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Thread: 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$\displaystyle \subseteq$R and p$\displaystyle \in$R. Prove that p is an accumulation point of A $\displaystyle \cup$B if and only if p is an accumulation point of A or p is an accumulation point of B: (A$\displaystyle \cup$B)'=A' $\displaystyle \cup$ B'.


    Now since we assumed the above is true,

    Conclude that the closure of a union is the union of the closures:
    (A $\displaystyle \cup$ B)^cl=A^cl $\displaystyle \cup$ 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 $\displaystyle A\subseteq\overline{A}~\&~ B\subseteq\overline{B} $.
    The union of two closed sets is a closed set.
    That forces $\displaystyle \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$\displaystyle \notin$A^cl and that p$\displaystyle \notin$ B^cl. That is, there exists $\displaystyle \epsilon$$\displaystyle _1$>0 such that A has no point in (p-$\displaystyle \epsilon$$\displaystyle _1$,p+$\displaystyle \epsilon$_1) and there exists $\displaystyle \epsilon$$\displaystyle _2$>0 such that B has no point in (p-$\displaystyle \epsilon$$\displaystyle _2$,p+$\displaystyle \epsilon$_2). Let $\displaystyle \epsilon$$\displaystyle _3$=min{$\displaystyle \epsilon$$\displaystyle _1$,$\displaystyle \epsilon$$\displaystyle _2$}, and clearly $\displaystyle \epsilon$$\displaystyle _3$>0 and satisfies the requirement.

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

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