Let $A$ and $B$ be a non-empty subsets of $\mathbb{R}$ suppose that $a \leq b$ for every $a\in A$ and for every $b\in B$

Question

Let $A$ and $B$ be a non-empty subsets of $\mathbb{R}$ suppose that $a \leq b$ for every $a\in A$ and for every $b\in B$ prove that there is a real number $c$ such that $c \geq a$ for every $a \in A$ and $c\leq b$ for every $b \in B$

My idea:

given that $a \leq b$ by Archimedes property in $\mathbb{R}$ , $\exists n \in N$ such that

$n(b-a) \geq 1 \Rightarrow nb\geq na+1$

Also $\exists$ a unique integer $m $ such that $m-1 \leq na<m $

$\therefore nb \geq na+1 \geq n >na$

or $na \leq m \leq nb$ or $a \leq \frac{m}{n} \leq b $

Right?

Answer 1

Take $c=\sup A$. Then $c \in \mathbb R$, since any element of $B$ is an upper bound of $A$.

Now $a\leq c$ for all $a\in A$ by definition of $c$. Also, $c \leq b$ for all $b\in B$, because if $b<c$ for some $b \in B$, we have $c-b>0$ and by the definition of $c$ we can find $a \in A$ s.t

$$ a + (c-b) > c$$

so

$$a>b$$

Which is a contradiction with the assumption.

Answer 2

Take $ c= sup\{a : a \in A \} $