The Axiom of Choice constrains every set of cardinal numbers which is linearly ordered by size to be well-ordered. By contrast, does ZF-without the Axiom of Choice (but with the Axiom of Foundation)-allow complete liberty in the ordering of cardinal numbers by size? More precisely: Is the following statement consistent with ZF? "If S is any linearly ordered set, there exists a set of cardinal numbers which is linearly ordered according to size and which is ordinally similar to S". Even without the Axiom of Foundation one can define a partial ordering among the sets of ZF. The set A precedes the set B if there exists an injective mapping of A into B. The set A strictly precedes the set B if there exists an injective mapping of A into B but no mapping of A onto B. One could then ask the same question about the linearly ordered subsets of this partially ordered class of sets.

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    $\begingroup$ I actually intend to address this (or rather "partial order" instead of "linear order") in my Ph.D., and I have the general layout prepared. Some abstract technical difficulties are currently holding me back, but I hope to overcome them at some point in the foreseeable future. The tentative answer, is yes, this is consistent. $\endgroup$ – Asaf Karagila Aug 21 '14 at 19:05
  • $\begingroup$ Oh @AsafKaragila, you're too modest. Your paper in Fund. Math. already has some results in this direction... $\endgroup$ – David Roberts Aug 21 '14 at 23:43
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    $\begingroup$ @David: But not more than what was known nearly 50 years ago. Yes, every partial order can be embedded into the cardinals of an extension. Here we reverse the quantifiers. $\endgroup$ – Asaf Karagila Aug 22 '14 at 1:54
  • $\begingroup$ Many thanks, Asaf, for your response. I am amazed because I was sure the answer would be "No". $\endgroup$ – Garabed Gulbenkian Aug 22 '14 at 20:30
  • $\begingroup$ @Asaf ah, of course, my mistake $\endgroup$ – David Roberts Aug 23 '14 at 22:02

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