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Math Help - closure of sets (topology)

  1. #1
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    closure of sets (topology)

    i don't know how to prove this..
    if it's false then i'm suppose give an example where the equality fails...

    So, the closure of the intersection of all Aalpha = intersection of all closure of Aalpha.

    the closure of the union of all Aalpha = union of all closure of Aalpha.

    I know for the last one it is false... but i can't find an example...

    Please help me out!

    Thank You!
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  2. #2
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    Quote Originally Posted by tomboi03 View Post
    i don't know how to prove this..
    if it's false then i'm suppose give an example where the equality fails...

    So, the closure of the intersection of all Aalpha = intersection of all closure of
    Aalpha.

    the closure of the union of all Aalpha = union of all closure of Aalpha.

    I know for the last one it is false... but i can't find an example...

    Please help me out!

    Thank You!
    The first statement is true but how you prove it depends entirely on how you DEFINE "closed
    set" and there are several equivalent ways. For example, it is possible to define "closed set" as
    being a member of a collection of all subset of the universal set U such that
    1) U is in the collection
    2) The empty set is in the collection
    3) The intersection of any family of sets in the collection is in the collection
    4) The union of any finite family of sets in the collection is in the collection.
    In this case, you theorem is just (3) in the definition of "closed"! (This particular definition is not
    often used.)

    What definition of "closed set" are you using?

    As for the second, yes it is true and you only need to give a single counter-example. Consider
    A_n= \left[\frac{1}{n}, 1- \frac{1}{n}\right] for all integers n> 1. That is
    A_2= \left[\frac{1}{2},1-\frac{1}{2}\right]= \{\frac{1}{2}\},
    A_3=\left[\frac{1}{3},1-\frac{1}{3}\right]= \left[\frac{1}{3}, \frac{2}{3}\right],
    A_4=\left[\frac{1}{4},1-\frac{1}{4}\right]= \left[\frac{1}{4}, \frac{3}{4}\right]
    You will need to show that each of those sets is closed, that the union of all such sets is (0, 1),
    and that (0, 1) is not closed.
    (Note that proving it is open is NOT proving it is closed.)

    Here [a,b] is the closed interval and (a, b) is the open interval in the real numbers.
    Last edited by HallsofIvy; February 17th 2009 at 07:02 AM.
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