Exercise 5.1.2

Proof. Suppose that $\kappa =|A|$ for a set $A$.

We claim that $A^{\text{∅}}=\left\{\text{∅}\right\}=1$ so that clearly

First consider any $f\in A^{\text{∅}}$. Suppose that $f\ne \varnothing$ so that there is a $(b,a)\in f\subseteq \varnothing \times A$. But then $b\in \varnothing$, which is a contradiction. Hence $f=\varnothing$. So if there are any $f\in A^{\text{∅}}$ then $f=\varnothing$ but are there any $f\in A^{\text{∅}}$? Clearly the empty set is a function from $\text{∅}$ to $A$ since it is vacuously true that for every $b\in \varnothing$ there is a unique $a\in A$ such that $(b,a)\in \varnothing$. Hence $\varnothing \in A^{\text{∅}}$ so that $A^{\text{∅}}=\left\{\text{∅}\right\}=1$.

We also claim that $|A^1|=|A|$ so that

To this end for any $f\in A^1$ define $F(f)=f(\varnothing )$, noting that $1=\left\{\text{∅}\right\}$, so that clearly $F:A^1\to A$. Now consider any $f,g\in A^1$ where $f\ne g$. Then it has to be that

so that $F$ is injective. Now consider any $a\in A$ and define $f\in A^1$ by $f(\varnothing )=a$. Then clearly

so that $F$ is surjective. Hence we’ve shown that $F$ is bijective. □