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Rudin Errata ? Appendix 1

  1. Sep 8, 2011 #1
    In Rudin PoMA , chapter 1
    in appendix 1 construction of real numbers as Dedekind Cut's is given.

    I feel there is an error in the definition of additive inverse of a 'cut' .

    given a cut [itex]\alpha[/itex] in rational numbers , its additive inverse is given by [itex]\beta[/itex] .

    a rational p belongs to [itex]\beta[/itex] , if there exists a rational number r>0 such that -p-r[itex]\notin[/itex] [itex]\alpha[/itex] .

    the additive identity 0* is the set of all negative rational numbers.

    No problem till this point.

    Then we are supposed to prove that [itex]\alpha[/itex] + [itex]\beta[/itex] = 0.

    For this ,part 1 of the proof is that any element of [itex]\alpha[/itex] + [itex]\beta[/itex] should be a negative rational.

    Here's Rudin's proof
    How does -s [itex]\notin[/itex] [itex]\alpha[/itex] follow from s [itex]\in[/itex] [itex]\beta[/itex] ? I feel this is printing error.(do you agree on this?)

    Anyway, the proof can be slightly changed to make it correct :-
    If r [itex]\in[/itex] [itex]\alpha[/itex] and s [itex]\in[/itex] [itex]\beta[/itex] ,
    then there is a rational number t>0 , such that
    - s - t [itex]\notin[/itex] [itex]\alpha[/itex],
    hence r < -s - t ,
    r + s < - t < 0.

    Am I right ?

    I ordered the book's 3rd ed in India, and I am discovering that the book has many errors . :cry:
    I checked this errata - http://math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_notes.pdf , but couldn't find the error I pointed out.

    *note :- if this isn't posted in right sub-forum,Please move this to appropriate forum, in which one is supposed to discuss errata*
  2. jcsd
  3. Sep 8, 2011 #2
    Rudin is correct. There exist a rational t>0 such that -s-t is not in [itex]\alpha[/itex]. But


    So -s is also not in [itex]\alpha[/itex].
  4. Sep 8, 2011 #3
    Oops. Thanks.

    I got confused between [itex]\notin[/itex] and [itex]\in[/itex] .


    Doubt resolved. Please lock the thread if the need be.
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