Analysis proof writing direction, tips or pointer (sup(A union B) smaller or equal min(supA, sup B)

Question

I've been working on a proof and I'm really trying to get the hang of writing rigorous proofs on my own. Therefore, even if intuitively I understand the statement, I feel like I can't find a simple and concise way to write it. So, I was wondering if someone could maybe give me pointers about where to start, maybe a trick or something or a point I'm missing without proving it for me.

So here is what I got so far:

Question:

Prove that if $A,B\subset \mathbb{R}$ are two bounded sets $ (A,B \neq \emptyset)$, then $\sup(A\cap B) \leq \min\{\sup A, \sup B\}$

Proof

Suppose $A$ and $B$ are two nonempty bounded subsets of $\mathbb{R}$

Then by definition of bounded set we have that

\begin{aligned}& \text{(i) }\exists \alpha \in \mathbb{R} : \forall a \in A, a \leq \alpha \\& \text{(ii) if }M \in \mathbb{R}\text{ is an upperbound of }A, \text{ then } M\geq \alpha \end{aligned}

and

\begin{aligned}&\text{(i) }\exists \beta \in \mathbb{R} : \forall b \in B, b \leq \beta \\& \text{(ii) if }N \in \mathbb{R}\text{ is an upperbound of }B, \text{ then } N \geq \beta \end{aligned}

and therefore

$$\exists \sup(A)\in \mathbb{R}, \sup(B) \in \mathbb{R}\\ \\ \\$$

Case 1:

If $A\cap B = \emptyset$ then, $\sup(A\cap B) = -\infty$ and since

$$\exists \sup(A)\in \mathbb{R}, \sup(B) \in \mathbb{R}\\ \\ \\$$

we have that

$\sup(A\cap B) = -\infty < x, \forall x \in \mathbb{R}$ and in particular,

$\sup(A\cap B) = -\infty < min(\sup A, \sup B)\\$

Case 2

Suppose $A\cap B \neq \emptyset$, then,

Since, $A \cap B := \left \{ x\in \mathbb{R} : x \in A \text{ and } x\in B \right \}$, then

$A\cap B \subset \mathbb{R}$

Moreover, since $\mathbb{R}$ has an order structure "<", we have that

(1) $\forall x \in A \cap \ B, x \leq a \leq b$ or $ x \leq b \leq a, \forall a \in A, \forall b \in B$

here I'm not so sure about this argument, is it enough with the definition of the set or do I need to explain more?

we have to show that in any cases, min(sup A, sup B) is an upperbound of $A \cap B$ which means that $sup(A \cap B)$ will have to be at least equal to the min of the supremums. By the thricotomy axiom, only one of those three cases are possible for any $x\in \mathbb{R}$

• if $sup A = sup B$, then by (1) and the definition of supremum: $\forall x \in A \cap B, x \leq sup A = sup B = min(sup A, sup B)$
• if $sup A < sub B$, then then by (1) and the definition of supremum: $\forall x \in A \cap B, x \leq sup A = min sup(A\cap B) $
• if $sup A > sub B$, then then by (1) and the definition of supremum: $\forall x \in A \cap B, x \leq sup B = min sup(A\cap B) \text{ }\square$

So, here is what I came with. I feel like my argument is more confusing than anything.

Plus, I'm not even sure if everything I say is true.

So, if someone could, without giving me the awnser, point out some flaws, tips, or directions, it would be very much appreciated.

Thank you!

Answer 1

These sorts of proofs about sups and infs are generally pretty brief if you look at them the right way -- no need to argue by cases.

Hint: Show that if $X$ and $Y$ are sets with $X\subset Y$, then $\sup X\le \sup Y$ (You may already know this.) Now apply this hint twice. If you need another hint , spoiler below:

If $a\le b$ and $a\le c$, then $a\le\min(b,c)$.