Let $S$ and $T$ be sets such that $|S|=|T|$, prove that $|P(S)|=|P(T)|$

Question

Let $S$ and $T$ be sets such that $|S|=|T|$, prove that $|P(S)|=|P(T)|$ where $P$ denotes a power set.

From the theorem: If $S$ is a finite set with $n$ elements, then the cardinality of $P(S)=2^n$ we can show that $|P(S)|=|P(T)|$ when $S$ and $T$ are finite. I'm not sure where to begin if $S$ and $T$ are infinite sets?

Answer 1

Since $|S|=|T|$, there exists a bijection $f:S \to T$. We can use this to construct a bijection $F:P(S) \to P(T)$.

Define $F(X)$ as follows, $F(X)=\{t \in T| t=f(s) \text{ for some } s \in X\}$.

This is clearly a function from $P(S)$ to $P(T)$, It remains to show that it is injective and surjective.

Surjective

Consider any set $Y \subseteq T$. For every element $y \in Y$, there is an element $s \in S$ such that $f(s)=y$ since $f$ is a bijection, so $F(f^{-1}(Y))=Y$.

Injective

Suppose there were two sets $X,Y \subseteq T$, $X \not=Y$ and $F(X)=F(Y)$. Consider an $s$ that is in one of these sets and not the other (WLOG $X$). Then there must be another element $s'$ in $Y$ such that $f(s)=f(s')$, since otherwise $f(s)$ would not be in $F(Y)$. But, $f$ was a bijection, so this cannot be the case.

Answer 2

Hint Let $f : S \to T$ be a bijection, with inverse $g$.

Can you think of any simple way of defining a simple function from $P(S)$ to $P(T)$? Can you do the construction in such a way that doing the same for $g$ gives an inverse?

Answer 3

$\;|S|=|T|\iff\;$ there exists a bijection $\;f:S\to T\;$, so define

$$F:P(S)\to P(T)\;,\;\;F(X):=\{\,f(x)\;:\;x\in X\,\}\;,\;\;X\in P(S)$$

Answer 4

Note that when $S$ and $T$ are finite, the bijection between $P(S)$ and $P(T)$ can be constructed by picking any particular bijective reordering in the corresponding subset lattices.

When they are infinite you need to use the Axiom of Choice (if you want to construct an explicit $f$), because it is not at all clear what to assign $p(x)\in P(S)$ to, in $P(T)$.