# Determine whether the given set is bounded and find its supremum and its infimum.

Let $$A = \{x \mid \Bbb R \mid x^2 \lt 2\}$$. Is the set $$A$$ bounded? If yes, find its supremum and its infimum.

Attempt: Notice that \begin{align*} A &= \{x \in \Bbb R \mid x^2 \lt 2 \} \\ &= \{x \in \Bbb R \mid x^2 - 2 \lt 0 \} \\ &= \{x \in \Bbb R \mid (x - \sqrt{2})(x + \sqrt{2}) \lt 0 \} \\ &= \{x \in \Bbb R \mid -\sqrt{2} \lt x \lt \sqrt{2} \} \\ &= (-\sqrt{2},\sqrt{2}). \end{align*} Hence, by definition, $$A$$ is bounded. Now, I claim $$\sup A = \sqrt{2}$$ and $$\inf A = -\sqrt{2}$$ which is true, indeed. For the supremum, let $$M$$ be the another upper bound of $$A$$ and suppose $$M \lt \sqrt{2}$$. Then, since $$A$$ is bounded, there exist an element of $$A$$, say $$r$$, such that $$M \lt r$$. A contradiction with the definition of an upper bound, that's $$M$$. Hence, we must have $$\sqrt{2} \le M$$ i.e. $$\sup A = \sqrt{2}$$.

Now, for the infimum, let $$m$$ be the another lower bound of $$A$$ and suppose $$-\sqrt{2} \lt m$$. Then, since $$A$$ is bounded, there exist an element of $$A$$, say $$s$$, such that $$s \lt m$$. A contradiction with the definition of a lower bound, that's $$m$$. Hence, we must have $$m \le -\sqrt{2}$$ i.e. $$\inf A = -\sqrt{2}$$.

Am I true ?

This is mostly right, but I think you should be a little more explicit about where you get the contradiction from. In particular, how do you know that there is an element $$r \in A$$ such that $$M? This doesn't come from the fact that $$A$$ is bounded, it's because $$M < \sqrt{2}$$.
• You could find an example of such an $r$ explicitly. For example, suppose $M< \sqrt{2}$ and let $r=\frac{M+\sqrt{2}}{2}$. Then $r^2 < 2$ so $r \in A$, but $M<r$ which contradicts the assumption that $M$ is an upper bound of $A$. Feb 27, 2021 at 2:10