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Math Help - Baby Rudin problem 1.5

  1. #1
    Member Mollier's Avatar
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    Baby Rudin problem 1.5

    Hi,

    the whole problem and solution is at: Solutions Baby Rudin 1.5 (work in progress) DaFeda's Blog. I'm sorry for posting an external link, just trying to avoid writing it all again. However, I do not think you even need to go there to answer my question.

    In my proof I write;

    If \alpha > \beta then -inf(A) > sup(-A). Since sup(-A)\geq -x for all x\in A this would imply that
    -inf(A) > -x , which is a contradiction since -inf(A) \geq -x for all x\in A.

    Can I use \geq and > in such a way? I have three different solutions available to me from different people, so I am not looking for another way of solving this.

    Hope someone has the time to take a look! Thanks.
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    Quote Originally Posted by Mollier View Post
    Since sup(-A)\geq -x for all x\in A this would imply that -inf(A) > -x , which is a contradiction since -inf(A) \geq -x for all x\in A.
    I don't see a contradiction here. The known fact that a\ge b (where a = -\inf(A) and b=-x) does not preclude that a>b. Moreover, the inequality \inf(A)<x (and, correspondingly, -\inf(A)>-x) can be strict for all x\in A. This happens when A does not have a minimum, e.g., when A=\{x\in\mathbb{R}\mid x>0\}.
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    Quote Originally Posted by emakarov View Post
    I don't see a contradiction here. The known fact that a\ge b (where a = -\inf(A) and b=-x) does not preclude that a>b. Moreover, the inequality \inf(A)<x (and, correspondingly, -\inf(A)>-x) can be strict for all x\in A. This happens when A does not have a minimum, e.g., when A=\{x\in\mathbb{R}\mid x>0\}.
    My thinking was that since there is some set A such that -inf(A)=-x for all x \in A, then saying that -inf(A)> -x would be a contradiction. I will look for a different approach.
    Thanks!
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    Member Mollier's Avatar
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    I think I can use the fact that since sup(-A) \geq -x then -sup(-A) \leq x which is the definition of inf(A) and so inf(A)=-sup(-A).
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  5. #5
    Guy
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    Quote Originally Posted by Mollier View Post
    I think I can use the fact that since sup(-A) \geq -x then -sup(-A) \leq x which is the definition of inf(A) and so inf(A)=-sup(-A).
    This shows that -\sup(-A) is a lower bound. Now, you must show that it is greatest among all lower bounds. So, suppose there is a lower bound of A that is greater and use this to get an upper bound of -A which is smaller than sup(-A). This contradicts the leastness of sup(-A).

    In general, to show something is a sup/inf, you need to check two things: (1) it is a upper/lower bound and (2) it is (nonstrictly) smaller/larger than any particular upper/lower bound.
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    Here is a different approach.

    Suppose that \lambda =\inf(A),  \left( {\forall a \in A} \right)\left[ { - a \leqslant  - \lambda } \right].
    Therefore, there exist \gamma=\sup(-A). Moreover \gamma \le -\lambda..

    So suppose that \gamma < -\lambda.
    That means \lambda < -\gamma which implies  \[\left( {\exists a' \in A} \right)\left[ {\lambda  \leqslant a' < -\gamma } \right].
    But that gives a contradiction:  \gamma < -a' . Do you see what & why?
    So \gamma= -\lambda
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    Member Mollier's Avatar
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    Quote Originally Posted by Guy View Post
    This shows that -\sup(-A) is a lower bound. Now, you must show that it is greatest among all lower bounds. So, suppose there is a lower bound of A that is greater and use this to get an upper bound of -A which is smaller than sup(-A). This contradicts the leastness of sup(-A).

    In general, to show something is a sup/inf, you need to check two things: (1) it is a upper/lower bound and (2) it is (nonstrictly) smaller/larger than any particular upper/lower bound.
    Let \gamma be a lower bound of A ( \gamma \leq x for all x\in A), and let
    \gamma > -sup(-A).
    Then x \geq \gamma > -sup(-A) and so -x \leq -\gamma < sup(-A) which makes -\gamma an upper bound of -A that is smaller than sup(-A). Contradiction.

    Something like this Guy?

    As for your solution Plato, I will take a good look at it later today. Gotta run to work!
    Last edited by Mollier; December 29th 2010 at 12:28 AM.
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    Quote Originally Posted by Plato View Post
    That means \lambda < -\gamma which implies  \[\left( {\exists a' \in A} \right)\left[ {\lambda  \leqslant a' < -\gamma } \right].
    So \gamma= -\lambda
    \lambda = inf(A) and inf(A) \leq a \; \forall a\in A.

    -\gamma = -sup(-A) and -sup(-A) \leq a \; \forall a\in A.

    So what you assuming is that,

    inf(A) < -sup(-A), but since a \geq -sup(-A) then how can there exists an a' in A such that inf(A) \leq a' < -sup(-A) ?

    Thanks!
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  9. #9
    Guy
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    Quote Originally Posted by Mollier View Post
    Let \gamma be a lower bound of A ( \gamma \leq x for all x\in A), and let
    \gamma > -sup(-A).
    Then x \geq \gamma > -sup(-A) and so -x \leq -\gamma < sup(-A) which makes -\gamma an upper bound of -A that is smaller than sup(-A). Contradiction.

    Something like this Guy?

    As for your solution Plato, I will take a good look at it later today. Gotta run to work!
    Yeah, that should be fine. I would be a little more explicit with the quantifiers, but the idea is there.
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