Let $A,B \subset \mathbb{R}$. Show that $\sup(A \cup B) = \max\{\sup A, \sup B\}$

Question

Let $A,B \subset \mathbb{R}$. Show that $\sup(A \cup B) = \max\{\sup A, \sup B\}$

Here's what I did so far:

Let $\sup A, \sup B$, and $\sup (A \cup B)$ denote upper bounds of $A,B$, and $A \cup B$, respectively. If $x \in A$, then $x \le \sup A$. If $x \in B$, then $x \le \sup B$. In either case, $x \in (A \cup B)$. So either $x \le \sup A$ or $x \le \sup B$. In either case, $x \le \max \{\sup A, \sup B\}$. So $\max \{\sup A, \sup B\}$ is the upper bound of $A \cup B$, which means $\sup (A \cup B) \le \max\{\sup A, \sup B\}$.

I do not know if there's a better way to go at it, but all I did was show the inequality. Can I do this, and show the reverse inequality? If so, how can I go about doing that?

Answer 1

Maybe you can reason it like this. Since $A \subset A \cup B \Rightarrow \sup(A)\leq \sup(A\cup B)$. Similarly, since $B \subset A \cup B \Rightarrow \sup(B)\leq \sup(A\cup B)$. Then $\sup(A\cup B)\geq \max\{\sup(A),\sup(B) \}$

Answer 2

What you did seems more than fine. You may want to add that $A$ and $B$ are non-empty, though.

The reverse is quite easy: $\sup A\leq \sup(A\cup B)$ and $\sup B\leq \sup(A\cup B)$, as the supremum can never decrease if you consider a larger set!