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Math Help - Proof: Interior points and closed sets.

  1. #1
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    Proof: Interior points and closed sets.

    I need help with another complex problem in a general topological space:

    Show that a set S is open if and only if each point in S is an interior point.
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    Quote Originally Posted by Porter1 View Post
    I need help with another complex problem in a general topological space:
    Show that a set S is open if and only if each point in S is an interior point.
    If each point of a set \mathcal{O} is an interior point then for all x \in \mathcal{O} there is an open set Q_x\subseteq \mathcal{O}.
    Prove that \bigcup\limits_{x \in \mathcal{O}} {Q_x }  = \mathcal{O}
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    Thats the thing, I dont know how to prove it.
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    Quote Originally Posted by Porter1 View Post
    Thats the thing, I dont know how to prove it.
    Please believe me, I mean you no disrespect.
    Have you ever considered that you may not have the necessary preparation to tackle this material?
    Is a retired chair of a mathematical sciences department, I have had this conversation with a countless number of students.
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    Quote Originally Posted by Porter1 View Post
    I need help with another complex problem in a general topological space:

    Show that a set S is open if and only if each point in S is an interior point.
    In the other direction to Plato's without all the details. Let  O be an open set in a topological space X. Suppose that x\in O was not an interior point, then every neighborhood N of x intersects X-O. Therefore, we see that x is a limit point of O'. But, we have that x is a limit point of O' that is not in O', so O' is not closed and thus O is not open. This, of course, is a contradiction.


    This is an indirect way. Of course it assumes you've defined closed set and limit point. If not, you should be able to see a more direct way.
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    Quote Originally Posted by Porter1 View Post
    I need help with another complex problem in a general topological space:

    Show that a set S is open if and only if each point in S is an interior point.



    By writing down two definitions the above statement follows quite easily:

    Def 1 x is interior point of S <=> there exists ε>0 and B(x,\epsilon)\subset S

    Def 2 S is open <=> for all ,x: xεS => there exists ε>0 and B(x,\epsilon)\subset S.

    And using the two definitions we have:

    S is open <=> (for all ,x : xεS => there exists ε>0 and B(x,\epsilon)\subset S) <=> (for all ,x : xεS => x is interior of S) <=> (each point of S is interior)
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xalk View Post



    By writing down two definitions the above statement follows quite easily:

    Def 1 x is interior point of S <=> there exists ε>0 and B(x,\epsilon)\subset S

    Def 2 S is open <=> for all ,x: xεS => there exists ε>0 and B(x,\epsilon)\subset S.

    And using the two definitions we have:

    S is open <=> (for all ,x : xεS => there exists ε>0 and B(x,\epsilon)\subset S) <=> (for all ,x : xεS => x is interior of S) <=> (each point of S is interior)
    This is not a metric space necessarily. So, while there are still neighborhoods the idea of an open ball is not present.
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