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Dedekind Cuts

  1. May 26, 2005 #1
    The question:

    Show D= {x: x [tex]\in[/tex] Q and (x [tex]\leq[/tex] or x^2 < 2)} is a dedekind cut.

    A set D c Q is a Dedekind set if

    1)D is not {}, D is not Q
    2) if r[tex]\in[/tex] D then there exists a s [tex]\in[/tex] D s.t r<s
    3) if r [tex]\in[/tex] D and if s [tex]\leq[/tex] r, then s [tex]\in[/tex] D.

    For the first case, D is not an empty set because x is equal to 0 or the sqrt of 2. But, how do I prove case 2,3. Do I have to use addition/multiplication to prove them?
  2. jcsd
  3. May 26, 2005 #2
    I'm not sure I understand your definition of D

    is it equivilant to [tex] D=\{x\in Q | x\le 2\}\cup \{x\in Q | x^2 < 2\}[/tex]
    Which means [tex]D= \{x\in Q | x\le 2\}[/tex] which seems to contradict 2).

  4. May 26, 2005 #3
    Sorry, D is actually
    D= {x: x [tex]\in[/tex] Q and (x [tex]\leq[/tex] 0 or x^2 < 2)}
  5. May 26, 2005 #4


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    1. Obviously, D is not empty- any negative number is in D. Obviously D is not all rational numbers, 2> 0, 22= 4> 2 so 2 is not in D.

    3. if r is in D and s< r then either:
    a) r< 0 in which case r is in D or
    b) 0< r< s so 0< r2< s2< 2 so r is in D.

    2. is the hard one. Obviously if r< 0, we can take s= 0. Ir r> 0, then r2< 2. Take d= 2- s2. Can you show that 0< (r+ d/4)2> 2?
  6. May 27, 2005 #5
    0< (r+ d/4)2> 2?
    if d= 2- s2
    0< r + (2- s2)/4 > 2
    Do I let r = sqrt(2) both plus and minus
    to show that
    0< r > 2, so this will confirm the fact that r<s?
  7. May 28, 2005 #6


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    Sorry, there was a misprint! I mean 0< (r+d/4)2< 2. (not > 2)

    Suppose r is the largest number in the set.

    It is obvious that (3/2)2= 2.25> 2 so 3/2 is not in this set. It is obvious that 1.42= 1.96 so 1.4 is in this set. Any possible maximum for the set must be greater than or equal to 1.4 and less than 1.5= 3/2: 1.4<= r< 3/2 and so d= 2- r2 must be less than or equal to 2- 1.96= 0.04. (r+ d/4)2= r2+ rd/2+ d2/16 so 2- (r+d/4)2= (2- r2)- rd/2- d2/16. 2- r2= d and since r< 3/2, rd/2< (3/4)d. d/16= d(d/16) and since d< 0.04, d/16< (.04/16)= (.01/4)= 0.0025. That is: 2- (r+d/4)2> d- (3/4)d- 0.0025d= d- (0.7525)d> 0 which means (r+ d/4)2.
    r+ d/4 is larger than r but (r+ d/4)2< 2 so r+ d/4 is still in the set contradicting the hypothesis that r is the maximum for the set. Therefore, the set has no maximum.
  8. May 28, 2005 #7
    For the third proof, do I go onto assume that s is equal or greater than r to prove that it's a contradiction?
  9. May 29, 2005 #8


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    No, just prove exactly what's given: if s less than or equal to r, then it must be in the set. That's exactly what I did in my first response:

    Suppose r is in this set. There are two possibilities: r< 0 or r2< 2.
    (a) If r< 0 and s<= r, then s< 0 so s is in the set.

    (b) If 0<= r, r2< 2, and s< r then either
    (i) s< 0 so s is in the set
    (ii) s>= 0 so 0<= s2< r< 2 and s is in the set.
    In any case, if r is in this set and s< r, then s is in the set.
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