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> But that set must have a member whose value is closest to zero This is not true for real numbers with the usual "less-than" relation. In fact, considering re
by johnnny 4y ago
> But that set must have a member whose value is closest to zero
This is not true for real numbers with the usual "less-than" relation. In fact, considering real numbers, no open subset of R admits a lower member.
According to Zorn's lemma, there exist some order relation where your assertion is true, but I seem to remember from my math course a few (!) years ago that Zorn's lemma is equivalent to the axiom of choice.
- WastingMyTime89 4y agoAccording to the well-ordering theorem which is equivalent to the axiom of choice (and Zorn's lemma), you can well-order any set. That's usually when people start doubting the axiom of choice because, well, that's quite unintuitive. There is a joke about it which has already been posted here.
- BeetleB 4y agoYou can well order any set, but you need to have the "right" ordering to do so. As your parent is pointing out, the usual ordering on the reals is not a well ordering, and one cannot achieve a well ordering of the reals using that ordering.
- ablob 4y agoThis doesn't work for rational numbers either. Take any rational a/b, then 0 < a/(2b) < a/b is both closer to zero, and it is also a rational number.