I am considering the construction in [Peng—Shen—Wu] in which the authors show the consistency of a set $X$ such that there is a surjection from $X^2$ onto the power set of $X$ (henceforth $\mathscr{P}(X)$).

First, some terminology. Given sets $X,Y$, I shall write $|X|\leq|Y|$ to mean that there is an injection $X\to Y$, and $|X|\leq^*|Y|$ to mean that $X$ is empty or there is a surjection $Y\to X$.

Given a set $X$, the Hartogs number of $X$, denoted $\aleph(X)$, is the least ordinal $\alpha$ such that $|\alpha|\not\leq|X|$. Similarly, the Lindenbaum number of $X$, denoted $\aleph^*(X)$, is the least ordinal $\alpha$ such that $|\alpha|\not\leq^*|X|$. These ordinals always exist and are cardinals.

The construction in [Peng—Shen—Wu] uses permutation models, but the translation to $\mathsf{ZF}$ from $\mathsf{ZFA}$ is clear. I noticed that the set constructed with this process satisfies $\aleph(X)=\aleph^*(X)=\aleph_\omega$. My question is thus: Is having limit Hartogs/Lindenbaum number necessary?

Really, this is (at least) four questions: Is it consistent with $\mathsf{ZF}$ to have a set $X$ such that $|\mathscr{P}(X)|\leq^*|X^2|$ and...

  1. $\aleph(X)$ and $\aleph^*(X)$ are successor cardinals?
  2. $\aleph(X)=\aleph^*(X)$ is a successor cardinal?
  3. $\aleph(X)$ is a successor cardinal?
  4. $\aleph^*(X)$ is a successor cardinal?


Peng, Yinhe; Shen, Guozhen; Wu, Liuzhen, A surjection from square onto power (ArXiv).

  • $\begingroup$ It seems to me that you can arrange this $X$ in a way that the Feferman–Levy symmetric system can be done, making the $\aleph_\omega$ into $\aleph_1$. $\endgroup$
    – Asaf Karagila
    Jun 25, 2023 at 18:33

1 Answer 1


In the Feferman-Levy model $M$ for $\mathbb{R}$ being a countable union of countable sets, there is $X$ with $|\mathcal{P}(X)| =^* |X^2|$ and $\aleph(X)=\aleph^*(X)=\omega_1.$

In this model, we can express $\mathbb{R}$ as an increasing union $\mathbb{R}=\bigcup R_n,$ each satisfying $R_n = \mathbb{R}^{L(R_n)}$ and $L(R_n)$ a $\text{Col}(\omega, \omega_n)$-generic extension of $L.$ Let $A_n=\{(m,r) \in n \times \mathbb{R}: r \in R_m\}$ and $A = \bigcup_{n<\omega} A_n.$ Note that $M \models V=L(\mathbb{R}, A).$

Define $X_n$ to be the set of reals in $L(R_{n+1})$ which are Cohen over $L(R_n),$ and let $X = \bigcup X_n.$ It is immediate that $\omega_1 \le \aleph(X) \le \aleph^*(X).$

Claim. $|X^2|=^*|\mathbb{R}|=^*|\mathcal{P}(X)|.$

Proof of Claim: Since $X_n$ is comeager in $L(R_{n+1}),$ we have a surjection $f: X^2 \rightarrow \mathbb{R}$ by $(x,y) \mapsto x-y.$ This implies $|X^2| \ge^* |\mathbb{R}|.$

From the order on $X \subset \mathbb{R},$ we can construct a surjection from $\mathcal{P}(X)$ onto $X^2,$ so $|\mathcal{P}(X)| \ge^* |X^2|.$

Now we will construct a surjection from $\mathbb{R}$ to $\mathcal{P}(X).$ Fix $S \subset X.$ There is $p \in R_n,$ $\alpha \in \text{Ord},$ and a formula $\varphi$ such that $S =\{r \in \mathbb{R}: \varphi(r, p, A, \alpha)\}.$ For $m \ge n,$ $S \cap X_m$ is an open subset of $X_m$ (in the Cantor space topology), since for $c \in X_m,$ we have $c \in S$ iff there is $i$ such that

$$L(R_m) \models ``(c \restriction i, \emptyset) \Vdash_{\text{Add}(\omega, 1) \times \prod_{j>m} \text{Col}(\omega, \omega_j)} `` \varphi^{L(\mathbb{R}, \dot{A}_{j>m})}(G \restriction 1, \check{p}, \check{A}_m \cup \dot{A}_{j>m}, \check{\alpha}).""$$

We can thus encode $S$ by $n,$ an enumeration of $S \cap R_n,$ and the constructed sequence of Borel codes for $\langle S \cap X_m: m \ge n \rangle,$ so $|\mathbb{R}| \ge^* \mathcal{P}(X).$ This completes the proof of the Claim.

Suppose $\aleph^*(X)>\omega_1.$ This would imply $\Theta=\aleph^*(\mathbb{R})=\aleph^*(\mathcal{P}(X)) > \omega_2.$ But $M$ is a symmetric submodel of $L^{\prod \text{Col}(\omega, \omega_n)},$ where $\mathbb{R}^M$ is countable and $\omega_2^M=\aleph^L_{\omega+1}=\omega_1.$ So there is no surjection from $\mathbb{R}^M$ to $\omega_2^M,$ contradiction. We conclude $\aleph(X)=\aleph^*(X)=\omega_1$ and $\Theta=\aleph^*(X^2)=\omega_2.$

The inequality $\aleph^*(X)<\aleph^*(X^2)$ is of independent interest, considering it's a ZF theorem that for any infinite $S,$ $\aleph^*(S^2) \le \aleph^*(S)^+,$ by a variation of the proof of Lemma here: https://math.stackexchange.com/a/3982905/210610.


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