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Math Help - Closure, Boundary, and Neighborhoods

  1. #1
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    Closure, Boundary, and Neighborhoods

    Prove that:
    Let E be a subset of R^d and x an element of R^d. Then:
    c) x ∈ ∂E if and only if every neighborhood of x contains points of E and points of the complement of E.

    (parts a and b of this theorem state: (a) x ∈ int(E) if and only if there is a neighborhood of x that is contained in E; (b) x ∈ E if and only if every neighborhood of x contains a point of E ... I've proven parts a and b though)
    Last edited by xsavedkt; September 5th 2011 at 05:24 PM.
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    Re: Closure, Boundary, and Neighborhoods

    Quote Originally Posted by xsavedkt View Post
    Prove that:
    Let E be a subset of Rd and x an element of Rd. Then:
    c) x ∈ ∂E if and only if every neighborhood of x contains points of E and points of the complement of E.
    (parts a and b of this theorem state: (a) x ∈ E◦ if and only if there is a neighborhood of x that is contained in E; (b) x ∈ E if and only if every neighborhood of x contains a point of E ... I've proven parts a and b though)
    I hate to tell you, but what you have posted is incomprehensible.
    The notation is completely non-standard.
    What in the world is Rd?
    You write that "c) x ∈ ∂E if and only if every neighborhood of x contains points of E and points of the complement of E."
    But that is the very definition of a boundary point.
    So what is the actual question?
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    Re: Closure, Boundary, and Neighborhoods

    Quote Originally Posted by Plato View Post
    I hate to tell you, but what you have posted is incomprehensible.
    The notation is completely non-standard.
    What in the world is Rd?
    You write that "c) x ∈ ∂E if and only if every neighborhood of x contains points of E and points of the complement of E."
    But that is the very definition of a boundary point.
    So what is the actual question?
    I made some minor changes to the notation. Rd was supposed to mean R^d (R being the real number line)... sorry!
    However, you said that the question is the very definition of a boundary point.... that's not true though, in the definition of a boundary, we don't talk about the neighborhood of x. And a boundary POINT is made possible by the definition of a boundary and then proving this theorem (at least that's how it seems like to me). I think the question is to prove that every boundary point's neighborhoods contain a point in E and the complement in E... does this make more sense?

    Thanks for trying to help...
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    Re: Closure, Boundary, and Neighborhoods

    Quote Originally Posted by xsavedkt View Post
    you said that the question is the very definition of a boundary point.... that's not true though, in the definition of a boundary, we don't talk about the neighborhood of x.
    That is total nonsense!
    For the last 100 years, a boundary point x of a set E has been defined as: if O is an open set containing x then O contains a point of E and a point of E complement.
    Now we cannot be faulted if you are burdened with non-standard definitions. Maybe you should bring this to the instructor's attention.
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    Re: Closure, Boundary, and Neighborhoods

    Quote Originally Posted by Plato View Post
    That is total nonsense!
    For the last 125 years, a boundary point x of a set E has been defined as...
    Maybe the instructor is using one of the other equivalent definitions of "boundary".

    I learned it as X(bar) toss X.
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    Re: Closure, Boundary, and Neighborhoods

    Quote Originally Posted by TheChaz View Post
    Maybe the instructor is using one of the other equivalent definitions of "boundary".

    I learned it as X(bar) toss X.
    The definition of boundary with this instructor is: the set E \ E◦ is called the boundary of E and is denoted ∂E.
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    Re: Closure, Boundary, and Neighborhoods

    Quote Originally Posted by Plato View Post
    That is total nonsense!
    For the last 100 years, a boundary point x of a set E has been defined as: if O is an open set containing x then O contains a point of E and a point of E complement.
    Now we cannot be faulted if you are burdened with non-standard definitions. Maybe you should bring this to the instructor's attention.
    Hmm, that is interesting. First off, our instructor didn't really talk about a "boundary POINT" anyways but just defined boundary as "the set E \ E◦ is called the boundary of E and is denoted ∂E"... Second, based on the definition you presented, I'm taking O to be any neighborhood of x and therefore by this definition, every neigh of x contains a point in E and E's complement. So yeah, you would be right... I'm just getting more and more confused I guess, I will have to bring it up for sure.
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    Re: Closure, Boundary, and Neighborhoods

    The statement that you are trying to prove is essentially that two or more of the common equivalent definitions of a boundary of a set are actually equivalent. See Boundary (topology) - Wikipedia, the free encyclopedia
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