Below is the proof I wrote. Can you please let me know if this is correct? Also, any suggestions for improving my proof writing style would be appreciated (I am new to proofs and still learning):  Without loss of generality, assume $\sup (A) \geq \sup (B)$. We know $\forall \ b \in B,\ \sup(B) \geq b$ $\implies \ \sup (A) \geq b \implies \sup (A)$ is an upper bound for $B$ (and for $A$ by definition). Now we know that $c \in A \cup B \iff c \in A \text{ or } c \in B \implies \not\exists \ c \in A \cup B \ \text{  s.t.  } \ c &gt; \sup (A)$. Thus, we have established that $\max\{\sup(A), \sup(B) \}$ is an upper bound for $A \cup B$. It also follows that $\sup (B)$ is not an upper bound for $A \cup B$ if $\max \{ \sup (A),\sup (B) \} \neq \sup (B)$ because this implies that $\exists \ a \in A \in A \cup B \text{ s.t } a&gt;\sup (B)$.   We now show that  $\max\{\sup(A), \sup(B) \}$ is the least upper bound for $A \cup B$. Suppose $ \sup(A \cup B) &lt;  \max\{\sup(A), \sup(B) \} = \sup(A)$. This is a contradiction since $A \subseteq A \cup B \implies \sup (A) \leq \sup (A \cup B)$.  Therefore, $\sup (A \cup B) = \max \{ \sup(A), \sup (B) \}$.
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?
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