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Thread: Connected subset

  1. #1
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    Connected subset

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

    I know from a previously proved theorem that if $\displaystyle S$ is a connected subset of $\displaystyle \mathbb{R}$ and $\displaystyle a<b$, $\displaystyle a,b \in S$, then $\displaystyle [a,b] \subseteq S$. And it seems intuitive that $\displaystyle a$ 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 $\displaystyle S$ 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:
    $\displaystyle ( - \infty ,a),( - \infty ,a],(a,b),[a,b),(a,b],[a,b],(a,\infty ),[a,\infty )$.
    Only the last two fit the description of $\displaystyle S$.
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  5. #5
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    OK, then. If $\displaystyle S$ is a connected subset of $\displaystyle \mathbb{R}$ and $\displaystyle S$ is bounded below, but not above, then $\displaystyle S$ is either $\displaystyle \left[ a, \infty \right)$ or $\displaystyle \left( a, \infty \right)$ for some $\displaystyle a \in \mathbb{R}$.
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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 $\displaystyle S$ 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 $\displaystyle S$ is bounded below, $\displaystyle S$ has a greatest lower bound, say $\displaystyle a$. Since $\displaystyle S$ is not bounded above, I claim that $\displaystyle S=(a, \infty)$ or $\displaystyle S=[a, \infty)$.
    Case 1: Suppose $\displaystyle a,x \in S$ such that $\displaystyle a \neq x$. Since $\displaystyle a=g.l.b.(S)$, $\displaystyle a<x$. Now since $\displaystyle S$ is unbounded, there is some $\displaystyle s \in S$ such that $\displaystyle s>x$. but since $\displaystyle a,s,x \in S$, by previously proved lemma, $\displaystyle [a,x] \in S$ and $\displaystyle [x,s] \in S$. Henc e $\displaystyle S=[a, \infty)$.
    Case 2: Suppose $\displaystyle a \notin S$ and suppose $\displaystyle x \in S$, $\displaystyle x>a$.

    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 $\displaystyle A$ is connected then the closure, $\displaystyle \overline{A}$, is also connected.
    So no limit point can be separated from the set.
    Thus both $\displaystyle (a,\infty)~\&~[a,\infty)$ are connected.
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  9. #9
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    Thank you, Plato!
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