Prove/disprove: Suppose R is a total order on A and $B \subseteq A$. Every element of $B$ is either the smallest element or the largest element

Question

Suppose $R$ is a total order on $A$ and $B \subseteq A$. Then every element of $B$ is either the smallest element of $B$ or the largest element of $B$

I believe that the proposition is incorrect.

My attempt at disproving it:

If there are more than $2$ elements in $B$, then it would imply that

1. either there are more than $2$ largest elements

2. or there are more than $2$ smallest elements

But we know that if there is smallest or largest element, it must be unique. Hence proposition is incorrect.

Will this explanation suffice?

Answer 1

To disprove this, you must produce a counterexample. An abstract argument can provide intuition, but it cannot be a proof. So you can pick a totally ordered set, say $A=\mathbb Z$, and let $B=\{0,1,2\}$. There is an element that is not the smallest or the largest.

You could also be fancier and pick a dense total order with a subset that has no smallest or largest element.