Let $A = [1,3) \cup (5,8)$. Find the supremum and infimum of $A$.
Firstly, I claim that $\sup A = 8$ and $\inf A = 1$. Now, by definition, $8$ is the upper bound of $A$. My next step is use this theorem, but I think there is a simple proof of this question without any theorem:
Let $A$ and $B$ be bounded subsets of $\Bbb R$. Then, $\sup (A \cup B) = \sup\{\sup A, \sup B \}$ is holds.
That is, I divide the set $A$ into two sets, say $A_1 = [1,3)$ and $A_2 = (5,8)$, then I prove that $\sup A_1 = 3$ and $\sup A_2 = 8$, and then use the above theorem, and done.
But, my question is, does there exist a simple proof that without any theorem? How about the infimum proof? Does there exist theorem that says a similar argument same as the supremum theorem above?
Thanks in advanced.
EDIT: I've did the approach of proof of supremum as @zkutch post below. Please correct me if I wrong. Here it is:
Claim that $\sup A = 8$. By definition, $8$ is the upper bound of $A$. We want to prove that $8$ is the least upper bound of $A$. Suppose for contradiction, that $8$ is not the least upper bound of $A$. Then, there exist a number $\epsilon \gt 0$ such that $8-\epsilon$ is also an upper bound of $A$. To contradict this, we exhibit $x \in A$ such that $8-\epsilon \lt x \lt 8$. Since $0 \lt \frac{\epsilon}{2} \lt \epsilon$, we can see that $x = 8 - \frac{\epsilon}{2}$ satisfies $8-\epsilon \lt x \lt 8$.
Now, since $8-\epsilon$ is (by assumption) a upper bound of $A$ and $5+\epsilon \in A$, then we have $5+\epsilon \le 8-\epsilon$, showing that $x \in A$. Thus, $8-\epsilon$ is not an upper bound of $A$, a contradiction. Hence, $\sup A = 8$.