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Thread: Rudin 1.9, least upper bounds in Q

  1. #1
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    Rudin 1.9, least upper bounds in Q

    we have 2 sets which are subsets of Q, A, where p^2<2 and B where p^2>2. how are the members of B the upper bounds of A? aren't the rationals close to the sqrt(2) the upper bounds of A and not the members of B? I can see that members of A can't be the upper bounds of A because then there would be members of A outside of that upper bound and there is no rational number sqrt(2) so that A has no least upper bound in Q?
    Last edited by professor25; May 4th 2017 at 11:02 PM.
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    Re: Rudin 1.9, least upper bounds in Q

    Use the definition for upper bound. Example: $(-1,1)$
    Has 17 as an upper bound because 17 is greater than every element of the interval. -17 is a lower bound because it is less than every element in the interval. Same with 7 and -7, 6 and -6, 2 and -2, 1 and -1. In fact, the set of upper bounds is $[1,\infty)$ while the set of lower bounds is $(-\infty,-1]$. For any element in the set of upper bounds, it is greater than every element in the interval $(-1,1)$.

    Back to your example, is there any value in B that is not greater than or equal to every value in A? Is there an element of $\mathbb{Q}$ that is an upper bound that is not in B?
    Last edited by SlipEternal; May 5th 2017 at 02:26 AM.
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    Re: Rudin 1.9, least upper bounds in Q

    well the whole set of B is greater than or equal to every value in A. there is no element of Q that is an upper bound that is not in B. there are infinite elements in B close to sqrt(2) such that r<q in B such that there is no least upper bound in Q
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    Re: Rudin 1.9, least upper bounds in Q

    The definition of upper bound does not state anything about "closeness". For a number to be an upper bound of a set, it is only required that it be greater than every element of that set. Does that answer your questions?
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    Re: Rudin 1.9, least upper bounds in Q

    Quote Originally Posted by professor25 View Post
    well the whole set of B is greater than or equal to every value in A. there is no element of Q that is an upper bound that is not in B. there are infinite elements in B close to sqrt(2) such that r<q in B such that there is no least upper bound in Q
    Have a look at this webpage on Rudin's book.
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    Re: Rudin 1.9, least upper bounds in Q

    Thank you sir!
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