Let $A = \{x \in \mathbb{Q} \mid x>0, x^2<2\}$ and $B=\{x \in \mathbb{Q} \mid x>0, x^2>2\}$ are subsets of the ordered set $\mathbb{Q}$.
- Prove that upper bounds of $A$ are exactly the members of $B$.
- What is the set of lower bounds of $B$?
My thoughts: $A = \{x \in \mathbb{Q} \mid x>0, x^2<2\} = (0, \sqrt{2}) \cap \mathbb{Q}$ and $B=\{x \in \mathbb{Q} \mid x>0, x^2>2\} = (\sqrt{2}, +\infty)\cap\mathbb{Q}$.
Can we say that upper bounds of $A$ are exactly the members of $B$? Because $\pi \not\in B$ but $\pi$ is an upper bound of $A$.
How to proceed further? Please help.
I believe that the question requires you to restrict your universe to $\mathbb{Q}$. Suppose that $x\in \mathbb{Q}$. Then, either $x^2>2$ or $x^2<2$. If $x^2<2$, then for $n\in \mathbb{N}$ satisfying $n>\frac{2|x|+1}{2-x^2}$, we have that $x+\frac{1}{n}\in A$. It is also easy to see that if $x\leq 0$, then $x$ cannot be an upper bound of $A$. This implies that no element in $\mathbb{Q}\setminus B$ can be an upper bound of $A$. Particularly, no element of $A$ can be an upper bound of $A$. If $x\in B$, for any $y \in A$, we have that $x^2-y^2>0$. This implies that $x-y>0$ and hence $x$ is an upper bound of $A$. Thus, we have shown that $\{x\in \mathbb{Q}| x\text{ is an upper bound of }A\}=B$.