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Math Help - Supremum not an element of set, prove infinite

  1. #1
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    Supremum not an element of set, prove infinite

    Let A be a nonempty subset of the Reals bounded above. Suppose sup A exists and that sup A is not an element of A. Show that A contains a countably infinite subset. In particular, A is infinite.

    I'm reasonably sure that the easiest way to prove this is using an epsilon argument, but I don't really know where to start on here.

    Here's my scrambled thoughts:

    Let a = sup A
    x be an element of A
    x <= a

    For any epsilon > 0 there is a-epsilon < x < a
    how do a I show that the interval from a-epsilon to x is countably infinite?

    Thanks
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    Re: Supremum not an element of set, prove infinite

    Quote Originally Posted by jsmith90210 View Post
    Let A be a nonempty subset of the Reals bounded above. Suppose sup A exists and that sup A is not an element of A. Show that A contains a countably infinite subset. In particular, A is infinite.

    I'm reasonably sure that the easiest way to prove this is using an epsilon argument, but I don't really know where to start on here.

    Here's my scrambled thoughts:

    Let a = sup A
    x be an element of A
    x <= a

    For any epsilon > 0 there is a-epsilon < x < a
    how do a I show that the interval from a-epsilon to x is countably infinite?

    Thanks
    I think the best way to do this is to do the easy proof that if A=\{x_1,\cdots,x_n\} with x_1\leqslant\cdots\leqslant x_n then \sup A=x_n and so \sup A\in A. So, for \sup A\notin A one MUST have that A is not finite. Sound feasible? Or do you want to do the more direct proof?
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    Re: Supremum not an element of set, prove infinite

    I see the logic in your proof, but if possible, I'd like a more direct proof
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    Re: Supremum not an element of set, prove infinite

    Quote Originally Posted by jsmith90210 View Post
    Let A be a nonempty subset of the Reals bounded above. Suppose sup A exists and that sup A is not an element of A. Show that A contains a countably infinite subset. In particular, A is infinite.
    Suppose \alpha=\sup(A)~\&~\alpha\notin A..
    \left( {\exists x_1  \in A} \right)\left[ {\alpha  - 1 < x_1  < \alpha } \right] WHY?

    If n\ge 2 then \left( {\exists x_n  \in A} \right)\left[ {n_{n-1} < x_n  < \alpha } \right]. WHY?

    How does that prove it?
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    Re: Supremum not an element of set, prove infinite

    Should that be xn-1 < xn? And I don't know how that proves it, that's the part I'm struggling with
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    Re: Supremum not an element of set, prove infinite

    Quote Originally Posted by jsmith90210 View Post
    Should that be xn-1 < xn? And I don't know how that proves it, that's the part I'm struggling with
    1) if x<\alpha then it is not an upper bound. So?

    2) we have x_1<x_2<\cdots<x_n<\cdots<\alpha.
    That is an infinite collection.
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