- Let S be an ordered set. Given any E as a strict subset of S show that sup E is unique. That is if
a = sup E and b = sup E
show that a = b.
Thank you!
a = sup E and b = sup E
show that a = b.
Thank you!
Here is a proof for partial orders.
By the definition of supremem we know that $e\le a$ for every $e\in E$, and we also know that if $u\in S$ is any element such that $e\le u$ for every $e\in E$, then $a\le u$. In words, $a$ is an upper bound for $E$, and if $u$ is any upper bound for $E$, then $a\le u$. In particular, since $b$ is an upper bound for $E$, we know that $a\le b$. But the entire argument works just as well if you interchange $a$ and $b$, so it must also be the case that $b\le a$. What property of partial orders then allows you to conclude that $a=b$?
If $\langle S,\le\rangle$ is a linear order, the argument can be simplified a little, because then we know right from the beginning that if $a\ne b$, then either $a or $b. Without loss of generality assume that $a. Then $b$ is an upper bound for $E$ that is strictly smaller than $b$, contradicting the hypothesis that $b=\sup E$.