Show $ \inf A =\sqrt{5}$

Question

Let $A=\{x \in \mathbb{R}:x^2 > 5 \ \text{and}\ x \ \text{is a positive irrational}\}$. I want to prove that $\inf A =\sqrt{5}$. Since $\sqrt{7} \in A$, the set is non-empty and $x>\sqrt{5}$ so it is bounded below by $\sqrt{5}$. So $\inf A \geq \sqrt{5}$. Now I argue by contradiction, assume $\inf A>\sqrt{5}$. By the density of the irrationals in the reals, there exists an irrational $x_0$ in the interval $(\sqrt{5},\inf A)$ which then means that $x_0^2 >5$, so $x_0 \in A$ and $x_0 \leq \inf A$ which contradicts the fact that $\inf A$ is a lower bound of $A$. However, I am not sure if I can just refer to the density of irrationals, is there anything more sophisticated here?

Answer 1

Let us define a sequence $$a_n=\sqrt{5}+\frac1n.$$ Then $a_n\in A$ for all $n\in \mathbb{N}$. since $\inf A>\sqrt{5}$ and $a_n\to \sqrt{5}$, then $\inf A=\sqrt5$.