Let $I\ne\emptyset $, and $\forall_{i\in I}\ \emptyset\ne A_{i}\subseteq \Bbb R$, and $\forall_{i\in I}\exists\ c\in\Bbb R$ which is upper bound of $A_{i}$.
Prove that $sup\bigcup_{i\in I} A_{i}=sup\{supA_{i}, i\in I\} $.
My attempt:
Let $A=\bigcup_{i\in I} A_{i}$
$c=supA_{i}\ for \ i=1, \ or\ i=2, \ or\ ...\ or\ i=i$
$a\in A\Rightarrow a\in A_{i}\ for \ i=1, \ or\ i=2, \ or\ ...\ or\ i=i\ \Rightarrow a\le supA_{i}\le c $
Now I'm looking for $a$ such that $a>c-\varepsilon$.
By definition $c$ will be $supA$ if
$\forall_{\varepsilon>0}\exists\ _{a\in A}\ c-\varepsilon<a$
but a$\in A_{i}\subseteq A $ so $ a \in A$
Is it all?
I do not understand this: $c=supA_i$ for $i=1,$ or $i=2$, or ... or $i=i$.
I would do it like this:
If either of the sups is infinite, the result is immediate. Otherwise
let $A=\bigcup_{i\in I} A_{i},\ c_i=\sup A_{i}\ $ and $B=\{\sup A_{i}, i\in I\}.$
We need to prove that $\sup A=\sup B.$
Suppose $\alpha=\sup A$. Then, there is an $a_i\in A_i$ such that $\alpha-\epsilon<a_i.$ Since $a_i<c_i$, we have $\alpha-\epsilon<c_i\le \sup B\Rightarrow \sup A\le\sup B.$
Now suppose $\beta =\sup B$. Then, there is a $c_i\in B$ such that $\beta-\epsilon/2<c_i$. For this $c_i$ there is an $a_i\in A_i$ such that $c_i-\epsilon/2<a_i.$ It follows that $\beta-\epsilon<a_i\Rightarrow \sup B\le \sup A$