Prove that $\sup{A} = \inf{B}$ if and only if for each $\delta > 0$ there exists $x \in A$ and $y \in B$ such that $x + \delta > y$:

Question

Let $A$ and $B$ be non-empty subsets of $\mathbb{R}$ such that for all $x \in A$ and $y \in B$ we have $x \leq y$. Prove that $\sup{A} = \inf{A}$ if and only if for each $\delta > 0$ there exists $x \in A$ and $y \in B$ such that $x + \delta > y$:

We are given that $\sup{A} \leq \inf{B}$. We are also given that $a = \sup{A}$ if and only if $a$ is an upper bound for $A$ and there exists $x \in A$ such that for each $\delta > 0$ , $x + \delta > a$.

I have a proof that I think works. First I prove the only if direction.

Suppose $\sup{A} = \inf{B}$. Since $x + \delta > \sup{A}$ for all $\delta > 0$ , $x + \delta > \inf{B}$. Since $\inf{B}$ is the greatest lower bound for $B$, $x + \delta$ can't be a lower bound for each $\delta > 0$. Thus there exists $y \in B$ such that $x + \delta > y$.

Now we prove the if direction.

Suppose there exists $x \in A$ and $y \in B$ such that $x + \delta > y$ for each $\delta >0$. Since $\inf{B} \leq y$ for each $y \in B$ we have $x + \delta > \inf{B}$. Since $\inf{B} \geq \sup{A}$ , $\inf{B}$ is an upper bound for $A$. This gives us our characterization of $\sup{A}$ stated above. Thus $\inf{B} = \sup{A}$. Any critique would be awesome. Thank you.

Answer 1

($\Rightarrow$) I think it would be better this way: Fix an arbitrary $\delta>0$, there exists $x\in A$ s.t. $x+\delta>\sup A=\inf B$, then $x+\delta$ cannot be a lower bound for $B$ and the rest is your conclusion.

($\Leftarrow$) Since $x$ and $y$ depends on $\delta$, you should write "Suppose that for any $\delta>0$, there exists $x\in A,y\in B$ s.t. $x+\delta>y$" then write your argument and finish with "...then $\inf B$ is an upper bound for $A$, moreover $\inf B\leqslant \sup A$ (by definition of $\sup$); thus $\inf A=\sup B$".

Answer 2

More succinctly:

($\Rightarrow$) $\forall\epsilon_1>0$, $\exists x\in A$ such that $x+\epsilon_1>\sup(A)$. $\forall\epsilon_2>0$, $\exists y\in B$ such that $y-\epsilon_2<\inf(B)=\sup(A)=L$. $\forall \delta>0$, $\exists \epsilon_1, \epsilon_2>0$ such that $\epsilon_1+\epsilon_2<\delta$. Clearly, $y-x=(y-L)+(L-x)<\epsilon_1+\epsilon_2<\delta $. (Although after $x+\epsilon_1>\sup(a)$ you can jump to the conclusion that $x+\epsilon_1>$ some $y$, if you like.)

($\Leftarrow$) We know that $\inf(B)\geq \sup(A)$. $\inf(B)>\sup(A)\Rightarrow \exists r>0$ such that $y-x>r$ $\forall x\in A, y\in B$. This immediately gives rise to a contradiction.