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Math Help - conclude that the closure of a union is the union of the closures.

  1. #1
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    conclude that the closure of a union is the union of the closures.

    For this question, we are allowed to assume the following is true and has been proven:

    Let A,B \subseteqR and p \inR. Prove that p is an accumulation point of A \cupB if and only if p is an accumulation point of A or p is an accumulation point of B: (A \cupB)'=A' \cup B'.


    Now since we assumed the above is true,

    Conclude that the closure of a union is the union of the closures:
    (A \cup B)^cl=A^cl \cup B^cl. How would I go about proving this?
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  2. #2
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    The closure of a set is the ‘smallest’ closed that contains the set.
    We know that A\subseteq\overline{A}~\&~ B\subseteq\overline{B} .
    The union of two closed sets is a closed set.
    That forces  \overline{A\cup B}\subseteq\overline{A}\cup \overline{B} . Why and how?

    Can you do the converse?
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  3. #3
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    After working on this for a little, is this proof correct?

    Assume p \notinA^cl and that p \notin B^cl. That is, there exists \epsilon _1>0 such that A has no point in (p- \epsilon _1,p+ \epsilon_1) and there exists \epsilon _2>0 such that B has no point in (p- \epsilon _2,p+ \epsilon_2). Let \epsilon _3=min{ \epsilon _1, \epsilon _2}, and clearly \epsilon _3>0 and satisfies the requirement.

    Then Assume p \inA^cl \cup B^cl.
    case 1: p \inA^cl which means that every neighborhood of p contains a point of A. Hence every neighborhood contains a point of A \cupB .

    case 2: p \inB^cl which means that every neighborhood of p contains a point of B. Hence every neighborhood contains a point of A \cupB .
    Thus, (A \cupB)^cl=A^cl \cupB^cl
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