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Thread: [SOLVED] Help finishing my proof for: Prove that inf A = -sup(-A)

  1. #1
    Member ilikedmath's Avatar
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    Exclamation [SOLVED] Help finishing my proof for: Prove that inf A = -sup(-A)

    Let A be a nonempty subset of R (the reals) and let -A = {-x : x in A}.
    Prove that inf A = -sup (-A).

    My work:

    Proof: Let A be non empty subset of R and
    -A = {-x : x in A}. If -x is in -A then
    -x < sup (-A) by the definition of supremum. This implies x > -sup (-A), and so
    -sup (-A) is a lower bound of A.


    Now that I look back on my work so far, I think I am wrong in assuming right away that a supremum exists. Do I have to first prove the existence of a supremum than show it is equal to the infimum? I am really stuck here.
    Any help is greatly appreciated. Thank you for your time!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ilikedmath View Post
    Let A be a nonempty subset of R (the reals) and let -A = {-x : x in A}.
    Prove that inf A = -sup (-A).

    My work:

    Proof: Let A be non empty subset of R and
    -A = {-x : x in A}. If -x is in -A then
    -x < sup (-A) by the definition of supremum. This implies x > -sup (-A), and so
    -sup (-A) is a lower bound of A.

    your work is incomplete. showing it is a lower bound is not enough. you need to show it is the greatest lower bound. there are ways to do that elegantly. i will use the fact that $\displaystyle x \le y \text{ and } y \le x \implies x = y$ for $\displaystyle x,y \in \mathbb{R}$.

    Let $\displaystyle A$ and $\displaystyle -A$ be as defined. By the definition of infimum, we have that $\displaystyle \inf A \le x$ for all $\displaystyle x \in A$. so that $\displaystyle - \inf A \ge -x$ for all $\displaystyle -x \in -A$. but that means $\displaystyle - \inf A$ is an upper bound for $\displaystyle -A$. thus, since the supremum is the least upper bound, we must have $\displaystyle - \inf A \ge \sup (-A)$ ........(1)

    Also, $\displaystyle \sup (-A) \ge -x$ for all $\displaystyle -x \in -A$ by the definition of supremum. but that means $\displaystyle - \sup (-A) \le x$ for all $\displaystyle x \in A$, so that $\displaystyle - \sup (-A)$ is a lower bound for the set $\displaystyle A$. since the infimum is the greatest lower bound, we must have that $\displaystyle - \sup (-A) \le \inf A \implies \sup (-A) \ge - \inf A$ ...........(2)

    By (1) and (2) we have $\displaystyle - \inf A = \sup (-A)$, as desired

    Now that I look back on my work so far, I think I am wrong in assuming right away that a supremum exists. Do I have to first prove the existence of a supremum than show it is equal to the infimum? I am really stuck here.
    Any help is greatly appreciated. Thank you for your time!
    yes, we can assume the supremum and infimum exist. they are $\displaystyle \infty$ and $\displaystyle - \infty$ in the extreme cases
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  3. #3
    Member ilikedmath's Avatar
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    thanks!

    thanks for the help!
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