Completeness Axiom Proof Let $A_1$, $A_2$, $A_3$... be a collection of nonempty sets, each of which is bounded above. 
(a.) Find a formula for sup($A_1$$\bigcup$$A_2$). Extend this to sup($\bigcup^{n}_{k=1}$$A_k$). 
(b.) Consider sup($\bigcup_{k=1}^{\infty}$$A_k$). Does the formula in (a) extend to the infinite case?  
For (a), I know that since each set A is bounded above, they all have a supremum. For (a) I want to say that sup($A_1$$\bigcup$$A_2$) would just be the greater supremum of the two sets $A_1$ and $A_2$?
Then for (b) I don't think it would extend to the infinite case, but not sure how to prove this
 A: For $(a)$ you're correct. Now, how would you prove this? You need to show that if $\sup A_1=x$ and $\sup A_2=y$, then 


*

*$\max\{x, y\}$ is $\ge$ ever element of $A_1\cup A_2$, and

*if $z$ is $\ge$ every element of $A_1\cup A_2$, then $z\ge \max\{x, y\}$.
Can you figure out why each of these statements is true?

For $(b)$, it might be best to try some examples - families of sets $(A_i)_{i\in\mathbb{N}}$ where each $A_i$, and the whole union $B=\bigcup A_i$, are easy to visualize. For instance, what if we have $A_i=[i, i+1)$? What if we have $A_i=[{i\over i+1}, {i+1\over i+2})$? Is there always a greatest sup? Is the sup of the union somehow "built out of" the sups of the $A_i$s?
A: For reference to other folks, this is Exercise 1.3.4 in Stephen Abbot's Understanding Analysis book.
Now, I believe the answer to (b) is "yes", IF it exists. And here's how it would generalize:
Let $S = \{x \in \mathbb{R} | \exists k \in \mathbb{N} (x = \sup A_k) \}$. Then $\sup S = \sup \bigcup_{k = 1}^{\infty} A_k$.
We do this by proving that each is an upper bound of the other "underlying set". This would show that $\sup S \geq \sup \bigcup_{k = 1}^{\infty} A_k$ and $\sup \bigcup_{k = 1}^{\infty} A_k \geq \sup S$.
First, let $s_1 = \sup S$ and $s_2 = \sup \bigcup_{k = 1}^{\infty} A_k$. Then $s_1 \geq \sup A_k$ for all $k \in \mathbb{N}$. Thus, for arbitrary $a_k \in A_k$, $s_1 \geq a_k$. Then, for arbitrary $z \in \bigcup_{k = 1}^{\infty}$, $z \in A_k$ for some $k \in \mathbb{N}$. So $s_1 \geq z$ , which means that $s_1$ is an upper bound of $\bigcup_{k = 1}^{\infty} A_k$. Thus, $s_1 \geq s_2$.
Now, since $s_2 \geq z$ for all $z \in \bigcup_{k = 1}^{\infty} A_k$, then in particular $s_2 \geq z$ for all $z \in A_k$ for arbitrary $k \in \mathbb{N}$. Therefore, $s_2$ is an upper bound of $A_k$, so $s_2 \geq \sup A_k$. Thus, $s_2 \geq y$ for all $y \in S$, so $s_2$ is an upper bound of $S$, which implies $s_2 \geq s_1$.
Note that the proof requires the existence of $s_1$.
