# Question regarding well-ordering theorem

The well-ordering theorem states that every set can be well-ordered. That is, there is a total order on the set in which every non-empty subset has a least element. I was wondering whether it's also true that for every set you can define a total order such that for every non-empty subset there is a greatest element and would that be equivalent to the well ordering theorem?

Thanks!

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Is it also true there is necessarily an order that has both a least and greatest element? – Serpahimz Jan 12 '13 at 23:52
Curious how three different answers speak of "reversing" the order rather than "inverting" the order. The latter is in a sense technically correct, but the former seems to be the prevailing informal usage. – Michael Hardy Jan 13 '13 at 0:32
If every subset of a linearly ordered set has both a least element and a greatest element, then the set is finite. So the proposed theorem would be true only of finite sets. – Michael Hardy Jan 13 '13 at 0:33

## 3 Answers

Any claim about an order is immediately applicable on its reverse order by "reversing" the statement. Minimals are now maximals, and vice versa.

Any well-order can be reversed to obtain an order in which every non-empty set has a maximal element; and vice versa. So the axiom of choice is equivalent to this principle as well.

Equally the axiom of choice is equivalent to the dual Zorn's lemma:

If $(P,\leq)$ is a partially ordered set in which every chain is bounded from below, then there is a minimal element.

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Yes, of course. Just reverse the order given by the well-ordering theorem.

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Yes: such an order is simply the reverse of a well-order, so every set has the one if and only if it has the other: just replace $\le$ by $\ge$.

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Thanks Brian. The answer is so simple I feel quite foolish for not seeing it right away. I'm going to blame the fact it's quite late here to make myself feel a little better.. – Serpahimz Jan 12 '13 at 23:25
@Serpahimz: You’re welcome. (And we’ve all had that experience, I’m afraid.) – Brian M. Scott Jan 12 '13 at 23:28