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It is a surprising fact that the statement $$A > B \implies \mathcal{P}(A) > \mathcal{P}(B)$$ is undecidable in $\mathrm{ZFC}$.

I want to ask a similar question: Assume $A > B$. Does it follow that $A^2 > B^2$?

Remark: $\mathcal{P}(A)$, as usual, denotes the power set of $A$. Also, we define $A > B$ to mean: there exists an injection $B\to A$ but no bijection between these two sets. $A^2$ is an abbreviation for $A\times A$.

Stefan Mesken
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For all $A$ of infinite cardinality we have $A \sim A^2$, i.e. there is a bijection $A \to A^2$ and hence, assuming $A > B$ are both infinite: $$ A^2 \sim A > B \sim B^2 $$ and thus $$ A^2 > B^2. $$

The fact that we have $A \sim A^2$ for all infinite $A$ is covered in any introductory book on set theory. [E.g. Kunen's Set Theory, Jech's Set Theory, Schindler's Set Theory: Exploring Independence and Truth, ...] It follows, for example, from a basic observation of Gödel's pairing function and doesn't rely on the axiom of choice as long as we are only talking about well-orderable sets $A,B$. However, for general $A$ that don't admit a wellorder in $\mathrm{ZF}$, we may have $A \not \sim A^2$.

Stefan Mesken
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The statement $|A| > |B| \implies |\mathcal P(A)| > |\mathcal P(B)|$ is not provable in ZFC in general, but it certainly is provable in many specific cases, including for $|A|$ finite.

For $|B|$ finite, you can prove $|A| > |B| \implies |A|^2 > |B|^2$ easily as well, while for all infinite $A$, we have $|A| = |A|^2$ in ZFC, so that $|A| > |B| \implies |A|^2 > |B|^2$ trivially.

Mees de Vries
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    Other than the finite case, there's absolutely nothing you can say in general. Sure, there is always a larger set with a larger power set, but that's not saying much. Assuming ZFC, given any cardinal $\kappa$, we can find a cardinal-preserving extension to the universe where $2^\kappa=2^{\aleph_0}$. – Asaf Karagila Oct 12 '17 at 18:23
  • Perhaps "many" is a big exaggeration, but the example I had in mind was the one I would guess most people are most familiar with, namely $|\mathbb R| > |\mathbb N|$. – Mees de Vries Oct 13 '17 at 11:02
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While it wasn't explicitly asked, I feel it's worth mentioning, as there was some question about this in the comments.

Theorem. $\sf ZF$ proves that $A<B\to A^2<B^2$ implies the axiom of choice.

Proof.

Recall Tarski's lemma, if $\lambda$ is an well-ordered cardinal and $x$ is any set, then $|X|+\lambda=|X|\cdot\lambda$ implies that $|X|$ is comparable with $\lambda$.

Lemma 1. Suppose that $x=|X|$ and $y=|X|$ are cardinals of non-empty sets such that $x^2=x$ and $y^2=y$, then $(x+y)^2=xy$.

Proof of Lemma 1. First note that if $x^2=x$, then $x+x=x+1=x$. Next, we have the following: $$(x+y)^2=x^2+2xy+y^2=x+xy+y.$$ But since $x+1=x$ and $y+1=y$, we get: $$xy=(x+1)(y+1)=xy+x+y+1=xy+x+y=(x+y)^2.\quad\square$$

Next, suppose that $X$ is any set, then $X^\omega$ satisfies that $|(X^\omega)^2|=|X^\omega|$, and $\lambda=\aleph(X^\omega)$—the least infinite cardinal which cannot be injectively mapped into $X^\omega$—satisfies this also. Let $A=X^\omega+\lambda$ and $B=X^\omega\times\lambda$.

If $|A|<|B|$, then by the assumption $|A|^2<|B|^2$. But this contradicts Lemma 1. Therefore $|A|=|B|$, and then by Tarski's lemma it has to be the case that $X^\omega$ can be well-ordered, and thus $X$ can be well-ordered as well.$\quad\square$

Asaf Karagila
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