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

Question

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 ?

Answer 1

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 $M<r$? This doesn't come from the fact that $A$ is bounded, it's because $M < \sqrt{2}$.