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Suppose $\kappa$ is an infinite von Neumann cardinal (well ordered by $\in$), and take ${<}$ a well-order on $\kappa$. Does there necessarily exists a subset $X\subset\kappa$ of full size (in bijective correspondence with $\kappa$ as a set) on which $<\upharpoonright X^2=\in\upharpoonright X^2$ (i.e. ${<}$ and ${\in}$ agree on $X$)?

Attempt:

Taking $(\kappa, <)\approx_f(\alpha, \in)$, where $f$ is an order-isomorphism and $\alpha$ is an ordinal yields $\alpha\in\kappa$ or $\alpha=\kappa$ (since $\kappa$ is the least ordinal order-isomorphic to a well order on $\kappa$). I feel something like $X=f^{-1}(\alpha)$ should work, but I don't see how I'm using that $\kappa$ is infinite in such a proof, and the statement is clearly false for finite cardinals. I guess this problem could be put "the collection of fixed points of $f$ has the same cardinality as $\kappa$"?

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1 Answer 1

up vote 3 down vote accepted

Marcel, the standard argument for this problem uses choice. One argues first for regular cardinals, by an easy recursion, and then for singular cardinals, where some care is needed. A second proof uses partition calculus, the Erdős-Dushnik-Miller theorem. Both these proofs use somewhat more sophisticated machinery than one would expect.

One can prove the result without using choice; there is a metamathematical trick for this, but I asked in MathOverflow for a combinatorial proof. Clinton Conley found a very pretty argument, and you can read the details here.

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Thank you! It is also nice to know this appears as a problem in Kunen. I thought it might but didn't spot it skimming through. –  Marcel T. Nov 6 '12 at 2:49

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