I've substantially rewritten this question, since the original was rather badly written.

Based on theorems about finite sets and/or finite cardinal numbers, we can make conjectures about infinite sets and/or infinite cardinal numbers. Some of these will be correct (e.g. "no infinite set is equinumerous with its powerset") and some of these will be incorrect (e.g. "no infinite set is equinumerous with one of its proper subsets," which is completely wrong!). In fact, some of them will be independent of ZFC (think: $\kappa < \nu \Rightarrow 2^\kappa < 2^\nu$.)

My question is, has anyone ever compiled a "wish-list" of sentences that are independent of ZFC, but which are intuitively plausible on the basis of theorems and/or intuitions about finite sets and/or finite cardinal numbers? And if so, has anyone ever suggested an axiom intended to prove as much of the wish-list as possible?

I'm especially interested in sentences and axioms of this sort that disprove the continuum hypothesis.

For instance, for finite cardinal numbers $\kappa$ and $\nu$ such that $\kappa < \nu$, the "gap" between $\kappa$ and $2^\kappa$ is smaller than the gap between $\nu$ and $2^\nu$. We can formalize this as follows. Write $[\kappa,\kappa')$ for the set of all cardinal numbers between $\kappa$ and $\kappa'$, including $\kappa$ but not including $\kappa'$. Then it is a theorem of ZFC that for all finite cardinal numbers $\kappa$ and $\nu$ satisfying $\kappa < \nu,$ we have that $\mathrm{ord}[\kappa,2^\kappa) < \mathrm{ord}[\nu,2^\nu),$ where $\mathrm{ord}$ denotes order-type. Dropping the requirement that $\kappa$ and $\nu$ be finite, we obtain a sentence that is independent of ZFC, which could therefore be put on the wish-list, and which disproves the continuum hypothesis.

  • $\begingroup$ @Brian: Well, I suppose you have to read the last couple of questions about $\sf GCH$ to have the context for that. $\endgroup$ – Asaf Karagila May 1 '13 at 12:09
  • $\begingroup$ @Asaf: Ah, now I see. I answered the How badly can GCH fail question on a Kindle while lying in bed; the identity of the OP didn’t register. $\endgroup$ – Brian M. Scott May 1 '13 at 12:11
  • $\begingroup$ At Brian & Asaf, thank you for your answers on that other thread. $\endgroup$ – goblin May 1 '13 at 12:15
  • $\begingroup$ This MathOverflow question may interest you. $\endgroup$ – Asaf Karagila May 2 '13 at 14:18
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    $\begingroup$ Not exactly what you are looking for, but: (Sh:666) arxiv.org/abs/math/9906113. Saharon Shelah. On what I do not understand (and have something to say): Part I. Fundamenta Mathematicae, 166, (2000), 1-82. $\endgroup$ – Andrés E. Caicedo May 3 '13 at 6:33

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