Let $A$ and $B$ be non-empty subsets of $\mathbb{R}$ with $a\leq b$ for all $a\in A, \space b \in B$ and suppose there exists a unique number $\alpha$ such that $a\leq \alpha \leq b$ for all $a\in A, \space b \in B.$ Then we have $\alpha = \sup A = \inf B$.
Proof: Suppose $\alpha < \sup A.$ If $S = \sup A\in A$ we are done so assume this is not the case. Choose $k = \dfrac{\alpha + S}{2}$ so $\alpha < k$ and $k < S.$ Now we can choose some $A\ni a \geq k$ which yields a contradiction since $\alpha <a$. One can similarly show that $\alpha \leq \inf B.$ We conclude that $\sup A \leq \alpha \leq \inf B$ and since $\alpha$ is unique it follows that $\alpha = \sup A = \inf B$.
Does my proof seem sound? If not, what do I have to adjust?
Best Answer
Your argumentation is correct but I would add a bit more detail to the last sentence: Assume that $\sup A\neq \alpha \neq \inf B$ so we would obtain the inequality $\sup A < \alpha < \inf B$. Then we could choose a number
$t = \dfrac{ \alpha + \inf B}{2}$ (analogous to how you chose $k$ in your proof) so $\sup A < \alpha < t < \inf B$ which would contradict the uniqueness of $\alpha.$ The same applies for $\sup A$. This makes your last conlcusion more clear in my opinion.
In general, it's good to try keeping the proof concise whereby one should not leave out details that make it "hard" to follow.