Show that $|\mathcal{P_0}(S)| = |\mathcal{P_0}(T)|$

Question

Let $\mathcal{P_0}(S)$ denote the collection of all countable subsets of $S$. Given that $|S| = |T| = c$, show that $|\mathcal{P_0}(S)| = |\mathcal{P_0}(T)|$.

Answer 1

Since $|S|=|T|=c$, there exists a bijection $f: S\to T$. Then consider a map $F: \mathcal P_0(S)\to \mathcal P_0(T)$ defined by $$F(A)=\{y\in T\,|\,f(x)=y, x\in A\}$$ for all $A\ne\emptyset$ and $A\in \mathcal P_0(S)$, and $F(\emptyset)=\emptyset$.

• $F$ is injective: If $A,B\subseteq S$ but $A\ne B$. Then there exists, without loss of generality, some $z\in A\backslash B$. Then by definition, $f(z)\in F(A)$ but $f(z)\notin F(B)$, meaning that $F(A)\ne F(B)$.
• $F$ is surjective:For each $B\subseteq T$, notice that $F(f^{-1}(B))=B$ and $f^{-1}(B)\subseteq S$.

Together, we have shown that $F$ is bijective, and the desired result holds.