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Math Help - Connected subset

  1. #1
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    Connected subset

    If S \subseteq \mathbb{R} is bounded below but not bounded above, then S is either \left( a,\infty \right] or  \left[ a,\infty \right) for some a \in \mathbb{R}.

    I know from a previously proved theorem that if S is a connected subset of \mathbb{R} and a<b, a,b \in S, then [a,b] \subseteq S. And it seems intuitive that a is the greatest lower bound for this set, but I am not sure how to begin to prove it.
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  2. #2
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    Surely you left something out.
    Did you mean that S is connected?
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  3. #3
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    I thought the same thing, but I am not sure. I have rewritten the problem as it was presented to me, but I feel sure that it must also mean S is a connected subset.
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  4. #4
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    Quote Originally Posted by tarheelborn View Post
    I thought the same thing, but I am not sure. I have rewritten the problem as it was presented to me, but I feel sure that it must also mean S is a connected subset.
    Surely. Otherwise it is false.

    There is a well known theorem: The only nontrivial connected sets of real numbers are:
    ( - \infty ,a),( - \infty ,a],(a,b),[a,b),(a,b],[a,b],(a,\infty ),[a,\infty ).
    Only the last two fit the description of S.
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  5. #5
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    OK, then. If S is a connected subset of \mathbb{R} and S is bounded below, but not above, then S is either \left[ a, \infty \right) or \left( a, \infty \right) for some a \in \mathbb{R}.
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  6. #6
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    I am trying to prove the two parts of that theorem described in the last post. I believe I need to show that since S is bounded below, it has a greatest lower bound and that said bound doesn't actually need to be included in the interval. But I am not sure how to get that started.
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  7. #7
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    I have gotten this far with the proof:


    Since S is bounded below, S has a greatest lower bound, say a. Since S is not bounded above, I claim that S=(a, \infty) or S=[a, \infty).
    Case 1: Suppose a,x \in S such that a \neq x. Since a=g.l.b.(S), a<x. Now since S is unbounded, there is some s \in S such that s>x. but since a,s,x \in S, by previously proved lemma, [a,x] \in S and [x,s] \in S. Henc e S=[a, \infty).
    Case 2: Suppose a \notin S and suppose x \in S, x>a.

    Now I am not sure how to move on to say that everything approaching a is in S but a is not in S.
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  8. #8
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    I will help with this point.
    If A is connected then the closure, \overline{A}, is also connected.
    So no limit point can be separated from the set.
    Thus both (a,\infty)~\&~[a,\infty) are connected.
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  9. #9
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    Thank you, Plato!
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