# An infinite cardinal agrees with all its well-orders on sets of full size.

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

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