# A surjection from square onto power: Is limit Hartogs/Lindenbaum number necessary?

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?

References

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

• 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 $\aleph_1$. Jun 25, 2023 at 18:33

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.