Working with subsets, as opposed to elements. Especially in algebraic contexts, we can often work with subsets, as opposed to elements. For instance, in a ring we can define
$$A+B = \{a+b\mid a \in A, b \in B\},\quad -A = \{-a\mid a \in A\}$$
$$AB = \{ab\mid a \in A, b \in B\}$$
and under these definitions, singletons work exactly like elements. For instance, $\{a\}+\{b\} = \{c\}$ iff $a+b=c$.
Now suppose we're working in an ordered ring. What should $A \leq B$ mean? I can think of at least two possible definitions.

*

*For all $a \in A$ and $b \in B$ it holds that $a \leq b$.


*There exists $a \in A$ and $b \in B$ such that $a \leq b$.
Also, a third definition was suggested in the comments:


*For all $a \in A$ there exists $b \in B$ such that $a \leq b$.
Note that according to all three definitions, we have $\{a\} \leq \{b\}$ iff $a \leq b$. That's because "for all $x \in X$" and "there exists $x \in X$" mean the same thing whenever $X$ is a singleton set.
What's the natural thing to do here? (1), (2), or something else entirely?
Note that our earlier definitions leveraged existence. For example: $$A+B = \{x\mid \exists a \in A, b \in B : a+b=x\}.$$

Since we're talking about ordered rings, maybe the ordering could be applicable to each comparison, too. i.e., 
$$ a_n \le b_n \forall a \in A,b \in B$$
if you were to apply this to the sets of even integers (greater than 0) and odd integers it might look like 1<2, 3<4, etc. Of course, cardinality comes into play since you need n-elements to exist in both sets. Maybe the caveat could be $n=min([A],[B])$. 
