# Prove that if $A$ is infinite and $B$ is countable, then $|A + B| = |A|$ (assume AC)

Here $$A + B := \{(0, a) \ | \ a \in A\} \cup \{(1, b) \ | \ b \in B \}$$.

Under the axiom of choice, we have that there exists an injection $$f: B \to A$$.

Because $$g: A \to A + B: a \mapsto (0, a)$$ is injective, it is enough to find an injection from $$A + B$$ to $$A$$ (Schröder-Cantor-Bernstein). I'm stuck on finding this injection.

Pick an enumeration of a countable subset $$S$$ of $$A$$, say $$x_1,x_2,\dots$$. Enumerate $$B$$ as $$y_1,y_2,\dots$$. Now enumerate $$S + B$$ by interlacing $$(0,x_i)$$ and $$(1,y_i)$$. Do you see how to use this enumeration of $$S + B$$ to construct a bijection between $$S + B$$ and $$S$$? If so, then the rest of $$A + B$$ can be handled by the mapping $$(0,x) \mapsto x$$.
If $$+$$ were replaced by $$\cup$$ then you would need to worry about $$S$$ and $$B$$ having nonempty intersection, but this can also be handled by a slightly different argument. However, this argument works pretty much exactly as written when $$B$$ is finite rather than countably infinite.