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A question occurred to me as I was considering an earlier question of mine, and wasn't closely enough related for me to feel I could to include it.

Let's say that a partial order $\langle S,\preceq\rangle$ is

  • well-chained if each chain of $\langle S,\preceq\rangle$ is well-ordered by $\prec,$ and
  • well-founded if every non-empty subset of $S$ has a $\preceq$-minimal element (though not necessarily a $\preceq$-least element, of course).

Clearly, well-founded partial orders are well-chained. Also, well-chained sets are certainly well-founded in $\mathsf{ZFC},$ but it isn't clear to me that this must be the case in the absence of Choice. Is it known whether/how much Choice is needed to prove that well-chained partial orders are well-founded?

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This is exactly Dependent Choice.

First note that $\sf DC$ is equivalent to well-foundedness being equivalent with no decreasing sequences. Then note that no decreasing sequences is almost the same as well-chained.

Why just almost? Because it is slightly weaker without assuming $\sf DC$. But here it doesn't matter, exactly because assuming $\sf DC$ fails gives us a tree without maximal elements and without infinite branches. Reversing the tree order gives the counterexample needed (since every chain is finite).

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  • $\begingroup$ Excellent! I knew $\mathsf{DC}$ was sufficient for the equivalence of well-foundedness with no decreasing sequences, and I wondered a while ago if it was necessary. Two questions with one answer! $\endgroup$ – Cameron Buie Sep 13 '15 at 16:51

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