I'm teaching myself calculus and stumbled upon a series of problems that confuse me. Can anybody be so kind and show me an exemplary approach to this exercise? Thanks a lot!
Show for an index set $I$ and sets $A_i \subseteq \mathbb{R}$ for $i \in I$:
(a)
If $\bigcup_{i\in I} A_i$ is bounded from above, we have
sup$\Big(\bigcup_{i\in I} A_i\Big) =$ sup{sup $A_i: i\in I$}.
(b)
If $\bigcap_{i\in I} A_i \neq \emptyset$, we have sup$\Big(\bigcap_{i\in I} A_i \Big) \leq$ inf{sup $A_i: i \in I$}.
(c)
If $\emptyset \neq A \subseteq \mathbb{R}$ is bounded from above, we have sup $A = -$inf$(-A)$, with $-A = $ {$-a:a\in A$}.
a) For any $x\in\displaystyle\bigcup_{i\in I}A_{i}$, then there exists some $i\in I$ such that $x\in A_{i}$, so $x\leq\sup A_{i}$, and hence $x\leq\sup\{\sup A_{i}: i\in I\}$, this shows $\sup\left(\displaystyle\bigcup_{i\in I}A_{i}\right)\leq\sup\{\sup A_{i}: i\in I\}$.
Fix an $i\in I$, for any $x\in A_{i}$, we have $x\in\displaystyle\bigcup_{i\in I}A_{i}$, so $x\leq\sup\left(\displaystyle\bigcup_{i\in I}A_{i}\right)$, this proves at first $\sup A_{i}\leq\sup\left(\displaystyle\bigcup_{i\in I}A_{i}\right)$, and then $\sup\{\sup A_{i}: i\in I\}\leq\sup\left(\displaystyle\bigcup_{i\in I}A_{i}\right)$.
b) Try to argue that for $A\subseteq B$, one has $\sup A\leq\sup B$.
c) Try to argue by $\sup A\leq-\inf(-A)$ and $-\inf(-A)\leq\sup A$. For example, if $x\in A$, then $-x\in-A$, hence $-x\geq\inf(-A)$, so $x\leq-\inf(-A)$, this shows $\sup A\leq-\inf(-A)$.