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Let $A$ be a nonempty subset of real numbers which is bounded below. Let $-A$ be the set of of all numbers $-x$, where $x$ is in $A$. Prove that $\inf A = -\sup(-A)$

So far this is what I have

Let $\alpha=\inf(A)$, which allows us to say that $\alpha \leq x$ for all $x \in A$. Therefore, we know that $-\alpha \geq -x$ for all $x \in -A$. Therefore we know that $-\alpha$ is an upper bound of $-A$. $\ \ \ \ $

Now let $b$ be the upper bound of $-A$. There exists $b \geq-x \implies-b \leq x$ for all $x \in A$. Hence, \begin{align} -b & \leq \alpha\\ -\alpha & \leq b\\ -\alpha & = - \inf(A) = \sup (-A) \end{align}

By multiplying $-1$ on both sides, we get that $\inf(A) = -\sup (-A)$

Is my proof correct?

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    $\begingroup$ +1. For showing your work. People appreciate if you show your work. And your answer is right. Good work! $\endgroup$ – user17762 May 15 '13 at 5:31
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    $\begingroup$ how did you get $-b \leq \alpha$? $\endgroup$ – Real Hilbert Nov 2 '13 at 15:41
  • $\begingroup$ See math.stackexchange.com/a/947049/589. $\endgroup$ – lhf Oct 31 '14 at 17:54
  • $\begingroup$ "Therefore, we know that $-\alpha \geq −x$ for all $x \in -A$." Consider $A = \{1, 3, 5\}$ then $\alpha = \mathrm{inf} A = 1$ and $-A = \{-1, -3, -5\}$ and you get a negation to much. With these variables $-1 \geq −x$ for all $x \in \{-1, -3, -5\}$ is not true. $\endgroup$ – Björn Lindqvist Jun 19 '18 at 0:09
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Let $A$ be a nonempty subset of real numbers which is bounded below. Let $−A$ be the set of of all numbers $−x$, where $x$ is in $A$. Prove that $\inf{A}=−\sup{(-A)}$.

I think that the purpose of this question is to show you why it is not required to include the existence of infimum into the Axiom of Completeness.

Here, i will show the existence of infimum of $A$.

$\forall a\in A, \exists x\in\mathbb{R}\text{ such that } x\leq a \implies -x\geq-a \implies -x \text{ is an upper bound for }-A.$

Here, $x$ is an arbitrary lower bound for $A$.

By axiom of completeness, $\exists y\in \mathbb{R} \text{ such that }y=\sup{(-A)}, \text{i.e. }-a\leq y\leq -x \text{ , }\forall a\in A,$ which implies that $x\leq -y \leq a$.

Here, $-y$ is a lower bound for $A$ and $-y$ is at least any lower bound for $A$, which means that $-y = \inf{A}$.

Hence, $-\sup{(-A)} =\inf{A}$.

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Aside from dropping a $-$ sign in your last line, that's good work. (Also, I'd say "let $b$ be an upper bound," instead, but it's clear that's what is meant.)

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Not a direct answer to your question, but here is essentially your proof written down in an alternative style, which might be helpful. (See, e.g., EWD1300 for more details on this proof style.)$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\inf}[1]{\text{inf}(#1)} \newcommand{\sup}[1]{\text{sup}(#1)} $

First, let's make our definitions explicit. The simplest definitions of $\;\inf{\cdot}\;$ and $\;\sup{\cdot}\;$ I know are \begin{align} \tag{0} z \leq \inf{A} \;\equiv\; \langle \forall a : a \in A : z \leq a \rangle \\ \tag{1} \sup{B} \leq z \;\equiv\; \langle \forall b : b \in B : b \leq z \rangle \end{align} for any real $\;z\;$, lower-bounded non-empty $\;A\;$ and upper-bounded non-empty $\;B\;$.


Armed with these definitions, let's try to lower-bound both sides, starting with the most complex side, so the right hand side: for any $\;z\;$,

$$\calc z \leq -\sup{-A} \op=\hints{negate both sides and swap them}\hint{-- preparing for the definition of $\;\sup{\cdot}\;$} \sup{-A} \leq -z \op=\hints{definition $\ref{1}$}\hints{-- allowed because $\;-A\;$ is upper-bounded and non-empty,}\hints{because $\;A\;$ is lower-bounded and non-empty}\hint{by the definition of $\;-A\;$} \langle \forall b : b \in -A : b \leq -z \rangle \op=\hint{definition of $\;-A\;$} \langle \forall b : -b \in A : b \leq -z \rangle \op=\hint{substitute $\;a := -b\;$; simplify using arithmetic} \langle \forall a : a \in A : z \leq a \rangle \op=\hints{definition $\ref{0}$}\hint{-- allowed because $\;A\;$ is lower-bounded and non-empty} z \leq \inf{A} \endcalc$$ In other words, $\;-\sup{-A}\;$ and $\;\inf{A}\;$ have the same lower bounds, and therefore they are equal.


This proof uses the following principle: $$ x = y \;\equiv\; \langle \forall z :: z \leq x \;\equiv\; z \leq y \rangle $$

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your method was good but you didn't check it . i try to proof it well : ( sorry my english is not well because i am not a english person ) $a=inf(A)$ so $a\le x$ for all $x\in A$. Therefore, $−a\le −x$ for all $-x\in -A$.

Now let $b$ be the upper bound of $−A$. There exists $b\ge−x$ $−b\le x$ for all $x\in . A$. Hence, $−b−a−a≤a≤b=−inf(A)=sup(−A)$ By multiplying $−1$ on both sides, we get that $inf(A)=−sup(−A)$

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Correct me if wrong:

$A\subset \mathbb{R}$, and $cA=${$y|y=cx, x \in A$}.

Let $c <0$.

Show that $\sup (cA)= c \inf (A)$.

Then

$cx \le M$ $\iff$ $x \ge M/c$, i.e.

$M$ is an upper bound of $cA$

$ \iff$

$M/c$ is a lower bound of $A$.

1)Let $M= \sup (cA)$, then

$x \ge \sup (cA)/c$, $x \in A$.

Hence $\inf A \ge \sup (cA)$. (Definition of $\inf A$).

2) Let $M/c =\inf A$, then

$cx \le M = c \inf A$.

Hence $\sup (cA) \le c \inf A$ . (Definition of $\sup$)

Finally 1) and 2):

$\sup (cA)=c \inf A$.

($c=-1$ in the original problem)

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