6
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ 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

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.