Proof on infimum and supremum

Question

Sorry for the poor photo quality. Can someone please tell me does my proof of this question valid or not?

Answer 1

Here's a way to prove the statement:

Show that $-\sup(A)$ is a lower-bound for $-A$

For all $a \in A$, $a \leq \sup(A)$ (by definition). Thus, $-a \geq -\sup(A)$.

Since $-a \geq \sup(A)$ for all $a \in A$, $-\sup(A)$ is a lower-bound for $-A$.

Show that $- \sup(A)$ is the greatest lower-bound for $-A$. That is, show that any other lower bound must be less than (or equal to) $-\sup(A)$.

Let $\alpha$ be an arbitrary lower bound of $-A$. Then, for all $a \in A: -a \geq \alpha$. Thus, for all $a \in A, a \leq -\alpha$. Thus, $-\alpha$ is an upper bound for $A$.

However, $\sup(A)$ is the least upper bound of $A$. Thus, $-\alpha \geq \sup(A)$. Thus, $\alpha \leq -\sup(A)$.

Thus, every lower bound $\alpha$ of $-A$ satisifies $\alpha \leq -\sup(A)$. Thus, $-\sup(A)$ is the greatest lower bound of $-A$.