Isn't every totally ordered set well-ordered?

Question

A well order is a total order on a set $S$ with the property that every non-empty subset of $S$ has a least element. But surely it follows from the definition of a total order that any non-empty subset will always have a least element because they are all comparable? I don't see how this is an additional property

Answer 1

You are thinking about finite subsets. And it is true, given a finite subset of a linearly ordered set, it is has a minimal (and maximal) element.

But what about infinite subsets? What about $\Bbb Q$, for example, as a subset of $\Bbb R$ or as a subset of itself?

And even more so, your argument if you look closely, should work for maximal. Every two elements are comparable, then there is a maximal element to every non-empty subset. But surely you can find linear orders without a maximal element, even well-orders without a maximal element, e.g. $\Bbb N$.

Answer 2

$\mathbb Z$ is totally ordered by $\leq$ but not well-ordered since there is no least element.

Answer 3

How about your favorite and most familiar totally ordered set, $\mathbb{R}$?

Answer 4

How about positive real numbers? Each two are comparable, but there is no the least one...

Answer 5

Take some total order $(X,<)$ and some strictly decreasing sequence $(a_n)_{n\in\mathbb N}\subseteq X$. Now tell me, what is the minimal element of $A:=\{a_n\mid n\in \mathbb N\}$?