Least upper bound of $\{x\in\mathbb{R}:x^2<2\}$

Question

I am currently studying the definition of the real numbers (i.e. the complete ordered field) and I want to make sure my understanding of the least upper bound is correct. Am I right to think that the least upper bound of $\{x\in\mathbb{R}:x^2<2\}$ is $\sqrt{2}$?

Answer 1

You certainly are correct. Since $A:=\{x\in\mathbb{R}:x^2<2\}=(-\sqrt{2},\sqrt{2})$ is a non-empty subset of $\mathbb{R}$ that is bounded above (by $\sqrt{2}$), it has a unique least upper bound $\text{sup}(A)$.

Since $\sqrt{2}$ is an upper bound, we must have $\text{sup}(A)\leq\sqrt{2}$. If $\text{sup}(A)<\sqrt{2}$, then there exists $x\in A$ with $x>\text{sup}(A)$. But since $x\in A$, we must have that $x\leq\text{sup}(A)$. Hence $\text{sup}(A)=\sqrt{2}$.

The same argument can be applied to show that for any $a,b\in\mathbb{R}$ with $a<b$, we have that $\text{sup}((a,b))=b$.