[johnnny]

> 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]

According 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]

You 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]

This 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.