Infinite, totally ordered and well ordered sets

It's quite easy to show that a finite set is well ordered iff it is totally ordered. Is the converse also true? That is: is it true that a set is infinite iff it admits a total order which is not a well order? (For the sake of brevity, I shall write t.o. and w.o. for total order and well order, respectively)

The if part follows from the previous observation (t.o.+finiteness implies w.o.), so the question becomes

Is it true that a set is infinite only if it admits a total order which is not a well order?

I know that every set admits a w.o. (and thus a t.o.); still, answering the question requires to prove (or to disprove) that t.o.≠w.o., and this is out of my reach.

The question supposes ZFC but every other set theory is accepted.

It is easy to prove that $$A$$ is finite if and only if there is an order $$<$$ on $$A$$, such that $$(A,<)$$ and the reverse order $$(A,>)$$ are both well-orders.

Now, note that the reverse order of a total order is again a total order, and so if $$A$$ is infinite, there is a well-ordering on $$A$$, $$<$$, such that $$(A,>)$$ is not a well-order itself. (Of course, one can show that this is true for any well-ordering of an infinite set.)

Just a remark on the Axiom of Choice, the above holds for $$\sf ZF$$ as well, although it is consistent with $$\sf ZF$$ that there are sets which do not admit any total orders, let alone well-orders.

But a curious thing is that if we change "there exists an order ..." to "every well-order $$<$$ satisfies ...", then any set which cannot be well-ordered satisfies the definition vacuously. And therefore the Axiom of Choice is equivalent to the statement "a set $$A$$ is finite if and only if for every well-order on $$A$$, the reverse order is also a well-order".

• Please participate here – Bill Dubuque Oct 31 '19 at 16:29

Not my area of expertise but I will suggest an answer.

If a $$S$$ is infinite, well order it. An initial segment will be the natural numbers with their usual order. Then reorder that initial segment using a bijection to the rational numbers to transfer the not-well-ordered numerical order on the rationals.

If this is wrong someone will point that out. Then I can delete it, or leave it as an instructive false start.

• Looks correct to me. – rawbacon Oct 28 '19 at 13:07
• How can we be sure that rearranging the order gives as an order? – Caffeine Oct 28 '19 at 13:14
• @GabrieleCassese You create a new ordering on the target space effectively ordered by the base space. The ordering may not be the same as the typical ordering on $\Bbb Q$. – Cameron Williams Oct 28 '19 at 13:15
• You can even exchange the initial segment $\Bbb N$ to $\Bbb Z$. – Berci Oct 28 '19 at 13:21
• @CameronWilliams I'm sorry, but I do not understand: how can we be sure that in this new order $\Bbb Q$ is not well ordered? After all, we know there is a order in which it is (thanks to AC) – Caffeine Oct 28 '19 at 13:29