The theorem:

Suppose there exist injective functions $A \to B$ and $B \to A$ between two infinite sets $A$ and $B$. Then there exists a bijection $A \to B$.


Let $f: A \to B$ and $g: B \to A$ be injective functions. Let $G$ be the bipartite and undirected graph with partition sets $A$ and $B$ and edges $ab$ when either $f(a) = b$ or $g(b) = a$ for all $a \in A$ and $b \in B$. Note that $f$ and $g$ each correspond to a matching of $A$ and $B$ respectively. Clearly, if $G$ has a perfect matching (which we'll prove), then there exists a bijection $A \to B$.

Because $f$ and $g$ are injective, each vertex in $G$ has at least degree $1$ and at most degree $2$. Therefore each component $C$ in $G$ is either a path or a cycle. If $C$ is a cycle then, because $G$ is bipartite, it follows that $C$ is even and thus has a perfect matching. On the other hand, if $C$ is a path, it may be finite or infinite. If $C$ is finite, then $C$ must be even, because the injectivity of $f$ and $g$ together imply that $|A \cap C| = |B \cap C|$, and so it has a perfect matching. Otherwise, if $C$ is infinite, then either $C$ has one vertex of degree 1 and all others of degree 2 or has all vertices with degree 2: in the first case we take the odd numbered edges (counting from the one edge incident with the vertex of degree 1) and in the second case we just take alternating edges starting wherever we want.

  • $\begingroup$ The path components are infinite. $\endgroup$ – confusedStudent Aug 5 '19 at 8:28

Yes, nice work -- the proof is correct. In fact, this proof does not rely on the axiom of choice (although this may not be obvious until you actually sit down and work out the set-theoretic details).

This same argument was described by John Conway and Peter Doyle in their 1994 paper, division by three, Section 4, "The Cantor-Schroder-Bernstein theorem". They color the edges red and blue, which helps clarify why the cycles and finite paths are of even size. Another way to help clarify this is to make the edges from $A$ to $B$ and $B$ to $A$ directed.

(I don't think that Doyle and Conway are the first to explain the Cantor-Schroder-Bernstein theorem in this way, but you may enjoy the paper more generally.)

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