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Math Help - Prove: Interior is an open set

  1. #1
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    Prove: Interior is an open set

    Let (X,d) be a metric space and let F be a subset of X. B(r,x) is the open ball of radius r about x.
    Definition: 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.

    Using these definitions, prove that int(F) is an open set.

    ============================

    Attempt:
    Let x E int(F), then there exists r>0 such that B(r,x) is contained in F.
    We need to prove that there exists some r' >0 s.t. B(r',x) is contained in int(F). How can we prove this?

    Any help is greatly appreciated!
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  2. #2
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    Quote Originally Posted by kingwinner View Post
    Let (X,d) be a metric space and let F be a subset of X. B(r,x) is the open ball of radius r about x.
    Definition: 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.

    Using these definitions, prove that int(F) is an open set.

    ============================

    Attempt:
    Let x E int(F), then there exists r>0 such that B(r,x) is contained in F.
    We need to prove that there exists some r' >0 s.t. B(r',x) is contained in int(F). How can we prove this?

    Any help is greatly appreciated!
    As a very brief hint, try taking r' = r/2 and using the triangle inequality.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    Let (X,d) be a metric space and let F be a subset of X. B(r,x) is the open ball of radius r about x.
    Definition: 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.

    Using these definitions, prove that int(F) is an open set.

    ============================

    Attempt:
    Let x E int(F), then there exists r>0 such that B(r,x) is contained in F.
    We need to prove that there exists some r' >0 s.t. B(r',x) is contained in int(F). How can we prove this?

    Any help is greatly appreciated!
    That is a strange definition for the interior. The way that I have most often seen it defined as is the union of all open subsets of the set.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    As a very brief hint, try taking r' = r/2 and using the triangle inequality.
    So you mean B(r/2,x) is contained in int(F)? How can we prove this? What is the intermediate point in the triangle inequality?

    thanks.
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  5. #5
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    Quote Originally Posted by kingwinner View Post
    So you mean B(r/2,x) is contained in int(F)? How can we prove this? What is the intermediate point in the triangle inequality?
    If y\in B(r/2,x) then B(r/2,y)\subseteq B(r,x)\subseteq F, because d(z,y)<r/2 and d(y,x)<r/2 together imply that d(z,x)<r, by the triangle inequality. Therefore B(r/2,x)\subseteq \text{int}(F).
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