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Math Help - Nested sets

  1. #1
    Senior Member Danneedshelp's Avatar
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    Nested sets

    Q:

    Consider the infinite expansion of \sqrt{2} =  <br />
1.414213562373095048801688724209698078569671875376  9 <br />
· · · = 1.x_{1} x_{2} . . . x_{n} . . . .
    Denote by r_{n} the truncation of the above representation after the first n decimal digits. So r_{n} = 1.x_{1}x_{2} . . . x_{n} ; for example r_{1} = 1.4,  r_{2} = 1.41,  r_{3} = 1.141, and so on.

    (a) Consider the indexed family of sets \{A_{n}\}_{n\in\mathbb{N}}, where A_{n} = (-\infty, r_{n} ), open intervals of \mathbb{R}. Find \bigcup_{n\in{\mathbb{N}}}A_{n} and \bigcap_{n\in{\mathbb{N}}}A_{n}. Justify your answers.

    b) Consider the indexed family of sets \{B_{n}\}_{n\in\mathbb{N}}, where A_{n} = [\frac{1}{n}, r_{n} ), open intervals of \mathbb{R}. Find \bigcup_{n\in{\mathbb{N}}}B_{n} and \bigcap_{n\in{\mathbb{N}}}B_{n}. Justify your answers.

    A:

    a) \bigcup_{n\in{\mathbb{N}}}A_{n}=(-\infty,\sqrt{2}) and \bigcap_{n\in{\mathbb{N}}}A_{n}=(-\infty,1.4].

    b) \bigcup_{n\in{\mathbb{N}}}B_{n}=(0,\sqrt{2}) and \bigcap_{n\in{\mathbb{N}}}B_{n}=[1,1.4).

    I do not know how to prove my answers. I have tried, and can't seem to reach a conclusion. Could someone please help me with either a or b? Is there a systematic whay to approach these kinds of problems? How do I use the Archimedean principle?

    I didn't really get this stuff down in my foundations course and now its my first HW question for real analysis course .
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  2. #2
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    Here is my attempt to prove part (a)

    To show \bigcup_{n\in{\mathbb{N}}}A_{n}=(-\infty,\sqrt{2}), first let x\in (-\infty,\sqrt{2}) . This means x<\sqrt{2}. Since \lim_{x \to\infty} r_n=\sqrt{2}, there exists m\in \mathbb N such that x<r_m (otherwise r_n \leq x \forall n\in \mathbb N implies that  \sqrt{2} \leq x , which is a contradiction). So,  x\in A_m \subseteq \bigcup_{n\in{\mathbb{N}}}A_{n} . It follows that  (-\infty,\sqrt{2}) \subseteq \bigcup_{n\in{\mathbb{N}}}A_{n} . To show there other inclusion, let  x\in \bigcup_{n\in{\mathbb{N}}}A_{n} . This means  x\in A_n for some n\in \mathbb N . That is,  x<r_n . Since  r_n<\sqrt{2}, x\in (-\infty,\sqrt{2}) . Thus, equality follows.

    I believe the second equality should be \bigcap_{n\in{\mathbb{N}}}A_{n}=(-\infty,1.4). That is, not including 1.4. The method of proof is the same as above, i.e., show inclusion in both directions. First, let x\in \bigcap_{n\in{\mathbb{N}}}A_{n} . This implies x<r_n \forall n\in \mathbb N , in particular,  x<r_1=1.4 . So, x\in (-\infty,1.4) . To show the other direction of the inclusion, let  x\in (-\infty,1.4) and note that x<1.4\leq r_n \forall n\in \mathbb N . This implies x\in \bigcap_{n\in{\mathbb{N}}}A_{n} . This proves the equality.

    I hope this helps a bit.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    [INDENT]
    Quote Originally Posted by Monkey D. Johnny View Post
    Here is my attempt to prove part (a)

    I believe the second equality should be \bigcap_{n\in{\mathbb{N}}}A_{n}=(-\infty,1.4). That is, not including 1.4. The method of proof is the same as above, i.e., show inclusion in both directions. First, let x\in \bigcap_{n\in{\mathbb{N}}}A_{n} . This implies x<r_n \forall n\in \mathbb N , in particular,  x<r_1=1.4 . So, x\in (-\infty,1.4) . To show the other direction of the inclusion, let  x\in (-\infty,1.4) and note that x<1.4\leq r_n \forall n\in \mathbb N . This implies x\in \bigcap_{n\in{\mathbb{N}}}A_{n} . This proves the equality.
    Why is 1.4 not included?

    Can a proof by contradiction be done for ( \Rightarrow)? I keep trying by assuming there exists an x not in the intersection, but I can't form a solid contradiction. I keep thinking, if I choose n+1 and take the limit, the limit will still be \sqrt{2} and thus, no such x can exist since, but I can't seem to formalize this idea and make it work for me.

    Aslo, for part b, how would I use the Archimedean principle?
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  4. #4
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    Quote Originally Posted by Danneedshelp View Post
    Q:

    Consider the infinite expansion of \sqrt{2} =  <br />
1.414213562373095048801688724209698078569671875376  9 <br />
• • • = 1.x_{1} x_{2} . . . x_{n} . . . .
    Denote by r_{n} the truncation of the above representation after the first n decimal digits. So r_{n} = 1.x_{1}x_{2} . . . x_{n} ; for example r_{1} = 1.4,  r_{2} = 1.41,  r_{3} = 1.141, and so on.
    (a) Consider the indexed family of sets \{A_{n}\}_{n\in\mathbb{N}}, where A_{n} = (-\infty, r_{n} ), open intervals of \mathbb{R}. Find \bigcup_{n\in{\mathbb{N}}}A_{n} and \bigcap_{n\in{\mathbb{N}}}A_{n}. Justify your answers.
    b) Consider the indexed family of sets \{B_{n}\}_{n\in\mathbb{N}}, where A_{n} = [\frac{1}{n}, r_{n} ), open intervals of \mathbb{R}. Find \bigcup_{n\in{\mathbb{N}}}B_{n} and \bigcap_{n\in{\mathbb{N}}}B_{n}. Justify your answers.

    A:

    a) \bigcup_{n\in{\mathbb{N}}}A_{n}=(-\infty,\sqrt{2}) and \bigcap_{n\in{\mathbb{N}}}A_{n}=(-\infty,1.4].

    b) \bigcup_{n\in{\mathbb{N}}}B_{n}=(0,\sqrt{2}) and \bigcap_{n\in{\mathbb{N}}}B_{n}=[1,1.4).

    I do not know how to prove my answers. I have tried, and can't seem to reach a conclusion. Could someone please help me with either a or b? Is there a systematic whay to approach these kinds of problems? How do I use the Archimedean principle?
    First, both of those answers are correct.

    Second, I cannot imagine what you mean by ‘proving’ these answers.
    It really is about knowing how to use union and intersection.

    BTW: Here is an easy way:  r_n  = \frac{{\left\lfloor {10^n \sqrt 2 } \right\rfloor }}{{10^n }}.
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Plato View Post
    First, both of those answers are correct.

    Second, I cannot imagine what you mean by ‘proving’ these answers.
    It really is about knowing how to use union and intersection.

    BTW: Here is an easy way:  r_n  = \frac{{\left\lfloor {10^n \sqrt 2 } \right\rfloor }}{{10^n }}.
    Well, as the question states, my teacher wants us to "justify" our answers. He went over something similar last semester in my foundations class. I guess he wants to show there is no other possible answer, by showing every arbitrary value we choose are within the union or intersection.
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  6. #6
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    1.4 is not included because if it were it would be in A_n for all n, in particular, n=1. But 1.4 \notin A_1 .

    I think the Archimedean principle is needed to show that  (0,\sqrt{2}) is a subset of the union of the  B_n's : Let x be an element of  (0,\sqrt{2}) , then there exists  m\in \mathbb N such that 1/m<x by the Archimedean principle. Next, we know that (\exists m'\in \mathbb N)(x<r_{m'}) , by the same argument used in part (a). Lastly, take  M=max(m,m') and note that  x\in B_M . Hence x is an element of the union of the  B_n's .
    Last edited by Monkey D. Johnny; August 28th 2009 at 01:58 AM.
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