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In my question, I denote by $|\cdot|$ the cardinality of any set. Moreover, if $f: X \to Y$, we denote by $\mathcal{P}f$ its direct image, i.e. $\mathcal{P}f(A)=\{f(a) : a \in A\}$. Let $X,Y$ be two sets.

Is it true that $|X| \le |Y| \iff |\mathcal{P}(X)| \le |\mathcal{P}(Y)|$?

My attempt. Assume first that $|X| \le |Y|$. Then there exists an injective function $f: X \to Y$. Hence we may easily check that the direct image $\mathcal{P}f: \mathcal{P}(X) \to \mathcal{P}(Y)$ is also injective, so that $|\mathcal{P}(X)| \le |\mathcal{P}(Y)|$.

[Actually, we should have $|X| < |Y| \implies |\mathcal{P}(X)| < |\mathcal{P}(Y)|$ because if the direct image $\mathcal{P}g$ of some function $g: X \to Y$ is surjective, then also $g$ is a surjection]

Conversely, assume that $|\mathcal{P}(X)| \le |\mathcal{P}(Y)|$ and suppose by contradiction that $|X|>|Y|$. Then by the above step we should have $|\mathcal{P}(X)| > |\mathcal{P}(Y)|$, absurd.

Is my argument correct? Furthermore, is it also true that $|\mathcal{P}(X)| < |\mathcal{P}(Y)| \implies |X| < |Y|$?

TheWanderer
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  • What is your definition of $\mathcal{P}f$, is it somthing like $\mathcal{P}f({x,y}) = {f(x),f(y)}$? – Afntu Jan 11 '24 at 12:13
  • Yes, the direct image by $f$. – TheWanderer Jan 11 '24 at 13:31
  • @spaceisdarkgreen Reading the answers of the various posts, it seems that $|\mathcal{P}(X)| \le |\mathcal{P}(Y)|$ simply implies $|X| \le|Y|$. Is it true or not? if true, can you give me a reference, please? – TheWanderer Jan 11 '24 at 14:07
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    @TheWanderer not sure where you’re getting that. The various posts should say that it is consistent with ZFC that $2^{\aleph_1}=2^{\aleph_0}$, which is a counterexample to that implication. (Also per the answer you accepted below.) – spaceisdarkgreen Jan 11 '24 at 15:39

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Your first argument (showing that $|X|\leq |Y| \implies|\mathcal{P}(X)|\leq |\mathcal{P}(Y)|$) is correct. (That is, assuming that your proof for the injectivity of the direct image function $\mathcal{P}f$ goes through $-$ this statement is, in any case, correct.)

Everything that follows, alas, is not. In fact, the reverse implication is independent from ZFC. By Easton's theorem, it is consistent with ZFC that, say, $|\mathcal{P}(\aleph_0)|=|\mathcal{P}(\aleph_1)|$, or even that all powersets of all infinite cardinals up to $\aleph_{\aleph_1}$ or anything of the like, have the same size.

As for your last question: That implication is just the contrapositive of the first, so it is true as well.

Tim Seifert
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  • Dear Tim, I edited my question adding the notion of direct image, according to the comment by @Afntu. Well, thank you for your answer. I do not understand how to use Easton's Theorem in my case. Is there an example where $|X|<|Y|$ but $\mathcal{P}(X)=\mathcal{P}(Y)$? – TheWanderer Jan 11 '24 at 13:41
  • I'm working with the category ${\bf Sets}$ of sets and functions and I'm working with NGB axioms for classes (see Borceux1 - Handbook of Categorical Algebra Vol. 1) – TheWanderer Jan 11 '24 at 13:56
  • @TheWanderer Well, it is also consistent that there are no such examples (most prominantly under the additional axiom GCH (the Generalized Continuum Hypothesis)). Since this statement is independent of the standard axioms of set theory, there really are no concrete examples to point at. Rather, the only possibility is to give consistency results as I alluded to in the answer. – Tim Seifert Jan 11 '24 at 13:57
  • @TheWanderer: There is no example, since it is consistent with ZFC that that is the case, but it is not provable, so there is no counterexample either. – Asaf Karagila Jan 11 '24 at 13:57
  • What a shame, for my computations I was hoping that $|\mathcal{P}(X)|\le |\mathcal{P}(Y)| \implies |X| \le |Y|$, but in view of your answers/comments it might not happen. – TheWanderer Jan 11 '24 at 13:59