This is Velleman's exercise 4.4.18.a:
Suppose $R$ is a partial order on $A$, $B_1 ⊆ A$, $B_2 ⊆ A$, $∀x ∈ B_1∃y ∈ B_2(x Ry)$, and $∀x ∈ B_2∃y ∈ B_1(x Ry)$. Prove that for all $x ∈ A$, $x$ is an upper bound of $B_1$ iff $x$ is an upper bound of $B_2$.
Here's my proof of it:
Proof. Let $a$ be an arbitrary element of $A$.
($\rightarrow$) Suppose $∀x ∈ B_1(xRa)$ and let $x$ to be an arbitrary element of $B_2$. Since $x$ is an arbitrary element of $B_2$ then from $∀x ∈ B_2∃y ∈ B_1(x Ry)$ we get $∃y ∈ B_1(x Ry)$. Now we choose some $x_0$ such that $∃x_0 ∈ B_1(xRx_0)$. Now from $∃x_0 ∈ B_1$ and $∀x ∈ B_1(xRa)$ we get $x_0Ra$. Since $x$ was arbitrary then we'll have $xRa$. Therefore if $∀x ∈ B_1(xRa)$ then $∀x ∈ B_2(xRa)$.
($\leftarrow$) Similar to the proof of the forward direction.
From ($\rightarrow$) and ($\leftarrow$) we get $∀x ∈ B_1(xRa) \iff∀x ∈ B_2(xRa)$.
Is my proof correct?
Thanks a lot.