Find the infimum (with proof) of the set $ X = \left \{ x \in \mathbb{R} : x^{2} < 3 \right \} $

Question

I know the answer is $-\sqrt{3}$ but I don't know how to give a formal proof.

Thanks in advance!

Answer 1

Notice that

$$ x^2 < 3 \Rightarrow (x+\sqrt{3})(x-\sqrt{3}) < 0 \Rightarrow -\sqrt{3} < x < \sqrt{3}.$$

Thus, $-\sqrt{3}$ is the infimum of $X$ because for any $y > -\sqrt{3}$ there exists a $x \in X$ such that $x < y$.

Answer 2

The infimum is the greatest lower bound.

Show it is a lower bound: if $x < -\sqrt 3$, then $x^2 > 3$ so $x \not\in X$.

Show it is the greatest lower bound: if $y$ is a lower bound such that $y > -\sqrt 3$, then for all $x \in X$, $-\sqrt 3 < y \leq x$. Let $x = \dfrac{y - \sqrt{3}}{2} \in (-\sqrt 3,y) \subset X$ to get a contradiction.