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Math Help - Proof check Accumulation point

  1. #1
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    Proof check Accumulation point

    Suppose b = Sup S and b is not in S. prove b is an accumulation point.

    if b is the supremum of S but not in S then b is in R/S.
    Since b is the supremum then for all s \in S, s < b.
    But, because b is not in S we can find a neighborhood across s(max) = c, such that s < c < b, where c \in S, but still less than b.

    is this ok?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by p00ndawg View Post
    Suppose b = Sup S and b is not in S. prove b is an accumulation point.

    if b is the supremum of S but not in S then b is in R/S.
    Since b is the supremum then for all s \in S, s < b.
    But, because b is not in S we can find a neighborhood across s(max) = c, such that s < c < b, where c \in S, but still less than b.

    is this ok?
    I don't understand why b\notin S is needed?

    Theorem: Let E\subseteq R such that \sup\text{ }E exists. Then \sup\text{ }E is a limit point of E.

    Proof: Suppose not. Then there exists a \delta>0 such that N'_{\delta}\left(\sup\text{ }E\right)\cap E=\varnothing, but by definition this means that there does not exists an x\in E such that \sup\text{ }E-\delta<x. This is a contradiction.
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    Quote Originally Posted by p00ndawg View Post
    Suppose b = Sup S and b is not in S. prove b is an accumulation point.

    if b is the supremum of S but not in S then b is in R/S.
    Since b is the supremum then for all s \in S, s < b.
    But, because b is not in S we can find a neighborhood across s(max) = c, such that s < c < b, where c \in S, but still less than b.

    is this ok?
    .

    Since b=SupS we have by definition that:

    1) for all ,x : x\in S\Longrightarrow x\leq b

    2) for all,ε>0 ,there exists a y belonging to S ,such that :

     b-\epsilon<y\leq b\Longrightarrow b-\epsilon<y<b+\epsilon \Longrightarrow |y-b|<\epsilon .................................................. .................................................. ..1

    And since b does not belong to S we have :

     y\neq b.................................................. .......................................2

    Hence from (1) and (2) we conclude ,that there exists a deleted neighbourhood round b containing yεS.

    So b is an accumulation point.
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    Quote Originally Posted by Drexel28 View Post
    I don't understand why b\notin S is needed?

    Theorem: Let E\subseteq R such that \sup\text{ }E exists. Then \sup\text{ }E is a limit point of E.

    Proof: Suppose not. Then there exists a \delta>0 such that N'_{\delta}\left(\sup\text{ }E\right)\cap E=\varnothing, but by definition this means that there does not exists an x\in E such that \sup\text{ }E-\delta<x. This is a contradiction.
    Drexel28 ,your theorem and your proof are both wrong.

    Your theorem should be:

    Theorem: Let E\subseteq R such that \sup\text{ }E exists .And SupE does not belong to E . Then \sup\text{ }E is a limit point of E

    And in your proof N'_{\delta}\left(\sup\text{ }E\right)\cap E=\varnothing ,does not imply :

    that there does not exists an x\in E such that \sup\text{ }E-\delta<x.

    But that:

    for all ,x : \neg (x\in E) or (SupE=x) or \neg |x-SupE|<\delta.

    Which is equivalent to:

    SupE\neq x\Longrightarrow\neg(x\in E\wedge |x-SupE|<\delta).

    And that shows that:

    Only if SupE\neq x, then there does not exists an x\in E such that \sup\text{ }E-\delta<x.

    And since SupE does not belong to E we have:

    SupE\neq x and the contradiction follows.
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