Verification of proof that $\sup(A) = - \inf(-A)$

Question

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 >− is not a lower bound of −.

Since >−, −< and thus − is not an upper bound of (by definition of =sup()). Thus there is ∈ so that −<. So >− and thus is not a lower bound of −.

The result follows.

The supremum of a set does not need to be an element of the set, so you are not justified in saying "since $\alpha \in A$", nor do you need that for the proof. — Joe, Jul 19 '21 at 11:07
Please, don't edit the post after receiving the answer.... now you make me looks like an idiot pointing out the mistake that you didn't make.... — Arctic Char, Jul 19 '21 at 11:22
Apologies. I believe the result is correct now and I will accept your answer. — learningmathematics, Jul 19 '21 at 11:26

score 3 · Accepted Answer · answered Jul 19 '21 at 11:07

3

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 $-A$.

answered Jul 19 '21 at 11:07

Arctic Char

16,972

Thanks. I have edited my post to include what you have said. But don't I still need to have $y \in -A$ to show that $- \alpha$ is a lower bound for $-A$? – learningmathematics Jul 19 '21 at 11:21
1

Yes, there's no mistake there. @learningmathematics – Arctic Char Jul 19 '21 at 11:26

Verification of proof that $\sup(A) = - \inf(-A)$

1 Answers1

Linked