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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 $\aleph(X) \leq |\mathcal P^2(X)|$.

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

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 $\sf GCH$.

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

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

$|\mathcal P(X)| + \aleph(X)$ 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).

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

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

Also the proof of $\sf AC$ from this in $\sf ZF$ is very simple, we start wit 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.

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.

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

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.

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.

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

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.

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.

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.

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 using this wavering scenario can solve the question I posed, but it might be of help?