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Does the axiom of choice allow a set well ordered both ways to have cardinality greater than countable?

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closed as unclear what you're asking by Andrés E. Caicedo, Eric Wofsey, Asaf Karagila Sep 9 '18 at 17:21

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    $\begingroup$ What's "both ways"? $\endgroup$ – Asaf Karagila Sep 9 '18 at 6:00
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    $\begingroup$ Do you mean an order $<$ such that $(S,<)$ and $(S,>)$ are both well orders? That is clearly impossible not only for uncountable sets, but even for infinite sets. Pick any well order on an infinite set. Then it contains the order of natural numbers as an initial segment, which is an infinite ascending chain. $\endgroup$ – A. Pongrácz Sep 9 '18 at 6:16
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    $\begingroup$ @A.Pongrácz: Yes, that's what I thought. But it could also be something akin to $\Bbb Z$, that can be split into a co-well-ordered part and a well-ordered part, in which case the axiom of choice is again irrelevant here. But, I guess we just have to wait for the OP to clarify. $\endgroup$ – Asaf Karagila Sep 9 '18 at 6:20