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Math Help - Proof of disconnected sets

  1. #1
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    Proof of disconnected sets

    I have to show that given any topological space Y, if a closed subset of Y is disconnected then it must be the union of two disjoint closed sets.

    i.e. C= U union V where U and V are closed nonempty sets, and their intersection is empty.

    My proof so far:

    Let C be a closed subset of Y. Assume C is disconnected. Then there exists two nonempty subsets U,V such that U does not equal C and V does not equal C. Im not sure if I can assume that U= C(V) (complement of V) and V= C(U). If so that would imply that they are disjoint. So I need to show that if U is relatively closed that the complement is relatively open. This implies that if V is relatively open and closed than U is also relatively open and closed. But this is where I get confused because I have to show that C=U union V where U and V are closed disjoint sets. I'm stuck.

    thanks!
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  2. #2
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    this proof may have been done in munkres or sheldon davis topology books. I don't remember how it goes, sorry.
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  3. #3
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    Quote Originally Posted by EricaMae View Post
    I have to show that given any topological space Y, if a closed subset of Y is disconnected then it must be the union of two disjoint closed sets.
    Imagine that S is disconnected and closed. But since S is disconnected it means there exists U,V, open sets in the topological space so that S\subseteq U\cap V and U\cap V = \emptyset. Now let A = S\cap U and B = S\cap V then A,B are closed and A\cup B = S (if we think of this geometrically).
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  4. #4
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    This is great, but there is one thing I'm a little confused about.

    S is contained in U intersect V, but U intersect V is empty? I'm not sure how to draw this geometrically.
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  5. #5
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    Quote Originally Posted by EricaMae View Post
    This is great, but there is one thing I'm a little confused about. S is contained in U intersect V, but U intersect V is empty? I'm not sure how to draw this geometrically.

    There are a couple of mistakes(typos?) in the TPH’s explanation.
    It should read that U & V are open and have nonempty intersections with S; S, U &V have an empty intersection; and S is a subset of U union V.
    Now you need to show that S \cap U\,\& \,S \cap V are both closed. That is easy because S is closed.
    And \left( {S \cap U} \right) \cap \left( {S \cap V} \right) = \emptyset again that is easy because U \cap S \cap V = \emptyset .
    All that is left is S \subseteq U \cup V\quad \Rightarrow \quad \left( {S \cap U} \right) \cup \left( {S \cap V} \right) = S.
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  6. #6
    Senior Member JaneBennet's Avatar
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    Looks like a recap of some basic definition is order here.

    To say that C is a disconnected subspace of Y means that there exist subsets S,T which are open in Y such that U=S\cap C and V=T\cap C are disjoint and C=U\cup V.

    Points to note:

    • U=S\cap C and V=T\cap C are disjoint, but S and T themselves need not be disjoint.

    • S and T are open in Y but U=S\cap C and V=T\cap C need not be open in Y. However, U and V are open in C considered as a subspace of Y.


    To show that U is closed, show that U^c=Y\setminus U is open.

    Let x\in U^c.

    Either x\in C or x\in C^c.

    If x\in C^c, then C^c is open and x\in C^c\subseteq S^c\cup C^c=U^c.

    If x\in C=U\cup V, then x\in V since x\notin U.

    Thus x\in T\cap C\subseteq T. Note that (S\cap C)\cap(T\cap C)=\O \Rightarrow T\subseteq(S\cap C)^c=U^c. Hence x\in T\subseteq U^c and T is open.

    In either case, there exists an open set E such that x\in E\subseteq U^c. This shows that U^c is open; hence U is closed.

    The proof of the closedness of V is similar.
    Last edited by JaneBennet; December 14th 2008 at 07:06 PM.
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