# Proof that infA=−sup(−A)\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,

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

Is my proof correct?

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