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Thread: supremum and infinum

  1. #1
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    supremum and infinum

    Theorem

    $\displaystyle E \subseteq \mathbb{R}, E \not = \varnothing$

    sup $\displaystyle E$ exists $\displaystyle \Leftrightarrow$ inf $\displaystyle (-E)$ exists

    Proof:

    $\displaystyle "\Rightarrow"$

    $\displaystyle supE$ exists

    Now we show that $\displaystyle -supE=inf (-E).$
    Show that
    1. $\displaystyle -sup E$ is a lower bound of $\displaystyle -E$.
    2. if $\displaystyle s$ is a lower bound of $\displaystyle -E \Rightarrow s \leq$ -sup $\displaystyle E.$

    1.
    $\displaystyle \because$ $\displaystyle sup E$ is an upper bound of $\displaystyle E$
    $\displaystyle \therefore x \leq$ $\displaystyle sup E, \forall \in E \Rightarrow -x \geq$ $\displaystyle -sup E, \forall x \in E$

    $\displaystyle \therefore -sup E$ is a lower bound of $\displaystyle -E$ (line added for correction 6:30 a.m. Sep 12)

    2. Suppose that $\displaystyle s$ is a lower bound of $\displaystyle -E$

    suppose not $\displaystyle \Rightarrow s >$ $\displaystyle -sup E \Rightarrow -s <$ $\displaystyle sup E$

    on the other hand

    $\displaystyle -x \geq s, \forall x \in E$ <------(Could someone tell me how this come about.)

    $\displaystyle \therefore x \leq -s$

    Hence,
    $\displaystyle -s$ is an upper bound of $\displaystyle E \rightarrow\leftarrow$

    By 1. & 2, $\displaystyle inf (-E) \exists$ & $\displaystyle inf (-E)$ = $\displaystyle sup E$
    Last edited by novice; Sep 12th 2010 at 05:32 AM.
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  2. #2
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    I donít follow what you posted.
    Suppose that $\displaystyle s = \sup (E)$ then if $\displaystyle y\in -E$ then $\displaystyle -y\in E$.
    This means that $\displaystyle -y\le s$ or $\displaystyle y \ge -s$.
    This shows that $\displaystyle - s \le \inf ( - E)$.
    Suppose that $\displaystyle - s < \inf ( - E)$ then $\displaystyle s>-\inf(-E)$.
    That means $\displaystyle \left( {\exists z \in E} \right)\left[ { - \inf ( - E) < z \le s} \right]$ or $\displaystyle -z<\inf(-E)$.
    What is wrong with that?
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  3. #3
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    Quote Originally Posted by Plato View Post
    I donít follow what you posted.
    Suppose that $\displaystyle s = \sup (E)$ then if $\displaystyle y\in -E$ then $\displaystyle -y\in E$.
    This means that $\displaystyle -y\le s$ or $\displaystyle y \ge -s$.
    This shows that $\displaystyle - s \le \inf ( - E)$. <----This part makes a whole world of a difference.
    Suppose that $\displaystyle - s < \inf ( - E)$ then $\displaystyle s>-\inf(-E)$.
    That means $\displaystyle \left( {\exists z \in E} \right)\left[ { - \inf ( - E) < z \le s} \right]$ or $\displaystyle -z<\inf(-E)$.
    What is wrong with that?
    My original post was from lecture notes of some school of mathematics I found on the web--I don't remember where I got it. There is a missing link in the proof which made it hard to understand.

    Since you said you could not follow it, you made me feel better.

    Your instructions are superb, which I am very glad that you are on board.
    I am not a math major, so I don't have access to a real math professor, but thanks to MHF for giving me access to a teacher like you.

    Thank you, sir.
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