$\sup$ and $\inf$ of $E=\{p/q\in\mathbb{Q}:p^2<5q^2 \text{ and } p, q > 0\}$

Question

I'd appreciate if you could please check to see if my proof is valid.

Find $\sup$ and $\inf$ of $E=\{p/q\in\mathbb{Q}:p^2<5q^2 \text{ and } p, q > 0\}$.

Solution:

$q^2 > p^2/5 \iff q > p/\sqrt{5} \iff 0<p/q<\sqrt{5} $. $\implies E = \mathbb{Q}\cap [0, \sqrt{5})$.

(i) Supremum. Since E is bounded, there exists an $M\ge y$ for all $y\in E$. Since $\sqrt{5}>y$ for all $y\in E$, $\sqrt{5}$ is an upper bound. But $\sqrt{5}\not\in E$. Suppose $s:=\sup E\ne \sqrt{5}$. Then for some $\varepsilon>0$, $\exists y_0\in E$ such that $s-\varepsilon <y_0 \le \sup E < \sqrt{5}$. Thus $\sqrt{5}-s>0$. Set $b:=\frac{1}{\sqrt{5}-s}\in \mathbb{R}$. By Archimedean Principle, $1\cdot n > \frac{1}{\sqrt{5}-s}\iff s < \sqrt{5} - 1/n < \sqrt{5}$. But $s\ge \sqrt{5}-1/n$ since $\exists y_1 \in E$ such that $\sqrt{5}-1/n < y_1 < \sqrt{5}$ (by density of rationals), and $y_1\le s<\sqrt{5}$. We thus arrive at a contradiction, hence $\sup E=\sqrt{5}.$

(ii) Infimum. There exists a lower boundary $L$, so that $L\le y$ for all $y\in E$. In particular, $0\ge L$, and $0\in E$, hence $0=\inf E$.

Answer 1

There is a mistake in part (ii): as pointed out in the comments, $0$ is not in $E$.

What you wrote in part (i) is true, but more complicated than it needs to be. Here are a few suggestions to simplify your proof:

1. Delete the first sentence. The variable $M$ is not used anywhere in the rest of the proof.
2. Delete the sentence that starts with "Then for some $\varepsilon > 0 \dots$". Your argument makes no use of $\varepsilon$ and $y_0$.
3. If you know that the rationals are dense in the reals, you can deduce directly from the inequality $\sup E < \sqrt{5}$ that there exists $y \in \mathbb{Q}$ such that $\sup E < y < \sqrt{5}$. There is no need to use the Archimedean principle.