Prove that $\sup A = \sup B.$

Question

Consider two sets $A, B ⊆ R$ such that $A$ is not a subset of $B$ and $B$ is not a subset of $A.$ If $\sup(A\setminus B)∈B$ and $\sup(B \setminus A) ∈ A$, prove that $\sup A = \sup B.$

My attempt $(A \setminus B)$ and $(B \setminus A)$ are non empty ($\because$ $A$ is not a subset of $B$ and $B$ is not a subset of $A.$)

we know that $A\setminus B \subset A$

so, $\sup(A\setminus B) \leq \sup(A)$

we know that $B\setminus A \subset B$

so, $\sup(B\setminus A) \leq sup(B)$

Also $\sup(A\setminus B)∈B \implies \sup(A\setminus B)\leq \sup B$

$\sup(B\setminus A)∈A \implies \sup(B\setminus A)\leq \sup A$

How do I complete the proof?

Answer 1

It’s easier to prove the contrapositive (or to prove it by contradiction).

If $\sup A\ne\sup B$, we may suppose without loss of generality that $\sup A<\sup B$. It follows immediately that there is some $b\in B$ such that $\sup A<b$, and clearly we must have $b\in B\setminus A$. But then $\sup(B\setminus A)\ge b>\sup A$, so $\sup(B\setminus A)\notin A$.