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Question Re. Simple epsilon proofs

  1. Aug 30, 2009 #1
    Here's an example that helps illustrate my question:

    Prove: A sequence in R can have at most one limit.

    Proof:

    Assume a sequence {xn}n[tex]\in[/tex]N has two limits a and b.
    By definition:

    -For any [tex]\epsilon[/tex]>0, there exists an N[tex]\in[/tex]N such that n[tex]\geq[/tex]N implies that |xn-a| < [tex]\epsilon[/tex]/2.
    -A similar argument can be made for the limit b.

    Thus:

    |a-b| [tex]\leq[/tex] |xn-a| - |xn-b| < [tex]\epsilon[/tex]/2 + [tex]\epsilon[/tex]/2 = [tex]\epsilon[/tex]

    Thus a=b.


    Now here's my question...in the step immediately following "By definition," the actual definition of a limit of a sequence shows that n[tex]\geq[/tex]N implies that |xn-a| < [tex]\epsilon[/tex], not that n[tex]\geq[/tex]N implies that |xn-a| < [tex]\epsilon[/tex]/2. So is it acceptable to put [tex]\epsilon[/tex] over any number when convenient in a proof? As in the above proof, it is convenient to put [tex]\epsilon[/tex]/2 instead of just [tex]\epsilon[/tex] so that in the final step the two add up to [tex]\epsilon[/tex] and show that a=b.

    Thanks.

    edit: sorry for the awkward formatting, if anything is unclear, let me know I'll explain it in words.
     
    Last edited: Aug 30, 2009
  2. jcsd
  3. Aug 30, 2009 #2
    Yes, if the limit exists, then you can replace epsilon with any positive number (so epsilon/2 is of course positive), because that's simply what epsilon typically denotes. Many people want the last step of a proof to be just epsilon, but often times it doesn't matter. In this case, it doesn't matter whether we have epsilon or 2*epsilon at the very end, since epsilon is arbitrary anyways. Note you have a typo in the leftmost inequality at the final step of the proof.
     
  4. Aug 30, 2009 #3
    Thanks for the reply, that's what I thought but wanted to hear it from someone who actually knew before I went on doing proofs on an assumption.
     
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