# Proof involving dedekind's cuts

## 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?

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## Answers and Replies

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.