If $A_1,A_2,\ldots,A_n$ are right bounded (left bounded) in an ordered field $F$, then so is their union Prove that if $A_1,A_2,\ldots,A_n$ are right bounded (left bounded) in $F$, then so
is
$$\bigcup_{k=1}^n A_k.$$
 A: As you can, for the practical purposes of the exercise, treat an ordered field as if it were the reals, this isn't very hard:  
If $A_i$ is bounded to the right by some $k_i$, then for all $x \in A_i$ we have 
$$x \leq k_i. \quad (1)$$
Next, let 
$$k:= \max \{k_i \} , \quad (2)$$ 
and note that for any $x \in \cup A_i$, there is an $i$ such that 
$$x \in A_i. \quad (3)$$
Hence, you have  
$$\begin{align}
x & \leq k_i \quad \text{ by (1) and (3)} \\
& \leq k \quad \text{ by (2)}\\ 
\end{align}$$
As $x$ was any element of $\cup A_i$, this was to be shown. 
A: Think of this geometrically first. It is given that $A_1$ is right bounded, so it lies to the left of some value, $k_1$. Similarly, $A_2$ lies to the left of some value $k_2$ and so on, but for the sake of intuition suppose there are only three sets (so the case $n=3$). If you draw it schematically, it is now clear that the union lies entirely to the left of the largest of the three values $k_1,k_2,k_3$. This indicates to you the general strategy, and the proof. 
Denote for every $A_i$ an upper bound for it by $k_i$. Thus, $a\le k_i$ for all $a\in A_i$. Take $M$ to be the maximum among $\{k_1,\cdots, k_n\}$, thus $k_i\le M$ for all $1\le i\le n$. Now, it follows that $M$ is an upper bound of $\bigcup_{i}^nA_i$, since for every $a\in \bigcup_{i}^nA_i$ holds that $a\in A_i$  for some $1\le i\le n$, and thus that $a\in k_i\le M$, so $a\in M$. 
Since the proof above only uses axioms of ordered fields, the general claim is proven. (You might want to figure out precisely which axiom is used where in the arguments above, to make sure you understand it). 
