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Math Help - example of open covering with no finite subcovering

  1. #1
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    example of open covering with no finite subcovering

    Let A=Q \cap [0, 1] where A is a subset of R.
    Give an example of an open covering of A which has no finite subcover.

    My Answer:
    Let p/q be any element of Q. Let I be the closes irrational number to p/q.
    Let U= ( (p/q)-I, (p/q)+I ) be an open set containing p/q.
    Let { U_i} be the set of all U_i for each (p/q)_i.
    Then the set { U_i} for all i, is an infinite open covering of A which has no finite subcovering of A.

    Is that right? I'm not sure there IS such an I because, for any rational p/q, aren't there an infinite number of irrationals infinitely close to p/q??
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  2. #2
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    Quote Originally Posted by 234578 View Post
    Is that right? I'm not sure there IS such an I because, for any rational p/q, aren't there an infinite number of irrationals infinitely close to p/q??
    Indeed, there is no "closest" irrational number to p/q, because for any irrational number i , you can find another one j , which is closer to p/q, that means |j-p/q|<|i-p/q|.
    This is because the irrational numbers are dense in R.

    Hint:
    First try to find an open covering of an open interval (a,b) , which has no finite subcovering.
    Then pick out two irrational numbers a<b in [0,1] and consider the open covering (-1,a), (a,b), (b,2) of A.
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