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Show that a subset of an ordered set has at most one smallest element and at most one largest element

This was one of the problems in Topology: A First Course by Munkres. But I don't see why a subset of an ordered set even has to have a smallest element or a largest element.

If we take $(0, 1) \subset \mathbb{R}$, we can see that $\sup (0,1) = 1 \ \ $and that $ \ \inf(0,1) = 0$, but $(0,1)$ doesn't contain a smallest element nor does it contain a largest element, so how does this question make sense?

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  • $\begingroup$ "at most one" means $\le 1$ $\endgroup$ – Mauro ALLEGRANZA Nov 19 '16 at 16:12
  • $\begingroup$ @MauroALLEGRANZA Whoops, that's embarrassing, $\endgroup$ – Perturbative Nov 19 '16 at 16:17
  • $\begingroup$ What does "largest element" mean? What would it mean if there were two of them? $\endgroup$ – Arthur Nov 19 '16 at 16:28
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You should be careful when you say only "ordered set". It could mean partially ordered, totally ordered or well-ordered set. However it really doesn't matter in this case.

But as pointed out in the comments at most one means it might not have the largest or least element. However you might be interested in Well-ordering theorem.

But to prove the claim consider the case that there are two (or more) largest elements $a$ and $a'$ of a set $A$. Then for every element $x$ of $A$: $$ x \leq a \text{ and } x \leq a'. $$ So $a' \leq a$ and $a \leq a'$. Now by antisymmetry $a' = a$.

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