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It is a folklore fact that within $\text{ZF}$ the generalized continuum hypothesis ($\text{GCH}$) implies the axiom of choice ($\text{AC}$), namely:

$$ZF+\forall \kappa\in Card~~~~~2^{\kappa}=\kappa^+\vdash AC$$

But note that $\text{GCH}$ just describes one of the most special cases of vast range of possible consistent behaviors of the continuum function $\kappa\mapsto 2^\kappa$.

Question: Which consistent behaviors of continuum function do imply axiom of choice? Precisely, for which non-trivial class functions $F:Card\longrightarrow Card$ do we have the following conditions:

(a) $Con(ZF)\Longrightarrow Con(ZF+ \forall \kappa\in Card~~~~~2^{\kappa}=F(\kappa))$

(b) $ZF+\forall \kappa\in Card~~~~~2^{\kappa}=F(\kappa)\vdash AC$

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If the power set of an ordinal is equipotent with an ordinal, then the axiom of choice holds.

Since it seems that you are using $Card$ to denote the $\aleph$-cardinals, this means that every such function $F$ would imply the axiom of choice.

This is an immediate consequence from Herman Rubin's theorem:

($\sf ZF$) If the power set of a well-ordered set is well-orderable, then the axiom of choice holds.

If we are talking about arbitrary cardinals, then you can prove the axiom of choice is implied (but not equivalent) to the following theorems:

  1. If $|A|<|B|\leq|\mathcal P(A)|$, then $|B|=|\mathcal P(A)|$. This is the equivalent of $\sf GCH$ when talking about arbitrary cardinals.

  2. If there exists an ordinal $\beta$, such that there is no set $X$ for which $\beta$ can be embedded (in the sense of partial orders) into the cardinal between $X$ and $\mathcal P(X)$. (Interestingly enough, you can replace $\mathcal P(X)$ by $X^2$ and obtain the same result.)

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Well this is not a solution to your question. But AC is a consequence of the statement that $\mathfrak{p}^2=\mathfrak{p}$ for all cardinal numbers $\mathfrak{p}$. Also one must be careful how one formulates the generalised continuum hypothesis, the formulation that implies choice is $$\mathfrak{p} \leq \mathfrak{q} \leq 2^{\mathfrak{p} }$$ implies $$\mathfrak{p} = \mathfrak{q} \ \ \ \ \text{or} \ \ \mathfrak{q} =2^{\mathfrak{p} }$$ Remember that without choice not every set is equipollent with a well ordered cardinal number.

For example (here is just a suggestion for a new hypothesis, it needs to be investigated) $$\mathfrak{p} \leq \mathfrak{q}_1 \leq 2^{\mathfrak{p} }$$ and $$\mathfrak{p} \leq \mathfrak{q}_2 \leq 2^{\mathfrak{p} }$$ implies $$\exists i (\mathfrak{p} = \mathfrak{q}_i \ \ \ \ \text{or} \ \ \mathfrak{q}_i =2^{\mathfrak{p} }) \ \ \text{or} \ \ \ \mathfrak{q}_1=\mathfrak{q}_2.$$

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  • $\begingroup$ "But AC is a consequence of the statement that $\frak p^2=p$ for all cardinal numbers $\frak p$." It is? $\endgroup$ – goblin May 31 '14 at 6:30
  • $\begingroup$ @user18921: For infinite $\frak p$, and of course this includes non-$\aleph$ cardinals. $\endgroup$ – Asaf Karagila May 31 '14 at 6:32

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