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Math Help - Proof: Intersection of infinitly many subsets empty?

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    Proof: Intersection of infinitly many subsets empty?

    Hello everybody,
    I was wondering if this infinite intersections of sets was empty.

    I have sets (-\frac{1}{n},\frac{1}{n}) for all n \in N. I was wondering if 0 was in the set of the intersection of all these subsets. I can prove that all negative or positive numbers are not, using the Archimedian property of R. However, this does not work for the number zero so I am puzzled if this was an element of the intersection. Intuitvely I would say no but I cannot prove it. However, If I said: 1 > 0 what is obviously true I could divide by any positive n to get: 1/n > 0 and any negative -n to get -1/n < 0. Consequently 0 would be in this intersection but I'm not sure at all... Thank you for any help!
    Best, Rafael
    Last edited by raphw; September 16th 2010 at 08:41 PM.
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    Quote Originally Posted by raphw View Post
    Hello everybody,
    I was wondering if this infinite intersections of sets was empty.

    I have sets (-\frac{1}{n},\frac{1}{n}) for n \in N. I was wondering if 0 was in the set of the intersection of all these subsets. I can prove that all negative or positive numbers are not, using the Archimedian property of R. However, this does not work for the number zero so I am puzzled if this was an element of the intersection. Intuitvely I would say no but I cannot prove it. Thank you for any help!
    Best, Rafael
    I have not studied this formally but my instinct is that 0 is the only element in the intersection, simply because it's in every set, and there is no other real number that is in every set.
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    Let x be any positive number. By the Archimedean property, there exist an integer, n, such that n> x and thus 1/n< 1/x. x is NOT in the set (-1/n, 1/n) for that n and so is not in the intersection of all such sets.

    Let x be any negative number. Then -x> 0 and as before there exist an integer, n, such that n> -x and thus, -1/n> x. x is NOT in the set (-1/n, 1/n) for that n and so is not in the intersection of all such sets.

    The third case, x= 0, is the easiest. For any positive integer, n, n> 0 by definition of "positive" and so -n< 0. It follows that 1/n> 0 and -1/n< 0. That is -1/n< 0< 1/n for all n. 0 is in every set of the form (-1/n, 1/n) and so is in their intersection. The intersection of the collection of all open sets of the form (-1/n, 1/n) is just {0} which is a closed set and is not open. That is a reason why, while a union of any collection of open sets is open, on intersections of finite collections of open sets is open.
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    Working from the title of the thread, this is true \bigcap\limits_{n = 1}^\infty  {\left( {0,\frac{1}{n}} \right)}  = \emptyset .
    As an example of infinite intersection being empty.
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