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: $$A+B = \{x\mid \exists a \in A, b \in B : a+b=x\}.$$
 A: This question is entirely subjective.  "should" and "natural" mean different things to different people.  When we move toward generality, we gain additional texture and details that we didn't have before.  Ideal numbers (i.e. ideals) have properties that ordinary numbers (i.e. ring elements) do not.  The "right" definition is the one that allows us to prove something interesting.  Absent any context, all of the above are right, and all of the above are wrong.
(1) treats sets as intervals.  $A\le B$ if $\sup A \le \inf B$.  This leads to a partial order rather than a total order.  Perhaps this is what you want, since after all this context is more general.  For nonoverlapping sets, this seems right, but for overlapping sets it's fairly conservative.
(2) is a strange variation of (1).  $A\le B$ if either $\inf A < \sup B$  or $\min A = \max B$ (provided both exist).  For overlapping sets, this will mean that both $A\le B$ and $B\le A$ hold, so the relation fails antisymmetry, which is strange.
(3) treats sets as points, namely their suprema.  $A\le B$ if $\sup A \le \sup B$.  This has the advantage of being a total order, on those sets that are bounded above.
(4) Yiorgos' suggestion is to consider the internal structure of sets; namely to decompose sets as sumsets.  This looks at entirely different properties of sets.
(5) A commonly used order on sets, not previously mentioned, is inclusion.  That is, $A\le B$ if $A\subseteq B$.  This is a specialization of Yiorgos' order, since he can take $C=\{0\}$.
However, there are a bazillion other orderings we can impose.  For example:
(6) Choose a prime $p$ in your ring, and define $A\le B$ if $\nu_p(\prod A)\le \nu_p(\prod B)$, where $\nu_p$ denotes the $p$-adic valuation, extending the definition to infinite products in some way.
None of these are worth anything in isolation, and any of them might be the most important partial order in the world if they happen to solve the problem you're working on.
A: Suggestion:
$$
A\le B \quad\text{iff}\quad \text{there exists a set on nonnegative elements $C$, s.t.}\quad
B=A+C.
$$
A: 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 $n=min([A],[B])$. 
