Results 1 to 4 of 4

Thread: Proof check Accumulation point

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    175

    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 $\displaystyle \in$ 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 $\displaystyle \in$ S, but still less than b.

    is this ok?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    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 $\displaystyle \in$ 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 $\displaystyle \in$ S, but still less than b.

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

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

    Proof: Suppose not. Then there exists a $\displaystyle \delta>0$ such that $\displaystyle N'_{\delta}\left(\sup\text{ }E\right)\cap E=\varnothing$, but by definition this means that there does not exists an $\displaystyle x\in E$ such that $\displaystyle \sup\text{ }E-\delta<x$. This is a contradiction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    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 $\displaystyle \in$ 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 $\displaystyle \in$ S, but still less than b.

    is this ok?
    .

    Since b=SupS we have by definition that:

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

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

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

    And since b does not belong to S we have :

    $\displaystyle y\neq b$.................................................. .......................................2

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

    So b is an accumulation point.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by Drexel28 View Post
    I don't understand why $\displaystyle b\notin S$ is needed?

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

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

    Your theorem should be:

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

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

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

    But that:

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

    Which is equivalent to:

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

    And that shows that:

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

    And since SupE does not belong to E we have:

    $\displaystyle SupE\neq x$ and the contradiction follows.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. accumulation point help
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: Feb 4th 2010, 11:30 AM
  2. Accumulation point
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Jan 27th 2010, 04:43 PM
  3. Inf A= 0 <=> 0 is an accumulation point
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Dec 7th 2009, 02:51 AM
  4. accumulation point
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 19th 2009, 02:11 PM
  5. Accumulation Point
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Aug 19th 2007, 01:53 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum