Prove that $\sup⁡(A\cup B)=\max\{\sup⁡(A),\sup⁡(B)\}$

Question

I want to prove $\sup⁡(A\cup B)=\max\{\sup⁡(A),\sup⁡(B)\}$, where $A,B\subset \mathbb R$ are non-empty and bounded sets from above. I have reviewed similar questions and answers, but I intended to prove directly using the following definition of least upper bound:

$\sup ( A) \in \mathbb R$ and $A\subseteq \mathbb R$ if

1. $\forall a\in A$, $a \leq \sup( A)$,
2. $\forall \varepsilon >0$, $\exists b \in A$ such that $\sup( A)-\varepsilon<b$.

Here's my proof:

1. We know $A \cup B=\{x \mid x\in A \text{ or }x\in B \} $,
2. $\forall a \in A$, $a \leq \sup⁡(A) $ and $\forall b \in B$, $b \leq \sup⁡(B) $, thus $\forall a \in A$ and $\forall b \in B$, $a,b\leq \max\{\sup⁡(A),\sup⁡(B)\}$.
3. $\forall \varepsilon>0 $, $\exists k_A \in A$ s.t. $\sup( A)-\varepsilon <k_A$ and $\exists k_B \in B$ s.t. $\sup( B)-\varepsilon <k_B$. $k_A\in A\cup B$, and $\sup(A)\geq \sup(B)$ implies $\sup(A)-\varepsilon=\max\{\sup⁡(A),\sup⁡(B)\}-\varepsilon <k_A$. After considering vice versa, we can write, $\exists k\in A\cup B$, $\max\{\sup⁡(A),\sup⁡(B)\}-\varepsilon<k$.

Since both properties are satisfied, we can conclude that $\sup⁡(A\cup B)=\max\{\sup⁡(A),\sup⁡(B)\}$.

Question: Is this proof correct? I have some bad feelings regarding my 3rd argument. Thank you!

Answer 1

Your proof looks fine after following the suggestions in the comments. But you may consider writing it in plain English to make it easier to understand.

Let $A, B \subseteq \mathbb{R}$ be bounded from above; i.e., there is some $x \in \mathbb{R}$ such that $a < x$ for all $a \in A$ and some $y \in \mathbb{R}$ such that $b < y$ for all $b \in B$. Let $C = A \cup B$.

The set $C$ must be bounded from above: take $z = \max\{a, b\} \in \mathbb{R}$; for any $c \in C$, if $c \in A$, then $c < x \le z$ and if $c \in B$, then $c < y \le z$.

Denote $m = \max\{\sup A, \sup B\}$; we want to show that $m = \sup C$. Then (1) $m$ needs to be an upper bound of $C$, and (2) $m$ cannot be larger than any upper bound of $C$ (equivalent to the second part of your definition).

To show (1) is easy: let $c \in C$ be arbitrary. If $c \in A$, then $c \le \sup A \le m$ and if $c \in B$, then $c \le \sup B \le m$. Either way, $c \le m$, as desired.

To show (2), suppose for the sake of contradiction that there is some $n < m$ that is an upper bound of $C$. Then $n < \sup A$ or $n < \sup B$. In the former case, there is some $a \in A$ such that $a > n$; since $a \in C$, this contradicts the fact that $n$ is an upper bound of $C$. A similar argument applies if $n < \sup B$. Therefore, $m$ must be the smallest upper bound of $C$, completing the proof.

Answer 2

Option:

$1)A \subset (A\cup B)$, and

$B\subset (A\cup B),$ hence

$\sup (A) \le \sup (A\cup B)$, and $\sup (B) \le \sup (A\cup B) $.

Thus

$\max ${$\sup (A) , \sup (B) $} $\le \sup (A\cup B).$

$2)a\le \max ${$\sup (A) , \sup (B) $}, and
$b\le \max ${$\sup (A) ,\sup (B) $},

Let $c \in (A \cup B),$ then
$c \le \max ${$\sup (A), \sup (B)$}, i e.

$\max ${$\sup(A),\sup (B) $}
is an upper bound for $c \in (A\cup B) $.

Since $\sup (A\cup B)$ is the least upper bound for {$c$}, we have

$\sup (A \cup B)$
$\le \max ${$\sup (A), \sup (B)$},

and we are done.