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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

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    $\begingroup$ $(0,1)$ is a subset of $\mathbb R$ with no least element. $\mathbb R$ is not well-ordered. $\endgroup$ – Akiva Weinberger Apr 30 '15 at 11:14
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    $\begingroup$ @columbus8myhw I thought all sets were well-ordered? Or do you mean well-ordered under $\leq$? $\endgroup$ – k_g Apr 30 '15 at 23:26
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    $\begingroup$ @k_g The second thing, sorry. (Though, strictly speaking, you need Choice for your first sentence.) $\endgroup$ – Akiva Weinberger Apr 30 '15 at 23:40
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    $\begingroup$ @k_g: There is an important distinction between "well-ordered" and "well-orderable". The former is a statement about a given order (and its domain, of course), and the latter is about a set. So when you talk about a set which is equipped with a standard order, e.g. a set of reals, we often like to think about that order if no other order is mentioned. So saying that $(0,1)$ is well-ordered is either a misnomer (you only mention a set, without an order) or a mistake (regarding the standard order of the real numbers). Anyway, "well-ordered" vs. "well-orderable" is an important distinction. $\endgroup$ – Asaf Karagila May 1 '15 at 4:01
  • $\begingroup$ @AsafKaragila OK thanks. I remember reading that it's consistent with ZFC that there's no explicit way to well-order the real numbers, so I guess the distinction is huge in this case. $\endgroup$ – k_g May 1 '15 at 4:11
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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. $\Bbb N$.

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    $\begingroup$ I'd be happy to hear about that downvote in words as well. $\endgroup$ – Asaf Karagila Apr 30 '15 at 13:12
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$\mathbb Z$ is totally ordered by $\leq$ but not well-ordered since there is no least element.

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

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  • $\begingroup$ Isn't $\Bbb R$ the Radin forcing poset, which is certainly not linear? :-) $\endgroup$ – Asaf Karagila Dec 22 '18 at 13:48
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How about positive real numbers? Each two are comparable, but there is no the least one...

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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 $A:=\{a_n\mid n\in \mathbb N\}$?

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