Lifting an equivalence relation on elements to a relation on sets If every element of $A$ is equivalent (by an equivalence relation $R$) to some element of $B$

$\forall x\in A\ldotp \exists y \in B\ldotp x\,R\,y$

and vice versa

$\forall y\in B\ldotp \exists x \in A\ldotp x\,R\,y$

then what can I deduce about the relationship between $A$ and $B$? I certainly can't say that they're equal, but they would seem to be equivalent in some sense. Perhaps one could say something to do with bijections here? Or perhaps equivalence classes?

My motivation for asking is that I have a theorem of the form outlined above, and I would like to find a simpler way to state it. Originally my $R$ was just straightforward equality ($=$), and then I could just deduce $A = B$. Now my $R$ is no longer $=$, but I'd still like to keep the statement of my theorem nice and succinct if I can.
 A: From your question it seems that the relation $R$ is defined in $A\cup B$. 
You can therefore say that from your assumptions, as sets, we have $A/R=B/R=(A\cup B)/R$
Indeed, any equivalence class in $(A\cup B)/R$ contains both an element of $A$ and an element of $B$. 
So the natural inclusions
$A/R\hookrightarrow (A\cup B)/R$ and $B/R\hookrightarrow (A\cup B)/R$ are onto.
A: This' neat :)
From the observation,
$
\;\;\; A \subseteq B
\\\equiv \forall a \in A. a \in B
\\\equiv \forall a \in A. \exists b \in B. b = a
$
We can generalize this preorder from any equivalence relation $\sim$ by defining
$$A \subseteq_\sim B :\equiv (\forall a \in A. \exists b \in B. a \sim b)$$
(Now this is indeed a preorder ---ie reflexive and transitive ... I couldn't prove antisymmetry.)
Anyhow, from this we can extend $\sim$ to sets by
$$A \approx B :\equiv A \subseteq_\sim B \land B \subseteq_\sim A$$
Now this is indeed an equivalence relation and it is induced by the given $\sim$. We may call this set equivalence by $\sim$-extensionality, for example.
I had fun doing this; hope it helps!

Edit: it'd be nice to relate $\approx$ with $\sim$; we can show
$$A \approx B \implies A/\sim \,= B/\sim$$
Indeed,
$\;\;\;X \subseteq_\sim Y
\\\equiv \forall x \in X. \exists y \in Y. x \sim y
\\\equiv \forall x \in X. \exists y \in Y. x/\sim \,=\, y /\sim
\\\equiv \forall x \in X. \exists y \in Y. x/\sim \,=\, y /\sim \,\in Y/\sim
\\\Rightarrow \forall x \in X. x/\sim \,\in Y/\sim
\\\equiv X/\sim \;\subseteq\, Y/\sim
$
and thus
$\;\;\; A \approx B
\\\equiv A \subseteq_\sim B \subseteq_\sim A
\\\Rightarrow A/\sim \subseteq B/\sim \subseteq A/\sim
\\\equiv A/\sim \,=\, B/\sim 
$
It'd be nice if the converse also held; we'll leave that as an exercise to the diligent reader/poster ;)
A: If $R$ is an equivalence relation on $X$ and $A,B\subseteq X$, then your statement is equivalent to: The partition of $X$ into equivalence classes is such that each equivalence class intersecting $A\cup B$ intersects both $A$ and $B$. I do not think much more can be said.
