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Proof involving dedekind's cuts

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Homework Statement


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


Homework Equations


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

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 [itex]a=a[/itex], [itex]b=a+\frac{1}{n+1}[/itex] or [itex]a=b-\frac{1}{n+1}[/itex], [itex]b=b[/itex] 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

  • #2
274
1
I have yet again taken another look and now I am thinking, since we have for any [itex]a\in A[/itex] and [itex]b\in B[/itex] that we have a unique [itex]x\in R[/itex] such that [itex]a\leq x\leq b[/itex] 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.
 

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