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Thread: 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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    MHF Contributor Drexel28's Avatar
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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 $\displaystyle A=\{x_1,\cdots,x_n\}$ with $\displaystyle x_1\leqslant\cdots\leqslant x_n$ then $\displaystyle \sup A=x_n$ and so $\displaystyle \sup A\in A$. So, for $\displaystyle \sup A\notin A$ one MUST have that $\displaystyle A$ 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 $\displaystyle \alpha=\sup(A)~\&~\alpha\notin A.$.
    $\displaystyle \left( {\exists x_1 \in A} \right)\left[ {\alpha - 1 < x_1 < \alpha } \right]$ WHY?

    If $\displaystyle n\ge 2$ then $\displaystyle \left( {\exists x_n \in A} \right)\left[ {n_{n-1} < x_n < \alpha } \right]$. 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 $\displaystyle x<\alpha$ then it is not an upper bound. So?

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