Let be $A,B \subseteq \Bbb{R}$, with $A \neq \emptyset $ and $B \neq \emptyset$ and $\Bbb{R}$ totally ordered by $\leq$, and $c,d \in \Bbb{R}$, with $c=\max (A)$ and $ d=\max (B)$. I must proof by contradiction the following:
"$\max (A \cup B ) \in \{c,d\}$"
Proof by contradiction: if $\max (A \cup B ) \notin \{c,d\}$ then $\max(A\cup B) \neq c \wedge \max(A \cup B) \neq d$, therefore $ \forall x \in (A \cup B) ( x \nleq \max(A) \wedge x \nleq \max(B))$, but if $x \in A \cup B$ then $x \in A \vee x \in B$, and if $x \in A$ then, by hypothesis $x \leq \max(A)$ and I have an absurd with $x \nleq \max(A)$, if $x \in B$ then $x \leq \max(B)$ and I have an absurd with $x \nleq \max(B)$
Is correct?
Thanks in advance!
Suppose it weren't. Then there is an $x\in A\cup B$ such that $x>c$ and $x>d$, but $x$ is either in $A$ or $B$. Can you see why this is a problem?