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Math Help - Closure Operation Properties? Kinda lost...

  1. #1
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    Red face Closure Operation Properties? Kinda lost...

    I'm sorta lost and sorta have an idea, so I don't know what to do really. The problem is to show that the closure operation has the following properties:
    (a) if E1 is a subset of E2, then closure(E1) is a subset of closure(E2)
    (b) closure(E1 U E2) = closure(E1) U closure (E2)
    (c) closure(E1 intersect E2) is a subset of closure(E1) intersect closure(E2)


    I do not need the full out proofs, but I do need some direction...

    on (a) I'm not sure how to show it at all, it's the one I'm totally lost on.

    (b) I get part of this one. I know you start with an element that is in the left side which leads to it being in E1 or E2 or (E1 U E2)', or the accumulations points of the union of E1 and E2 but do not know how to rewrite it so I can get the element in E1' and E2', or if (E1UE2)' is just E1' U E2.

    (c) I can get this once I can get b.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathgirl13 View Post
    I'm sorta lost and sorta have an idea, so I don't know what to do really. The problem is to show that the closure operation has the following properties:
    (a) if E1 is a subset of E2, then closure(E1) is a subset of closure(E2)
    (b) closure(E1 U E2) = closure(E1) U closure (E2)
    (c) closure(E1 intersect E2) is a subset of closure(E1) intersect closure(E2)


    I do not need the full out proofs, but I do need some direction...

    on (a) I'm not sure how to show it at all, it's the one I'm totally lost on.

    (b) I get part of this one. I know you start with an element that is in the left side which leads to it being in E1 or E2 or (E1 U E2)', or the accumulations points of the union of E1 and E2 but do not know how to rewrite it so I can get the element in E1' and E2', or if (E1UE2)' is just E1' U E2.

    (c) I can get this once I can get b.
    In what kind of space are we working?

    Note for the first one that if E\subseteq E_1 then every closed neighborhood which contains E_1 contains E. So the intersection of all neighborhoods which contain E has to contain the intersection of all the closed neighborhoods which contain E_1

    For the second one it looks like you got the one way, the other way comes from saying let x\in \overline{E_1}\cup\overline{E_2} then it is either an adherent point of E_1 or of E_2. Assume WLOG that it is one of E_1 then every neighborhood intersects E_1 and thus every neighborhood intersects E_1\cup E_2 and so x is an adherent point of E_1\cup E_2 or equivalently x\in\overline{E_1\cup E_2}
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