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