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

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?

• "at most one" means $\le 1$ – Mauro ALLEGRANZA Nov 19 '16 at 16:12
• @MauroALLEGRANZA Whoops, that's embarrassing, – Perturbative Nov 19 '16 at 16:17
• What does "largest element" mean? What would it mean if there were two of them? – Arthur Nov 19 '16 at 16:28

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 $a' = a$.