If two sets have the same cardinality, then so do their power sets. Converse can't be answered?

Question

_{The following is my rewrite of this proof for the following assertion :}

For infinite sets $A, B$, $|A| = |B| \Longrightarrow \require{cancel} \cancel{\Longleftarrow} |P(A)| = |P(B)|$.

$\bbox[2px,border:1px solid black]{\text{ Proof Strategem : }}$ We are given that $f :A \rightarrow B$, is a bijection. How do we construct a bijection between subsets of A and subsets of B? We need a rule whereby, if we take one subset of A, then we can use the rule to uniquely construct a unique subset of B. My intuition on this is to take all the elements of the subset of A, and map them under $f$ to elements of B, which will form a subset of B.

$\bbox[2px,border:1px solid black]{\text{ Proof : }}$ In mathematical notation, we define $g$ as: $g : P(A) \rightarrow P(B)$ by means of $\color{#009900}{g(S) = \{f(x) : x ∈ S\}} \; \forall S \subseteq A$.

$\bbox[2px,border:5px solid grey]{\text{ $g$ onto ? }}$ For all $T \subseteq B$, does there exist $S \subseteq A$ such that $\color{#009900}{g(S) = \{f(x) : x ∈ S\}} \; = T$?

Since $f^{-1}$ exists, define $S = \{f^{-1}(x) : x ∈ T\}.$ We must prove $\color{#009900}{\{f(x) : x ∈ S\}} = T$.

$\bbox[5px,border:1px solid grey]{T \subseteq g(S)}$ Take $y \in T \iff f^{-1}(y) ∈ S \iff \color{ #FF4F00}{f}[f^{-1}(y)] ∈ \color{ #FF4F00}g[S] \iff y \in g[S]. $

$\bbox[5px,border:1px solid grey]{{g(S)} \subseteq T}$ Take $y \in g(S)$ which means there exists $x \in S \ni f(\color{#C154C1}{x}) = y $.
Since $x \in S$ means there exists $t \in T \ni \color{#C154C1}{f^{-1}(t) = x}$, substitute this into the previous equation: $f(\color{#C154C1}{f^{-1}(t)}) = y \implies t = y $. Since $t \in T$, thus $y \in T$.

$\bbox[2px,border:5px solid grey]{\text{ $g$ 1-1 ? }}$ Suppose we have $G,H ⊆ A$ such that: $g(G) = g(H)$ which means $\{f(x) : x ∈ G\} = \{f(x) : x ∈ H\}$.

$\bbox[5px,border:1px solid grey]{\{f(x) : x ∈ G\} \subseteq \{f(x) : x ∈ H\}}$ Take $g \in G \implies f(g) ∈ \quad g(G) = g(H) \quad \implies f(g) ∈ g(H)$.
Thus there exists a $h ∈ H$ such that $f(g) = f(h)$
$\implies g = h$ because $f$ is a bijection. Since $g ∈ G$, thus $g ∈ H$.

We can prove the other direction by the same argument, just with $G$ and $H$ swapped around.

$1.$ In the proof for $g$ onto, how would you (fore)know (ie: divine or presage) to define $S = \{f^{-1}(x) : x ∈ T\}$?

$2.$ Would someone please explain the step $f^{-1}(y) ∈ S \iff \color{ #FF4F00}{f}[f^{-1}(y)] ∈ \color{ #FF4F00}g[S] $?

$3.$ Are there easier proofs? The result seems intuitive but the proof is complex.

$4.$ Are there any pictures?

$5.$ My course doesn't include $ZFC$. So what's the intuition why the converse is false?

Answer 1

1) Well as card(A) = card(B) we know there is a 1-1 bijection (yes, I know that is redundant) where $a \leftrightarrow b = f(a)$. So it's just plain common sense to look at $S = \{a\} \leftrightarrow g(S) = \{f(a)\}$. If theres a bijective relation between the elements of A and B then it's just plain common sense to imagine there is a relationship between the subsets of A and those subsets of B with the corresponding equivalent elements, isn't it?

2) If $f^{-1}(y) = x \in S \implies y = f(f^{-1}(y)) \in g(S) =\{f(x)|x \in S\}$

if $y = f(f^{-1}(y)) \in g(S) =\{f(x)|x \in S\}$ then $y = f(x)$ for some $x \in S$. so $f^{-1}(y) = x \in S$.

3) Easier proofs? How to proof card p(a) = card p(b)? is the same proof but easier to read. This is as basic a proof as it gets. It is in short "there is a 1-1 correspondence between A and B so there will be a 1-1 correspondence between the subsets of A and the subsets of B containing the corresponding elements". Can't get easier than that.

4) Pictures:

A = {1,2,3} B = {1x, 2x, 3x}. 1 <=> 1x, 2<=>2x, 3<=>3x.

P(A) = subsets of {1,2,3}. P(B) = subsets of {1x,2x, 3x}

Subset S = {2,3} $\subset$ A <=> Subset g(S) = {2x, 3x}. And so on.

5) Converse. Given that |P(A)| = |P(B)| $\not \implies$ |A| = |B|. Well we can't construct A, a superset, from a subset without the axiom of choice. I think. There is no counterexample. It just can't be proven it can be done.

Answer 2

The converse can not be answered in $ZF$. Let me explain: the general continuum hypothesis (GCH) states that if $\alpha$ is an infinite cardinal then there is not cardinal $\beta$ with $\alpha < \beta < 2^\alpha$. This hypothesis is independent of other axioms of set theory (you can not prove or disprove it). So if you accept it as another axiom then yes the converse of your problem it true (which is clear). On the other hand there are models of set theory, ZF, such that your problem fails to be true.