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Math Help - Closed and Open Subsets

  1. #1
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    Closed and Open Subsets

    If anyone helps me with these questions, I will be thankful.


    (a) Show that every closed set in a metric space can be expressed as an
    intersection of countably many open sets.



    (b) Show that every open set in a metric space can be expressed as an
    union of countably many closed sets. (Hint: take complements.)


    (c) Express the interval (0, 1) in R as a union of countably many open
    sets.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by selinunan View Post

    (c) Express the interval (0, 1) in R as a union of countably many open
    sets.
    (0,1) = \bigcup_{n\in\mathbb{N}} \left(\dfrac{1}{n+1},1\right)
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  3. #3
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    Thank you for your help.

    Will it be OK if I answer the part (a) like this?


    Let A_t be an open set in (M,d).

    Then, A_t = { d (a,x) < t for some a in A }

    So, ∩ { d (a,x) < 1/n for some a in A } = cl (A) which is the smallest closed set containing A.


    If it is true, how can I modify it for the answer of part (b)?
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  4. #4
    MHF Contributor kalagota's Avatar
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    i think, you should start with an arbitrary closed set and find a way that it can be expressed as an intersection of countably many open sets..

    what you did is you take an arbitrary open set and come up with a closed set related to that..
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  5. #5
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    Do you have an idea how I can manage to do it?

    Thank you.
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  6. #6
    MHF Contributor kalagota's Avatar
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    honestly, this is the first time i encountered such question.. so i'm still trying to device a proof for it..
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  7. #7
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    Here is a start. Suppose that Mis a closed set.
    \left( {\forall n \in \mathbb{Z}^ +  } \right) define <br />
O_n  = \bigcup\limits_{x \in M} {B\left( {x;\frac{1}{n}} \right)} .
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  8. #8
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    Thank you, but I couldn't understand how I can go on.
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  9. #9
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    Quote Originally Posted by selinunan View Post
    I couldn't understand how I can go on.
    Well each O_n is an open set being a union of balls and \left( {\forall n} \right)\left[ {M \subseteq O_n } \right].
    Thus M \subseteq \bigcap\limits_{n = 1}^\infty  {O_n }. To show equality suppose that y \notin M.
    Because of closure, \left( {\exists \delta  > 0} \right)\left[ {B(y;\delta ) \cap M = \emptyset } \right].
    \left( {\exists N} \right)\left[ {\frac{1}{N} < \delta } \right] so is it possible for y \in B\left( {x;\frac{1}{N}} \right) \subseteq O_N .
    If not then M = \bigcap\limits_{n = 1}^\infty  {O_n } .
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