How to prove $a^x\times a^y=a^{x+y}$ for cardinals?

Question

How can I prove this: $$a^x\times a^y=a^{x+y} $$ when $card(A)=a$ , $card(X)=x$ and $card(Y)=y$.

Answer 1

Assuming the axiom of choice, and assuming that $a$ is an infinite cardinal and that $x,y>0$: use the fact that if $\kappa$ and $\lambda$ are infinite cardinals then $\kappa + \lambda = \kappa \cdot \lambda$ ($= \mathrm{max} \{ \kappa, \lambda \}$). This reduces the problem to proving that $a^x \cdot a^y = a^{x+y}$, which you have likely already proved.

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Edit: I see you've updated your question to ask about the product instead of the sum.

In this case, try to find a bijection between the following sets:

• The set of functions $f : X \sqcup Y \to A$, where $\sqcup$ denotes the disjoint union operation; and
• The set of pairs of functions $(f_X : X \to A, f_Y : Y \to A)$.

The first of these sets has cardinality $a^{x+y}$, and the second has cardinality $a^x \cdot a^y$.

Bigger hint:

Consider the restrictions of a function $X \sqcup Y \to A$ to $X$ and $Y$, respectively.