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

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

    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 03:16 AM.
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  2. #2
    MHF Contributor

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