Solutions

We have that $-A \ne \varnothing$ because $A \ne \varnothing$ and there exists a mapping $f : A \to -A$ defined as $f(a) = -a$.

Let $\lambda$ be a lower bound of $A$. Also, let $-a \in -A$. It follows that $a \in A$ and

$$\lambda \le a \implies -a \le -\lambda.$$

Hence, $-A$ is bounded above.

We prove that $-\sup(-A)$ is the greatest lower bound of $A$ by contradiction. If $-\sup(-A) > \inf(A)$, then for $-a \in -A$,

$$-a \le \sup(-A) < -\inf(A)$$

and

$$\inf(A) < -\sup(-A) \le a,$$

which contradicts the definition of $\inf(A)$ since $a \in A$. Alternatively, if $-\sup(-A) < \inf(A)$, then for $a \in A$,

$$-\sup(-A) < \inf(A) \le a$$

and

$$-a \le -\inf(A) < \sup(-A),$$

which contradicts the definition of $\sup(-A)$ since $-a \in -A$.

b) Since $A$ is bounded below and $B$ is the set of all lower bounds, $B \ne \varnothing$. Also, $B$ is bounded above because otherwise there would exist $b \in B$ and $a \in A$ such that $a \le b$, which would contradict the definition of $B$.

Again for contradiction, suppose $\sup B \ne \inf A$. If $\sup B < \inf A$, then there exists $c$ such that

$$\sup B < c < \inf A.$$

But $c$ is then a lower bound of $A$ and thus belongs to $B$, which contradicts $\sup B < c$.

Likewise, if $\sup B > \inf A$, then there exists $b \in B$ such that

$$\inf A < b < \sup B.$$

But by definition $b$ is a lower bound of $A$, which contradicts $\inf A < b$.