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Thread: Just a matter of rigor

  1. #1
    Aug 2010

    Just a matter of rigor

    My teacher really takes account about the rigor of a proof. The following is the proposition and a proof given by myself. Can someone please check it whether it is rigorous enough?

    Proposition: Let (s_n) be a sequence that converges to s\in \mathbb{R}. Show that if s_n\ge a for all n>N, then s\ge a.

    Proof: Let \epsilon>0, then there exists N_0 such that n>N_0 implies |s_n-s|<\epsilon. Now take N_1=\text{max}\{N_0, N\} and n>N_1 implies
    Since x\le |x|, we must have
    Since the difference can be taken arbitrarily small, or even negative. This proves that s\ge a.

    I really doubt the rigor of this proof. And after I type in the proof, I doubt the validity.
    Last edited by Hinatico; Aug 28th 2010 at 04:16 AM.
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  2. #2
    MHF Contributor

    Aug 2006
    Here is an easy proof.
    If it were true that s<a then \varepsilon  = a-s.
    From that we get \left| {s_N-s} \right| < \varepsilon \, \Rightarrow \,s_N<a.
    What is wrong to that?
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