# Proof that $\inf A = -\sup(-A)$

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?

• +1. For showing your work. People appreciate if you show your work. And your answer is right. Good work! – user17762 May 15 '13 at 5:31
• how did you get $-b \leq \alpha$? – Real Hilbert Nov 2 '13 at 15:41
• – lhf Oct 31 '14 at 17:54
• "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. – Björn Lindqvist Jun 19 '18 at 0:09

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

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.)


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: $$x = y \;\equiv\; \langle \forall z :: z \leq x \;\equiv\; z \leq y \rangle$$

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 $inf(A)=−sup(−A)$

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)