I am trying to prove the proposition:
$\sup A <\sup B$ implies there is an upper bound of $A$ in $B$ ($A ,B \subseteq \mathbb{R}$).
Is my proof sensible?
If there was no $b \in B$ with $\sup{A} < b$:
$ \Rightarrow \sup A \geq b$ for all $b \in B$
$ \Rightarrow \sup A$ is an upper bound of B
as $\sup B \leq $ all upper bounds of $B$ this means we have a contradiction as $\sup A <\sup B$ by the proposition.
Can we conclude from this there must be a $b \in B$ such that $\sup{A} < b$. Making it an upper bound of $A$?
Best Answer
Your proof is quite correct though I would write it a bit differently:
Let $A,B$ be non-empty sets with $\sup(A) < \sup(B)$.
Suppose for a contradiction that $\forall b \in B: \lnot(\sup(A) < b)$
Then $\forall b \in B: b \le \sup(A)$, as we have a linear order. Which makes $\sup(A)$ an upperbound for $B$ and by minimality of $\sup(B)$ among all upperbounds of $B$, we get $\sup(B) \le \sup(A) (< \sup(B))$, a contradiction.
So $\exists b \in B: \sup(A) < b$ and as all $a \in A$ obey $a \le \sup(A)$, this newly found $b$ is indeed an upperbound for $A$.
Note also that the reverse almost but not quite holds: if $B$ contains a strict upperbound for $A$, so $\exists b \in B :\forall a \in A: a < b$ , then $\sup(A) \le b \le \sup(B)$, and we can do no better (not $\sup(A)< \sup(B)$) as $A = (0,1), B= (0,1]$ shows.