Given $A \neq \emptyset$ and bounded above define $-A:= \{-x: x \in A\}$
I am trying to prove that $\sup(A) = – \inf(-A)$ I have prove already that the $\inf(A) = – \sup(-A)$.
Proof attempt:
Since $A$ is non-empty and bounded above, there exists $\alpha = \sup(A)$. Therefore, for all $x \in A, \alpha \geq x$. if $x \in A$ and $x < \alpha$ then $x$ is not an upper bound for $A$.
I claim that $- \alpha = \inf(-A)$.
Let $y \in -A$ then there exists $x \in A$ such that $-x = y$. Then $\alpha \geq x, -\alpha \leq -x = y$. Thus $y \geq – \alpha$ for all $y \in -A$. Therefore $-\alpha$ is a lower bound for $-A$.
We need to show that any real number 𝑦>−𝛼 is not a lower bound of −𝐴.
Since 𝑦>−𝛼, −𝑦<𝛼 and thus −𝑦 is not an upper bound of 𝐴 (by definition of 𝛼=sup(𝐴)). Thus there is 𝑥∈𝐴 so that −𝑦<𝑥. So 𝑦>−𝑥 and thus 𝑦 is not a lower bound of −𝐴.
The result follows.
Best Answer
Several mistakes: (1) the supremum $\alpha$ might not be in $A$: think of $A = (0,1)$. (2) You need to show that any real number $y >-\alpha$ is not a lower bound of $-A$. But instead you assumed $y\in -A$.
The correct argument is that, since $y >-\alpha$, $-y < \alpha$ and thus $-y$ is not an upper bound of $A$ (by definition of $\alpha = \sup (A)$). Thus there is $x\in A$ so that $-y <x$. So $y>-x$ and thus $y$ is not a lower bound of $-A$.