Proof involving dedekind's cuts

[DeadOriginal]

Homework Statement

Show that for any non-empty sets A and B as in (P13') and for any n \in N,
there is a pair (a,b) with a\in A and b\in B such that |b-a|=b-a<\frac{1}{n}.

Homework Equations

Property (P13'). For any two non-empty sets A and B of rational numbers such that
Q = A\cup B, if all elements of A are less than all elements of B, then there is a unique
x\in R s.t. for any a\in A and any b\in B we have a\leq x\leq b.

This is called P13' because I am currently learning from Spivak and have finished chapter 1-5 where we get P1-P12. P13 is in a later chapter so I haven't seen it yet. P13' was given to me to help introduce me to Dedekind's cuts.

The Attempt at a Solution

I have considered the pairs like (a,b) where a=a, b=a+\frac{1}{n+1} or a=b-\frac{1}{n+1}, b=b but this isn't correct. My professor says there is a very easy proof but I have thought about this so much that my mind is clogged and I can no longer see clearly. Could someone please point me in the right direction?

Edit: Taking another look at the problem, I feel like maybe I could use epsilon delta to prove that the limit goes to 0 as a and b get arbitrarily close? Since |b-a|=b-a<1/n is talking about distance after all... hmm. Any comments on this?

 

[DeadOriginal]

I have yet again taken another look and now I am thinking, since we have for any a\in A and b\in B that we have a unique x\in R such that a\leq x\leq b then would it be possible for my to take say a certain x from one pair and subtract in from an x of another pair to show that it'll be less than 1/n? I'm not sure how I would make a proof from that.