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Homework Help: Prove: Set of rational numbers cannot be expressed as intersection of open sets

  1. Aug 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the set of rational numbers in the interval (0, 1) cannot be expressed as the intersection of a countable collection of open sets.

    2. Relevant equations



    3. The attempt at a solution

    This sounds like something requiring proof by contradiction. There must be a property of [tex]\mathbb{Q} \cap (0,1)[/tex] that I need, but I can't think of any. Can someone give me a hint?
     
  2. jcsd
  3. Aug 23, 2010 #2

    CompuChip

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    The property you are looking for, is most likely the fact that [itex]\mathbb{Q} \cap (0,1)[/itex] is dense in (0, 1).

    In intuitive terms: if you want a rational number to be in your intersection [itex]\bigcap_{i \in \mathcal I} A_i[/itex] (with [itex]\mathcal I[/itex] some countable index set), you need an open interval containing it in every Ai. But every such interval also contains non-rationals. Since you can find a non-rational arbitrarily close to your rational number, you can never get rid of it.
     
  4. Aug 23, 2010 #3

    Dick

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    I don't think it's that simple. If A_i=(1/2-1/i,1/2+1/i) the intersection is {1/2}. No irrationals there.
     
    Last edited: Aug 23, 2010
  5. Aug 23, 2010 #4

    HallsofIvy

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    Should this be "an intersection of a countable collection of open intervals"?


     
  6. Aug 23, 2010 #5

    CompuChip

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    Hmm, I forgot about that. So my first idea would be that you can set
    [tex]A_i = \bigcup_{q \in \mathbb{Q}} \left( (q - 1/i, 1/2 + 1/i ) \cap (0, 1) \right) [/tex]
    and be done with it? But then probably, there is some catch like "an infinite union of opens is not necessarily open".

    Isn't that impossible right away?
    If you intersect (0, 1) with a proper subinterval S, I can immediately point out infinitely countably many rational numbers in its complement which are not in [itex](0, 1) \cap S[/itex].

    I'll shut up now and let somebody answer who actually thought about this :)
     
  7. Aug 23, 2010 #6
    @azure kitsune:

    Do you know Baire's Theorem (also known as the Baire Category Theorem)?
     
  8. Aug 23, 2010 #7
    It does look convincing, and I had to do a lot of searching to convince myself that it wouldn't work. (Edit: although with a bit more thought I realised [tex]A_i=(0,1)[/tex] for all i)
    No, any union of opens is open. The catch is that the intersection of your sets isn't the rationals - there are lots of irrationals that get in as well.
     
    Last edited: Aug 23, 2010
  9. Aug 23, 2010 #8

    Office_Shredder

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    I think you can nail this with a measure argument. You have an intersection of open sets containing a dense subset, hence each of them is of full measure. In particular the complements have null measure
     
  10. Aug 23, 2010 #9

    Dick

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    Now I don't think that's true either. Take an enumeration q_k of the rationals and let A_i be the union of all the open intervals (q_k-1/2^(i+k),q_k+1/2^(i+k)). An open set containing a dense set doesn't have to have full measure. Petek has the right clue. You should be thinking about the Baire Category Theorem.
     
    Last edited: Aug 23, 2010
  11. Aug 25, 2010 #10
    Thank you everyone for responding.

    No, it is definitely "sets." The problem is copied directly from the textbook. (which is Apostol's Mathematical Analysis, chapter 3)

    No, I have never heard of it before, probably because I'm pretty new to analysis.

    Wolfram MathWorld tells me that the Baire Category Theorem states:

    "A nonempty complete metric space cannot be represented as the union of a countable family of nowhere dense subsets."

    But the set of rational numbers in (0,1) is not complete, right? I'm not sure how to use this theorem.
     
  12. Aug 25, 2010 #11
    Here's the version of the Baire Category Theorem that I had in mind:

    If {[itex]H_k: k \in N[/itex]} is a countable family of closed subsets of R whose union contains a non-void open set, then at least one of the sets [itex]H_k[/itex] contains a non-void open set.

    However, if this theorem hasn't been proved yet in your book, then the author must have some other solution in mind.
     
  13. Aug 27, 2010 #12

    Dick

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    A direct proof would be to show that if A_k is a countable family of open sets whose intersection contains only the rationals then the intersection of the A_k must contain an uncountable number of points. That's a contradiction. Show you can assume the family is nested. Then the general idea is to show that for each of the open intervals in one of the sets A_i, there is an A_n n>i that contains at least two open intervals contained the original interval whose size is less than half the original interval. So the intervals split endlessly. There are an uncountable numbers of ways of following this splitting down to a limit point.
     
    Last edited: Aug 27, 2010
  14. Mar 14, 2012 #13
    Quite right, the infinite intersection of open sets need not be open, as opposed to closed sets whereas the infinite intersection of closed sets **is** closed.

    Example:

    A(n) = (-1/n, 1/n)

    The infinite intersection as n goes from 1 to infinity is the point 0 and that is not open.

    :)
     
  15. Mar 14, 2012 #14
    This thread is over 2 years old. Please check the date of the thread before posting.
     
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