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  1. #1
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    Infimum

    Let D be a set of real numbers. Prove that if y is a lower bound of D and y also a limit point of D, then y = inf D.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by anlys View Post
    Let D be a set of real numbers. Prove that if y is a lower bound of D and y also a limit point of D, then y = inf D.
    Appreciate someone's help on this?
    For \xi to be a greatest lower bound of E two criteria must be met

    1. \xi\leqslant x~~\forall x \in E
    2. If \xi<x then x is not a lower bound of E

    Criterion 1. was given so we must just prove that if x is greater than \xi then it is not a lower bound of E

    Since \xi<x it follows that d(x,\xi)=D>0

    But by the Archimidean property we may pick a D_1\in\mathbb R such that 0<D_1<D. So now consider that since \xi is a limit point of E we may pick a x_1\in E such that x_1\in\mathcal{O}_{D_1}(\xi) (this is just the neighborhood of radius D_1). This implies that d(x_1,\xi)\leqslant D_1<D which in turn implies that \xi<x_1<x with x,x_1\in E which proves 2.
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  3. #3
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    Can you also show the proof by contradiction?
    So, supposed that a is not an infimum of D. So, there must exists a larger lower bound of D, let denote it as b.
    i.e. a < b <= x for all x in D

    So, there exists a delta > 0 such that (a, a + delta) intersect with D = empty set.

    Also, a does not belong to the set D because all elements of D are bounded below by b which is greater than a.

    Now, let delta = b-a > 0. This implies that for any sequence {X_n} in D\{a}, |X_n - a| >= b-a
    So, this means that {X_n} does not converge to a.

    So, we get that for any sequence {X_n} in D\{a}, it is not convergent to a. This implies that a is not a limit point. Contradiction!
    Is this correct?
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  4. #4
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    Quote Originally Posted by anlys View Post
    Can you also show the proof by contradiction?
    So, supposed that a is not an infimum of D. So, there must exists a larger lower bound of D, let denote it as b.
    i.e. a < b <= x for all x in D

    So, there exists a delta > 0 such that (a, a + delta) intersect with D = empty set.

    Also, a does not belong to the set D because all elements of D are bounded below by b which is greater than a.

    Now, let delta = b-a > 0. This implies that for any sequence {X_n} in D\{a}, |X_n - a| >= b-a
    So, this means that {X_n} does not converge to a.

    So, we get that for any sequence {X_n} in D\{a}, it is not convergent to a. This implies that a is not a limit point. Contradiction!
    Is this correct?
    Can someone please check my work? Thanks...
    Last edited by anlys; December 15th 2008 at 06:26 AM.
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  5. #5
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    infimum (Urgent)

    Quote Originally Posted by anlys View Post
    Can you also show the proof by contradiction?
    So, supposed that a is not an infimum of D. So, there must exists a larger lower bound of D, let denote it as b.
    i.e. a < b <= x for all x in D

    So, there exists a delta > 0 such that (a, a + delta) intersect with D = empty set.

    Also, a does not belong to the set D because all elements of D are bounded below by b which is greater than a.

    Now, let delta = b-a > 0. This implies that for any sequence {X_n} in D\{a}, |X_n - a| >= b-a
    So, this means that {X_n} does not converge to a.

    So, we get that for any sequence {X_n} in D\{a}, it is not convergent to a. This implies that a is not a limit point. Contradiction!
    Is this correct?
    Can someone please check my work? Thanks...
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by anlys View Post
    Can you also show the proof by contradiction?
    So, supposed that a is not an infimum of D. So, there must exists a larger lower bound of D, let denote it as b.
    i.e. a < b <= x for all x in D

    So, there exists a delta > 0 such that (a, a + delta) intersect with D = empty set.

    Also, a does not belong to the set D because all elements of D are bounded below by b which is greater than a.

    Now, let delta = b-a > 0. This implies that for any sequence {X_n} in D\{a}, |X_n - a| >= b-a
    So, this means that {X_n} does not converge to a.

    So, we get that for any sequence {X_n} in D\{a}, it is not convergent to a. This implies that a is not a limit point. Contradiction!
    Is this correct?
    What do you mean by "So, there exists a delta > 0 such that (a, a + delta) intersect with D = empty set."?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Sorry to take so long to get back to you. You can do this proof by contradiction. In the following we are talking about a metric space \chi with E\subset \chi.

    Assume \xi is a limit point of a set E, a lower bound of E, and there exists a x\in E such that \xi<x is a lower bound of E.

    So since \xi<x it follows that d_{\chi}(x,\xi)>0. Now because x is a lower bound of E it follows that there does not exist a x_1\in E such that \xi<x_1<x but this is a contradiction to \xi being a limit point because then there would be a neighborhood of \xi not containing a point of E

    EDIT: Someone informed me to the following typo: It should be \chi=\mathbb R^n...I use \chi since it is easier to write than \mathbb R^n
    Last edited by Mathstud28; December 17th 2008 at 02:28 PM.
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