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Math Help - Nonempty set bounded from below

  1. #1
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    Nonempty set bounded from below

    S is a nonempty set of real numbers bounded from below.

    Here is what I have done so far:
    x=inf S=glb S
    Since x=glb S, there cannot be an x+epsilon that is a lower bound. This implies there exists a elemt of S such that x-epsilon<a<x<x+epsilon.

    My problem is that I think I went wrong somewhere in the 2nd sentence. I can do the rest of my problem if I can get this cleared up, but I'm not entirely sure where I went wrong.
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    Please state the problem!
    We don't know what you are to prove,
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    Let S be a nonempty set of real numbers that is bounded from below and let x=inf S. prove that either x belongs to S or x is an accumulation point of S.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kathrynmath View Post
    Let S be a nonempty set of real numbers that is bounded from below and let x=inf S. prove that either x belongs to S or x is an accumulation point of S.
    here is a sketch of the proof

    if S has a minimum, then inf(S) = min{S} and so is in the set

    now suppose S does not have a minimum, and let x = inf(s)

    then there is some \epsilon > 0 so that (x + \epsilon )\in S. let s_1 = x + \epsilon. then by the denseness of \mathbb{R}, there is some s_2 \in \mathbb{R}, so that x < s_2 < s_1 and s_2 \in S. continue this logic by induction to create a sequence \{ s_n \} that converges to x
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    Quote Originally Posted by Jhevon View Post
    here is a sketch of the proof

    if S has a minimum, then inf(S) = min{S} and so is in the set

    now suppose S does not have a minimum, and let x = inf(s)

    then there is some \epsilon > 0 so that (x + \epsilon )\in S. let s_1 = x + \epsilon. then by the denseness of \mathbb{R}, there is some s_2 \in \mathbb{R}, so that x < s_2 < s_1 and s_2 \in S. continue this logic by induction to create a sequence \{ s_n \} that converges to x
    Ok, but was I on the right track at all in what I first wrote?
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    Quote Originally Posted by Jhevon View Post
    here is a sketch of the proof

    if S has a minimum, then inf(S) = min{S} and so is in the set

    now suppose S does not have a minimum, and let x = inf(s)

    then there is some \epsilon > 0 so that (x + \epsilon )\in S. let s_1 = x + \epsilon. then by the denseness of \mathbb{R}, there is some s_2 \in \mathbb{R}, so that x < s_2 < s_1 and s_2 \in S. continue this logic by induction to create a sequence \{ s_n \} that converges to x
    Ok I got that, but now I'm trying to figure out how to create a sequence that converges to x.
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    Quote Originally Posted by kathrynmath View Post
    Ok I got that, but now I'm trying to figure out how to create a sequence that converges to x.
    You donít need a sequence.
    Say that \sigma  = \inf (S). If \sigma  \in S you are done.
    So suppose that \sigma  \notin S.
    Hence \left( {\forall z \in S} \right)\left[ {\sigma  < z} \right].
    If \varepsilon  > 0 \Rightarrow \left( {\exists x \in S} \right)\left[ {\sigma  < x < \sigma  + \varepsilon } \right]\quad .
    That means that \sigma is an accumulation point of S.
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