Supremum and union $A_i \subset \mathbb{R}, i \in \mathbb{N}$
Is $$ \sup \{\sup A_i, i \in \mathbb N\} = \sup (\cup_i A_i)?$$
Thanks.
 A: Yes. 
Denote $A=\bigcup A_{i}$. Then for every $i$ we have $\sup A_{i}\leq\sup A$
as a consequence of $A_{i}\subseteq A$ so that $\sup\sup A_{i}\leq\sup A$.
Conversely if $x<\sup A$ then $x<a$ is true for some $a\in A$. Let's say that $a\in A_{i_{0}}$. 
Then $x<a\leq\sup A_{i_{0}}\leq\sup\sup A_{i}$ and
since this is true for every $x<\sup A$ we are allowed to conclude
that $\sup A\leq\sup\sup A_{i}$.
A: Firstly, since $\cup_i A_i \supset A_i$, we have $$\sup\cup_i A_i \geq  \sup A_i$$
then take $\sup$ again on rhs, we have $$\sup\cup_i A_i \geq  \sup\{\sup A_i, i \in \mathbb{N}\} \tag 1$$
To prove the equality, take any $\epsilon >0$, then by definition,  $\exists x \in \cup_i A_i$ such that $$x + \epsilon > \sup\cup_i A_i$$ 
Suppose $x \in A_{i_0}$, then $$\sup A_{i_0} + \epsilon \geq x + \epsilon > \sup\cup_i A_i $$
which means for any $\epsilon >0$, we can find one element in $\{\sup A_i, i \in \mathbb{N}\}$ such that this element plus $\epsilon$ is greater than $ \sup\cup_i A_i$. Combined with $(1)$, we've proven $$\sup\cup_i A_i = \sup\{\sup A_i, i \in \mathbb{N}\} $$
A: Every upper bound of $\bigcup A_i$ is an upper bound of $\{\sup A_i, i\in\Bbb N\}$. Indeed, if some $x$ is an upper bound of every $A_i$, then $x\geq \sup A_i$ for all $i$
Conversely, every upper bound $x$ of $\{\sup A_i, i\in\Bbb N\}$ is an upper bound of each $A_i$, that is, it is greater than or equal to every element of every $A_i$. Then $x$ is an upper bound of $\bigcup A_i$.
But note that for all of this makes sense, the set $\bigcup A_i$ must be upper bounded (unless you consider that the sup of a non bounded set is $\infty$). Note also that the set of indexes can be uncountable. The fact that the set of indexes is $\Bbb N$ is not used anywhere.
