Different formulation of the generalized continuum hypothesis in $\mathsf{ZF}$

Question

Write $A\le^* B$ if there is a surjection from $B$ onto $A$, or if $A$ is empty. The concept "$B$ is larger than $A$" could be formulated in four inequivalent ways, namely (thanks to @Alex Kruckman for correction)

$(1)$ $A\le B$ (there is an injection $A\to B$);

$(2)$ $B\not\le^* A\vee A=B$ (if there is a surjection $A\to B$ or $B$ is empty, then there is a bijection $A\to B$);

$(3)$ $A\le^* B$ ($A$ is empty, or there is a surjection $B\to A$);

$(4)$ $B\not\le A\vee A=B$ (if there is an injection $B\to A$, then there is a bijection $A\to B$).

Clearly we have $(1)\Longrightarrow (3)$, $(1)\Longrightarrow (4)$, and $(2)\Longrightarrow (4)$ in $\mathsf{ZF}$, but not any other implications.

Now, to express the idea "for every infinite cardinality of sets $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, either $\mathfrak{a}$ is larger than $\mathfrak{b}$, either $\mathfrak{b}$ is larger than $2^\mathfrak{a}$" in $\mathsf{GCH}$, we can make a combination of two statements listed above, one for $\mathfrak{b}$ and $\mathfrak{a}$, and one for $2^\mathfrak{a}$ and $\mathfrak{b}$. For example, the combination $(1)(1)$ reads

$(1)(1)$ For every infinite cardinality $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, we have $\mathfrak{b}\le\mathfrak{a}\vee2^\mathfrak{a}\le\mathfrak{b}$;

the combination $(1)(4)$ is

$(1)(4)$ For every infinite cardinality $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, we have $\mathfrak{b}\le\mathfrak{a}\vee\mathfrak{b}\not\le2^\mathfrak{a}\vee2^\mathfrak{a}=\mathfrak{b}$,

or

$(1)(4)$ For every infinite cardinality $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, we have $\mathfrak{b}\le2^\mathfrak{a}\Longrightarrow\mathfrak{b}\le\mathfrak{a}\vee\mathfrak{b}=2^\mathfrak{a}$,

and the combination $(4)(4)$ reads

$(4)(4)$ For every infinite cardinality $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, we have $\mathfrak{a}\not\le\mathfrak{b}\vee\mathfrak{b}=\mathfrak{a}\vee\mathfrak{b}\not\le2^\mathfrak{a}\vee2^\mathfrak{a}=\mathfrak{b}$,

which can be reformulated as

$(4)(4)$ For every infinite cardinality $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, we have $\mathfrak{a}\le\mathfrak{b}\le2^\mathfrak{a}\Longrightarrow\mathfrak{b}=\mathfrak{a}\vee\mathfrak{b}=2^\mathfrak{a}$.

It appears that $\mathsf{GCH}$ is usually stated either as $(1)(4)$ or as $(4)(4)$.

If I got it correctly, the combination $(4)(4)$ is enough to deduce $\mathsf{AC}$ in $\mathsf{ZF}$, hence every other combination according to Wikipedia. So I was wondering about the other weakest combinations $(3)(3)$, $(3)(4)$ and $(4)(3)$: are they also enough to deduce $\mathsf{AC}$?

$(3)(3)$ For every infinite cardinality $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, we have $\mathfrak{b}\le^*\mathfrak{a}\vee2^\mathfrak{a}\le^*\mathfrak{b}$;

$(3)(4)$ For every infinite cardinality $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, we have $\mathfrak{b}\le2^\mathfrak{a}\Longrightarrow\mathfrak{b}\le^*\mathfrak{a}\vee\mathfrak{b}=2^\mathfrak{a}$;

$(3)(4)$ For every infinite cardinality $\mathfrak{a}$ and every cardinality $\mathfrak{b}$, we have $\mathfrak{a}\le\mathfrak{b}\Longrightarrow2^\mathfrak{a}\le^*\mathfrak{b}\vee\mathfrak{b}=\mathfrak{a}$.

Perhaps related: How to formulate continuum hypothesis without the axiom of choice?

Answer 1

To prove $AC$ it suffices to show that for every $\alpha\in Ord$ $\mathcal{P}(\aleph_\alpha)=2^{\aleph_\alpha}$ is well ordered.

$(3)(3)$ We have that since $\aleph_{\alpha+1}\nleq^* \aleph_\alpha$ that $2^{\aleph_\alpha}\leq^* \aleph_{\alpha+1}$ must hold. We also have that either $2^{\aleph_\alpha}\leq^* \aleph(2^{\aleph_\alpha})$, and in such case $2^{\aleph_\alpha}$ is well ordered and we are done, or $2^{\aleph(2^{\aleph_\alpha})}\leq^* 2^{\aleph_\alpha}$. Since $\aleph_\alpha<2^{\aleph_\alpha}$ we have that $\aleph_{\alpha+1}\leq^*\aleph(2^{\aleph_\alpha})$ and so $2^{\aleph_{\alpha+1}}\leq^* 2^{\aleph(2^{\aleph_\alpha})}$. Putting the inequalities together we get: \begin{equation*} 2^{\aleph_{\alpha+1}}\leq^* 2^{\aleph(2^{\aleph_\alpha})}\leq^* 2^{\aleph_\alpha}\leq^*\aleph_{\alpha+1} \end{equation*} But this is absurd by Cantor's theorem. So we have that $2^{\aleph_\alpha}$ must be well ordered.

$(3)(4)$ We know that $2^{\aleph_\alpha}\leq 2^{\aleph_{\alpha+1}}$ and so we get $2^{\aleph_\alpha}\leq^* \aleph_{\alpha+1}\vee 2^{\aleph_\alpha}=2^{\aleph_{\alpha+1}}$. If $2^{\aleph_\alpha}\leq^* \aleph_{\alpha+1}$ then $2^{\aleph_\alpha}$ is well ordered and we are done. Otherwise if $2^{\aleph_\alpha}=2^{\aleph_{\alpha+1}}$ then $\aleph_{\alpha+1}<2^{\aleph_\alpha}$. So we get that either $\aleph_{\alpha+1}\leq^*\aleph_\alpha \vee \aleph_{\alpha+1}=2^{\aleph_\alpha}$. Clearly $\aleph_{\alpha+1}\leq^*\aleph_\alpha$ is false, so we have $\aleph_{\alpha+1}=2^{\aleph_\alpha}$.

$(4)(3)$ Since $\aleph_\alpha<\aleph_{\alpha+1}$ you get $2^{\aleph_\alpha}\leq^*\aleph_{\alpha+1}\vee \aleph_\alpha=\aleph_{\alpha+1}$. But $\aleph_\alpha\neq\aleph_{\alpha+1}$ so $2^{\aleph_\alpha}\leq^*\aleph_{\alpha+1}$ and we are done.