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Math Help - Clousre, Interior, Complement of sets

  1. #1
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    Clousre, Interior, Complement of sets

    Let (X,d) be a metric space and B(r,x) is the open ball of radius r about x.
    Definition: Let F be a subset of X. The interior of F, int(F), is the set of all x E F such that there is an r > 0 such that B(r,x) is contained in F.
    Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.
    Definition: Let F be a subset of X. F is called closed iff whenever (x_n) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F) The closure of F is the set of all limit points of sequences in F.
    Theorem: x is in the closure of F iff for all r>0, B(r,x) intersect F is nonempty.

    Let (X,d) be a metric space and E is a subset of X. Prove that

    (c means complement, E bar means the closure of E)
    =======================================

    I wrote all the relevant definitions in front of me, but I still can't figure out this one. I know that for proving equality of sets, I have to prove that A is contianed in B(x E A => x E B) and B is contianed in A (x E B => x E A). However, for this problem, I can write down the basic definitions for each direction, but I just can't figure out the connection between the two sets...

    Any help is greatly appreciated!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    Let (X,d) be a metric space and B(r,x) is the open ball of radius r about x.
    Definition: Let F be a subset of X. The interior of F, int(F), is the set of all x E F such that there is an r > 0 such that B(r,x) is contained in F.
    Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.
    Definition: Let F be a subset of X. F is called closed iff whenever (x_n) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F) The closure of F is the set of all limit points of sequences in F.
    Theorem: x is in the closure of F iff for all r>0, B(r,x) intersect F is nonempty.

    Let (X,d) be a metric space and E is a subset of X. Prove that

    (c means complement, E bar means the closure of E)
    =======================================

    I wrote all the relevant definitions in front of me, but I still can't figure out this one. I know that for proving equality of sets, I have to prove that A is contianed in B(x E A => x E B) and B is contianed in A (x E B => x E A). However, for this problem, I can write down the basic definitions for each direction, but I just can't figure out the connection between the two sets...

    Any help is greatly appreciated!
    Let x\in\left(\overline{E}\right)' then there exists some neighborhood N of x such that N\cap E=\varnothing\implies N\subseteq E' which of course implies that x\in \left(E'\right)^{\circ}.

    Conversely, let x\in\left(E'\right)^{\circ}, then there exists some neighorhood N of x such that N\subseteq E'\implies N\cap E=\varnothing and thus x\notin \left(E\cup D(E)\right)\implies x\in\left(E\cup D(E)\right)'=\left(\overline{E}\right)'
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    Quote Originally Posted by Drexel28 View Post
    Let x\in\left(\overline{E}\right)' then there exists some neighborhood N of x such that N\cap E=\varnothing\implies N\subseteq E' which of course implies that x\in \left(E'\right)^{\circ}.

    Conversely, let x\in\left(E'\right)^{\circ}, then there exists some neighorhood N of x such that N\subseteq E'\implies N\cap E=\varnothing and thus x\notin \left(E\cup D(E)\right)\implies x\in\left(E\cup D(E)\right)'=\left(\overline{E}\right)'
    What is D(E)? Is it the boundary of E?
    My textbook never really defined the "boundary" of E. Is it possible to prove the result without using "boundary" concepts?

    thanks.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    What is D(E)? Is it the boundary of E?
    My textbook never really defined the "boundary" of E. Is it possible to prove the result without using "boundary" concepts?

    thanks.
    It isn't the boundary. It is more commonly used in topological spaces than metric spaces...it is the derived set...the set of all limit points. And, there is no need to use it here. I just put it there for added insight.

    \text{Boundary of E}=\partial E
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    Quote Originally Posted by Drexel28 View Post
    It isn't the boundary. It is more commonly used in topological spaces than metric spaces...it is the derived set...the set of all limit points. And, there is no need to use it here. I just put it there for added insight.

    So the notation D(E) means the set of all limit points of sequences in E? But that's the definition of closure of E. So D(E) is equal to the closure of E?

    For the second part of the proof, I don't follow why x is not an element of E U D(E). Can you explain a little more, please?

    thanks.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    So the notation D(E) means the set of all limit points of sequences in E? But that's the definition of closure of E. So D(E) is equal to the closure of E?

    For the second part of the proof, I don't follow why x is not an element of E U D(E). Can you explain a little more, please?

    thanks.
    \bar{E}\ne D(E)!!! It is commonly known that \bar{E}=\left\{x\in X:B_{\varepsilon}(x)\cap X\ne\varnothing,\text{ }\forall\varepsilon>0\right\} and this is precisely E\cup D(E).
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