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Math Help - The Completeness Axiom

  1. #1
    MHF Contributor harish21's Avatar
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    The Completeness Axiom

    We have S defined as: S = {x|x in R, x>=0, x^2< c}

    How do I show that c+1 is an upper bound for S, and that S has a least upper bound denoted by b??

    I tried doing it this way:

    If c<1, we have x<1<1+c, i.e, 1+c is an upper bound for S

    If c>1, then x<sqrt(c)<c<c+1 and thus, c+1 is an upper bound for S.

    I am still unsure if this is the correct method!! Any suggestions?
    Last edited by harish21; February 2nd 2010 at 12:00 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by harish21 View Post
    We have S defined as: S = {x|x in R, x>=0, x^2< c}

    How do I show that c+1 is an upper bound for S, and that S has a least upper bound denoted by b??

    I tried doing it this way:

    If c<1, we have x<1<1+c, i.e, 1+c is an upper bound for S

    If c>1, then x<sqrt(c)<c<c+1 and thus, c+1 is an upper bound for S.

    I am still unsure if this is the correct method!! Any suggestions?
    Well...this is so easy...it's hard. I'm not entirely sure I know a good way of saying this. Suppose that there existed some \ell\in S such that c+1<\ell. Then, since c,\ell>0 we have that \left(c+1\right)^2<\ell^2<c which is clearly a contradiction.

    Now, since S has an upper bound c+1 we see that S\ne\varnothing (e.g. 0\in S), S is bounded from above, and that S\subseteq \mathbb{R}...therefore \sup\text{ }S exists! ta-da!
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  3. #3
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    Quote Originally Posted by harish21 View Post
    We have S defined as: S = {x|x in R, x>=0, x^2< c}
    How do I show that c+1 is an upper bound for S, and that S has a least upper bound denoted by b??
    Suppose that \left( {\exists t \in S} \right)\left[ {c + 1 < t} \right] \Rightarrow \quad \left( {c + 1} \right)^2  < t^2  < c
    What is wrong with that?
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