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Math Help - example of a set

  1. #1
    Member Jason Bourne's Avatar
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    example of a set

    Can anyone think of an example of a set B of real numbers such that B \subseteq [0,1], B is dense, countable and consists entirely of irrational numbers? I can't think of any.

    (Let X be a set and let \leq be a total order on X. We say that X is dense if for any two x,y \in X with x < y there exists z \in X such that x<z<y)
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  2. #2
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    Quote Originally Posted by Jason Bourne View Post
    Can anyone think of an example of a set B of real numbers such that B \subseteq [0,1], B is dense, countable and consists entirely of irrational numbers
    It is clearly true that such a set must exist.
    But I do not know how to give a ‘constructive’ proof of the set.
    I hope someone else can do it. I can show it using the axiom of choice.
    Here is an outline.
    The set of ordered pairs of rational numbers \left( {p,q} \right)\,,\,0 < p < q < 1\, is countable.
    Between any two numbers there is an irrational number.
    Thus using that we can produce the required set.
    Such a set is clearly dense in \left[ {0,1} \right] and must be countable.
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  3. #3
    Member Last_Singularity's Avatar
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    Quote Originally Posted by Jason Bourne View Post
    Can anyone think of an example of a set B of real numbers such that B \subseteq [0,1], B is dense, countable and consists entirely of irrational numbers? I can't think of any.

    (Let X be a set and let \leq be a total order on X. We say that X is dense if for any two x,y \in X with x < y there exists z \in X such that x<z<y)
    Consider the set B \subseteq [0,1] where the elements of B consist of \frac{p}{\pi q} where \frac{p}{q} is rational and between 0 and 1.

    Since we define 0 \leq \frac{p}{q} \leq 1, then 0 \leq \frac{p}{\pi q} \leq 1 as well. That verifies the first requirement.

    It is countable because there exists a bijection from the rationals of [0,1] (which are countable) to the set B. Specifically, the bijection just takes an element of the rationals of [0,1] and divides it by pi.

    It is irrational since there is no way to express any \frac{p}{\pi q},p \in Z, q \in Z as a ratio between two integers.

    And it is dense because the rationals are dense, so too will B

    QED

    Note that this could have worked with any irrational number, such as \sqrt{2} or e or whatever.

    Also note that if a set is dense, it cannot be finite. So by definition, it must be countably infinite or uncountable infinite anyway.
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    Quote Originally Posted by Last_Singularity View Post
    Consider the set B \subseteq [0,1] where the elements of B consist of \frac{p}{\pi q} where \frac{p}{q} is rational and between 0 and 1.

    Since we define 0 \leq \frac{p}{q} \leq 1, then 0 \leq \frac{p}{\pi q} \leq 1 as well. That verifies the first requirement.

    It is countable because there exists a bijection from the rationals of [0,1] (which are countable) to the set B. Specifically, the bijection just takes an element of the rationals of [0,1] and divides it by pi.

    It is irrational since there is no way to express any \frac{p}{\pi q},p \in Z, q \in Z as a ratio between two integers.

    And it is dense because the rationals are dense, so too will B

    QED

    Note that this could have worked with any irrational number, such as \sqrt{2} or e or whatever.

    Also note that if a set is dense, it cannot be finite. So by definition, it must be countably infinite or uncountable infinite anyway.
    you need to be careful here: if r_n=\frac{p_n}{q_n} \in [0,1], then \frac{r_n}{\pi} \in [0, 1/\pi] and therefore they can't be dense in [0,1]. to fix this problem you need to choose r_n \in [0, \pi].
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  5. #5
    Member Last_Singularity's Avatar
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    Thank you for the good catch, NonCommAlg

    Sorry for the mistake, Jason Bourne.
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  6. #6
    Member Jason Bourne's Avatar
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    Quote Originally Posted by Last_Singularity View Post
    Thank you for the good catch, NonCommAlg

    Sorry for the mistake, Jason Bourne.
    No worries, it's easily done. Thanks for all the help.
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