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Set exercise

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem is attached as a picture.


    2. Relevant equations
    ...


    3. The attempt at a solution
    I have been trying a lot to prove this without any really fruitful approach. At first I thought that the statement was false, or that you could at least construct a sequence of rationals such that V=(0,1) the following way:
    Let a1 be a real in (0,1). Since the rationals are dense there exists a rational number b1 such that d(a1,b1)<ε. Let this be the first rational number in the sequence. Now let a2 be another real. Because the rationals are dense there exists a rational b2 such that d(a2,b2)<ε/2 etc. etc. and by successive use of this method I could generate the whole (0,1) with my approach. But this fails because (0,1) is not countable. So I'm open for any other approach to this problem.
     

    Attached Files:

  2. jcsd
  3. Sep 12, 2013 #2

    Zondrina

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    Homework Helper

    Can you always construct a sequence from the enumeration that converges to an arbitrary ##x \in (0,1)##?

    Also, showing it's a proper subset isn't too bad. Can you think of an element in ##(0,1)## that isn't in ##U##?
     
  4. Sep 12, 2013 #3
    The statement that you are trying to prove is false (or, rather, not necessarily true) even after fixing it so that epsilon is fixed and the last line refers to V rather than U.

    Edit: The problem, as stated, is very poorly posed. Could you maybe copy it verbatim from the source?
     
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