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Math Help - Proof of closure relationship

  1. #1
    DMT
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    Proof of closure relationship

    Can someone help me with proving the following simple relationship:

    <br />
\bar A \setminus \bar B \subset \overline {{A \setminus B}}<br />
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  2. #2
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    If x \in \left( {\overline A \backslash \overline B } \right) then \left( {\exists O_x } \right)\left[ {x \in O_x \,\& \,O_x  \cap B = \emptyset } \right] for some open set.
    But any open set that contains x must also contain a point of A because x is in the closure of A.

    Now if Q is any open set containing x then Q \cap O_x is also open and as such must contain a point of A\backslash B.

    Can you finish now.
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  3. #3
    DMT
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    I think I solved it, but I did it differently, and I don't follow some of your explanation. But maybe I'm confused about something ... do elements of a set have to be either limit points or interior points? This seems to be implied in your explanation, but I thought this wasn't true? (for example the set of natural numbers in the usual topology of Reals).

    As I understand it the closure is the union of a set and it's derived set:

    <br />
\bar A = A \cup A^\prime<br />

    So that your second statement about any open set containing x also containing another point of A doesn't make sense to me ... unless you didn't mean "another" point, but included the possibility of x itself?

    I solved it like this. First I changed it to the equivalent:


    <br />
(A \cup A^\prime) \setminus (B \cup B^\prime) \subset (A \setminus B) \cup (A \setminus B)^\prime<br />

    <br />
(A \setminus B)^\prime = A^\prime \setminus B^\prime<br />

    If <br />
p \in (A \cup A^\prime) \setminus (B \cup B^\prime)<br />
then ...

    I just realized an error with what I was doing ... I wanted to write next:
     p \in A \setminus B or  A^\prime \setminus B^\prime

    But I see now I can't directly rule out  p \in A \setminus B^\prime or  A^\prime \setminus B

    And I'm concerned that my derived set equality above is not accurate, though it seemed straightforward a moment ago. I can at least show:
    <br />
(A \setminus B)^\prime \subset A^\prime \setminus B^\prime<br />
    Though that goes the wrong direction, so I would need equality.

    Well, now I've managed to thoroughly confuse myself, so I think I better take a closer look at your method.
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  4. #4
    DMT
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    Quote Originally Posted by Plato View Post
    If x \in \left( {\overline A \backslash \overline B } \right) then \left( {\exists O_x } \right)\left[ {x \in O_x \,\& \,O_x  \cap B = \emptyset } \right] for some open set.
    But any open set that contains x must also contain a point of A because x is in the closure of A.

    Now if Q is any open set containing x then Q \cap O_x is also open and as such must contain a point of A\backslash B.

    Can you finish now.
    I think I got this one now. Either x is in A or A, if it is in A, then you get directly A\B, otherwise you get (A\B) (since there has to be another point of A\B in the open set) ... but it seems to be the introduction of Q was not necessary ... I don't see what this adds. The first open set seems to work fine by itself.
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