# How close to being well-orderable does this make my powerset?

Let's work in a set theory without assuming AC (for instance, but not necessarily, ZF). Fix a set $k$ satisfying $k\times k \simeq k$, and consider its powerset $X = 2^k$. I have a technical condition that is satisfied whenever $X$ is well-orderable, but I can't tell what other examples there might be. Excuse the peculiar phrasing: I'm aiming for the most constructive means possible (by "well-orderable" I mean classically so: a total order such that every inhabited subset has a least element. This is equivalent to $X$ being a choice object).

Let $p\colon X \to I$ be a partition of $X$ into subsets $X_i=p^{-1}(i)\lt X$, $i\in I$ such that the images of all functions $f_i\colon X_i \to X$ have inhabited complement $\forall i \in I$. Assume I is minimal in the $\lt$-ordering of sets by cardinality (if $X$ is well-orderable, then $I$ is unique and $I\simeq cf(X)$). The condition is as follows:

For all partitions as just described, and all families of functions $(f_i\colon X_i \to X\mid i \in I)$, the surjection $$\coprod_{i\in I} X_i \setminus f_i(X_i) \to I$$ has a section.

The consequence of this that I want is that I can choose a point in the complement of every map $X \to X^I$, namely $X < X^I$. For well-orderable $X$ this is a corollary of a restriction of König's theorem to powers of well-orderable sets in place of arbitrary products of families.

I would like to know if there are examples that aren't well-orderable, or otherwise how close to be able to well-order $X$ this gets. Note in particular, that we could take every function $f_i$ to be constant at a fixed point $x\in X$, and so this case implies that we should be able to take a section of $X\setminus\lbrace x\rbrace \to I$ and hence of $p$, for $p$ and $I$ as above.

Note that I don't quite have the result that $X$ is a choice object (equivalently, classically well-orderable). I'm not demanding the existence of sections for arbitrary surjections, or more generally for total relation into $X$ (a fine distinction, I'm sure).

Edit: $X$ being a choice object is equivalent to saying that there is a choice function from the set $P_+X$ of inhabited subsets to $X$. I am only asking for a choice function on the subset of $P_+X$ given by those subsets of $X$ of strictly smaller cardinality, call it $P^{\lt X}_+X$.

• I would also be happy assuming that $k$ is regular, in the sense that if given a partition $k to J$ with properties as above, and $J$ is minimal, then $k \simeq J$ – David Roberts Nov 27 '15 at 1:07
• Do you want this to be true for all such $p:X\to I$, or only for some such $p:X\to I$? – Eric Wofsey Nov 27 '15 at 1:13
• No, just the ones satisfying the conditions as given before the grey box. If it were for all $p$, then clearly I get a choice object. – David Roberts Nov 27 '15 at 1:14
• Are you also assuming $X_i < X$? – François G. Dorais Nov 27 '15 at 13:53
• @Joel, David: The blog post is up. I'd link it here, but there is a link to this post and I don't want to create a non-well founded chain of links. :-) – Asaf Karagila Nov 29 '15 at 21:08

This destroys the inequality $\kappa < 2^{cf(\kappa)}$ resulting from König's theorem (I like to call this inequality König's corollary), for the notion of cofinality in the original question.