Measurable total order

MathOverflow Asked by Aryeh Kontorovich on November 5, 2020

Under what conditions on a metric space $X$, equipped with the Borel $sigma$-algebra, does there exist a measurable total ordering of the elements of $X$?

By "measurable total ordering" we mean that any initial segment $I_y:={x: x<y}$ is Borel-measurable.

Edit: We know that separability is sufficient for a measurable total order to exist.

Edit II: Vladimir Pestov (private communication) has shown that a measurable total order always exists; will post answer soon with link to full paper.

One Answer

As mentioned in the OP, Vladimir Pestov has answered the question affirmatively. See Appendix D here:

Correct answer by Aryeh Kontorovich on November 5, 2020

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