Does Well-Ordered Interval Power Set "WOIPS" principle , prove $\sf AC$ in $\sf ZFA$?

Question

Does $\sf ZFA + WOIPS$ prove $\sf AC$?

Where $\sf WOIPS$ is phrased as: for every infinite set $X$ the set of intervening cardinals between $|X|$ and $|\mathcal P(X)|$ is well-ordered.

In $\sf ZF$, I think $\sf AC$ follows because we'll have a well ordered class of cardinals. But, I suspect this might not the case for $\sf ZFA$?!

I posed this question originally in terms of $\sf GCH$ itself, but this turned to prove $\sf AC$ (Specker ; pp: 419,420). What I had in my mind is to start with a set $A$ of urelements such that $|A|=\aleph_1$, then construct $M=L_\kappa(A) \models \sf ZF$ for some countable $\kappa$, this way $M$ would see $A$ incomparable to any of its pure sets, the reason is because there are no injections available from $A$ to any of the pure sets in $M$ because all of those are externally countable and $A$ is not, on the other hand due to $M$ satisfying Replacement, then Hartogs are definable in $M$ and so any external injection from $\aleph(A)$ to $A$ cannot be an element of $M$. So $M$ sees $A$ as non-well-orderable. If $\sf GCH$ holds, then we'll have $|A| < |A|+ \aleph(A) \leq 2^{|A|}$ leading to $\aleph(A)= 2^{|A|}$, and so well ordering $A$ which cannot be. So we must have $|A| < |A|+ \aleph(A) < 2^{|A|}$. But, given the construction of $M$ then per the above argument all intervening cardinals between $|X|$ and $|\mathcal P(X)|$ are of the form $|A| + \aleph_\alpha$ for $\aleph_\alpha \geq \aleph(A)$, and those are well-orderable! If $M$ sees multiple incomparable subsets of $A$, then the set of intervening cardinals may not be well-orderable. But is this inevitable? I mean can we have a model in which $M$ shuns existence of such incomparable subsets of $A$?

More generally, per specifications of $M$ given above:

Can we have $M \models \sf WOIPS$?

Answer 1

I’ll prove the following: over Z set theory, AC is equivalent to cardinal trichotomy holding for the sets between $X$ and $\mathcal{P}^3(X)$ for all infinite $X.$ I suspect the superscript 3 is unnecessary, but that would require much more careful engineering (and I suspect this result satisfies the underlying philosophical motivation of your question).

The forward immediate implication follows from ZC proving the well-ordering theorem (by Zermelo’s original proof).

For the reverse implication: let $X’$ be non-well-orderable. Then $X=\mathcal{P}^3(X’)$ is non-well-orderable and satisfies $|X|=|X|^2.$ Let $\kappa=\aleph(X)$ be the Hartogs number (constructed directly, not through ordinals). Then $Y=|X| \times \kappa$ and $Z=|X|+\kappa^+$ are cardinalities between $X$ and $\mathcal{P}^3(X)$ which we’ll show to be incomparable.

The inequality $\kappa^+ \not \le Y$ follows from productivity of Hartogs numbers.

Suppose there is an injection $f: Y \rightarrow Z.$ For each $x \in X,$ there is a least $\alpha$ such that $f(x, \alpha) \in \kappa^+.$ This injects $X$ into $\kappa^+,$ providing a well-ordering of $X,$ contradiction.

Answer 2

The following question was posed by Guozhen Shen in comments.

Does the statement that there is at most one cardinal between an infinite cardinal and its powerset implies the axiom of choice in $\sf ZF$?

However before reaching to this, one can solve simpler versions before it, for instance a related statement is the following strong form of $\sf GCH$:

$\forall \text { infinite cardinals } \kappa, \lambda \, ( \kappa < \lambda \to 2^\kappa \leq \lambda)$

Its important to notice that those cardinals are Scott's cardinals, so they are not necessarily well orderable (meaning their elements are well-orderable).

The proof that this implies $\sf AC$ over $\sf ZF$ is actually very easy. Simply take $\kappa + \aleph(\kappa)$, if $\kappa$ is not well-orderable, then $\kappa < \kappa+\aleph(\kappa)$ [By definition of Hartogs] , so $2^\kappa \leq \kappa + \aleph(\kappa)$ but since $\kappa < 2^\kappa$, then this entails $2^\kappa \leq \aleph(\kappa)$ (Specker: Lemma 2.3), thus $\kappa$ is well orderable. A contradiction, so $\kappa$ must be well-orderable.

We notice that this proof is much simpler than the weaker form of $\sf GCH$ stated by absence of intervening cardinals between cardinalities of infinite sets and their powersets.

Similarly we can have a strong form of Guozhen Shen question, that of:

$\forall \text { infinite cardinals } \kappa, \lambda, \zeta: \kappa \leq \lambda < \zeta \to 2^\kappa \leq \zeta$

Also the proof of $\sf AC$ from this in $\sf ZF$ is very simple, we start with the same proof of the strong $\sf GCH$, but add here the cardinal $\kappa + (\aleph(\kappa))^+$ or what I'd label as $\kappa + \aleph_1(\kappa)$, clearly we have $\kappa < \kappa + \aleph(\kappa) < \kappa + \aleph_1(\kappa)$, and so we must have $2^\kappa \leq \kappa + \aleph_1(\kappa)$, and $2^\kappa \leq \aleph_1(\kappa)$ follows by Lemma 2.3.

However, the weaker statements of $\sf GCH$ and the maximally single interval powerset phrased in terms of intervening cardinals between sets and their powers (i.e.; Shen's question), are harder to prove. The proof of the first goes like that: Lets take $X$ to be an infinite stage $V_\alpha$ of the cumulative hierarchy of $\sf ZF$. Let it be some limit stage. So, here we'll have $X \times X \subset X$, and so $|X \times X|= |X|$, let $\aleph(X)$ be defined in the original Hartog manner as the set of all equivalence classes under order isomorphism of well-orderings of subsets of $X$, this is a subset of $\mathcal P^2(X)$, so we have $\aleph(X) \leq |\mathcal P^2(X)|$. So, we assume $X$ to be non-well orderable, and we conclude that this leads to intervening cardinals between sets and their powersets, thus violating $\sf GCH$.

Now, we have $|\mathcal P(X)| + \aleph(X) \geq |\mathcal P(X)| $ (Definition of $\geq$),

• if $|\mathcal P(X)| + \aleph(X) > |\mathcal P(X)| $ then

$|\mathcal P(X)| + \aleph(X)$ would intervene between $|\mathcal P(X)|$ and $|\mathcal P^2(X)|$ (Dedekind infinite cardinal characteristics; Specker: Lemma 2.3);

• if $|\mathcal P(X)| + \aleph(X) = |\mathcal P(X)| $ then $\aleph(X) < |\mathcal P(X)|$ (otherwise $\mathcal P(X)$ is well-orderable). But, by then we'll have $ |X|+\aleph(X) $ intervening between $|X|$ and $|\mathcal P(X)|$(Definition of Hartogs; Dedekind infinite cardinal characteristics; Specker: Lemma 2.3).

The above argument would constitute the nucleus of further proofs here, I describe this by $\aleph(X)$ causing a waver across $\mathcal P(X)$! That is, it induces an intervening cardinal across $|\mathcal P(X)|$, i.e. between cardinalities of $\mathcal P(X)$ and either of $X$ or $\mathcal P^2(X)$. For a Hartog $\aleph(X)$ we name the least $\mathcal P^n(X)$ that its injective to as the ceiling, while its floor denotes $X$ itself where it is not injective to. Potentially a Hartog would induce a wavering state across all single power intervals below its ceiling and above its floor.

Now, before solving Shen's question we can solve a simpler one, that if $\sf ZF$ proves $\sf AC$ to follow from the statement that no more than two intervening cardinals between $X$ and $\mathcal P^2(X)$ can exist if $X$ is infinite.

We repeat the same scenario done above but also bring $\aleph(\mathcal P(X)); \aleph(\mathcal P^2(X))$ into the picture, these would cause intervening cardinals that waver across $\mathcal P^2(X)$ and across $\mathcal P^3(X)$, the result of these wavers is that we'll have three cardinalities intervening between some set and its second powerset.

To be noted is that we used the theorem of $\sf ZF$ stating that if $|X \times X|=|X|$ then $|\mathcal P(X) \times \mathcal P(X)|=|\mathcal P(X)|$. This way we prove by induction that actually all stages $V_\alpha$ of the cumulative hierarchy to be bijective to their Cartesian products.

The hard Guozhen Shen question, I think can be solved by bringing more Hartogs into the picture. I'll use the notation $\aleph_{n+1}(X)$ to denote $(\aleph_n(X))^+$. So, we bring $\aleph_1(X)$, and $\aleph_1(\mathcal P(X))$, now the first is injective to $\mathcal P^3(X)$, and the second to $\mathcal P^4(X)$ (because both $X$ and $\mathcal P(X)$ are bijective to their Cartesian products: see Glazer's comment). These would cause wavers across $\mathcal P^3(X); \mathcal P^2(X)$. So, we have five Hartogs $\{\aleph(X), \aleph(\mathcal P(X)), \aleph(\mathcal P^2(X)), \aleph_1(X), \aleph_1(\mathcal P(X))\}$ causing wavers across three levels $\{\mathcal P(X), \mathcal P^2(X), \mathcal P^3(X)\}$ and four single power intervals below $\mathcal P^4(X)$, so we must have two intervening cardinals in some single power interval.

I'm not sure if using this wavering scenario can solve the question I posed, but it might be of help?