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Math Help - Proof of the Least Upper Bound(Supremum) of a Set

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    Proof of the Least Upper Bound(Supremum) of a Set

    I'm having a little bit of a problem with this; any help would be handy!

    Find, with proof the least upper bound of this set of real numbers!

    E = {6 - (2/(3 n + 1)) : n is a element of N}
    Where N is the natural Numbers
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  2. #2
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    Quote Originally Posted by SuperFuzzyGremlin View Post
    Find, with proof the least upper bound of this set of real numbers! E = {6 - (2/(3 n + 1)) : n is a element of N}
    Where N is the natural Numbers
    The sequence \left(6-\frac{2}{3n+1}\right) is increasing and bounded above. So it converges. RIGHT?
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    Soloution

    Claim Sup(E) = 6

    To prove this, first 6 must be found to be a valid upper bound for E. Then we must ascertain that their is no smaller upper bound then 6. Where y < 6, where y might be a smaller upper bound for E.

    Proof:

    Let x ∈ E with x>y
    \left(6-\frac{2}{3n+1}\right)>y
    \left(-\frac{2}{3n+1}\right)>y-6
    \left(-\frac{2}{3n+1}\right)>y-6
    -2>(y-6)(3n+1)
    n> \frac{1}{3}\left(\frac{2}{y-6}+1\right)

    Then pick n \in N
    such that n> \frac{1}{3}\left(\frac{2}{y-6}+1\right)
    This means that \left(6-\frac{2}{3n+1}\right)>y
    Hence y can not be an upper bound for E!!!

    Finally got it after pouring through a book for a very long while!!!
    Thanks for the help!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by SuperFuzzyGremlin View Post
    Claim Sup(E) = 6

    To prove this, first 6 must be found to be a valid upper bound for E. Then we must ascertain that their is no smaller upper bound then 6. Where y < 6, where y might be a smaller upper bound for E.

    Proof:

    Let x ∈ E with x>y
    \left(6-\frac{2}{3n+1}\right)>y
    \left(-\frac{2}{3n+1}\right)>y-6
    \left(-\frac{2}{3n+1}\right)>y-6
    -2>(y-6)(3n+1)
    n> \frac{1}{3}\left(\frac{2}{y-6}+1\right)

    Then pick n \in N
    such that n> \frac{1}{3}\left(\frac{2}{y-6}+1\right)
    This means that \left(6-\frac{2}{3n+1}\right)>y
    Hence y can not be an upper bound for E!!!

    Finally got it after pouring through a book for a very long while!!!
    Thanks for the help!
    this is incorrect. you assumed x > y. that is, y is not an upper bound. this is what you want to show, not assume. your choosing of n seems to be inconsequential. moreover, you ended up with exactly what you assumed, which means you've shown nothing, just went in a circle.

    I would proceed this way:

    Clearly 6 is an upper bound for E, since (i leave it to you to show this).

    To show that 6 = \sup E, we need to show that 6 is the least upper bound. Assume, to the contrary, that it isn't. Rather, let y = \sup E where y \in \mathbb R,~y < 6. Then, for all x \in E, we have x \le y < 6.

    Now, let y = 6 - \epsilon for some \epsilon > 0. Then we have, for all n \in \mathbb N, that

    6 - \frac 2{3n + 1} \le 6 - \epsilon < 6

    \Leftrightarrow 0 < \epsilon \le \frac 2{3n + 1} for all n.

    Which is absurd (why?). Hence our original assumption must have been in error and, in fact, \sup E = 6
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