not incident with any edge of color $c_i$, and give it the color $c_i$.
It is easy to see that this construction can be carried out, and the resulting coloring has the desired properties.
A similar argument shows that every $\aleph_\alpha$-regular graph is $1$-factorizable.
