Proof that there exists a set in $\mathbb{Q}$ which is bounded above but has no least upper bound The following Lemma 3 is from my introductory analysis course notes.

Lemma 1. There is no rational number $q$ such that $q^2=2$.
Lemma 3. There exists a nonempty set $S\subseteq \mathbb{Q}$ which is bounded above in $\mathbb{Q}$, but has no least upper bound in $\mathbb{Q}$.
Proof. Let $S:=\{q\in\mathbb{Q}:q^2<2\}$. For $q\in S$, we have $-2<q<2$ so certainly $S$ is bounded above ($2$ is an upper bound in $\mathbb{Q}$ for $S$). To see that it has no least upper bound in $\mathbb{Q}$, we fix an upper bound $k=\frac{a}{b}\in\mathbb{Q}$ for $S$. We must show that there is an upper bound $l$ in $\mathbb{Q}$ for $S$ such that $l<k$. 
We first need to establish that $k^2>2$. To do this, we suppose $k^2\leq 2$ and seek a contradiction. By Lemma 1, we can not have $k^2=2$ so we must have $k^2<2$. So for every $N\in\mathbb{N}$, we have $\frac{N^2a^2}{N^2b^2}=\frac{a^2}{b^2}<2$. Let
  $$
\delta:=2-k^2>0
$$
  Fix $N\in\mathbb{N}$ such that
  $$
N\geq\max\left\{2a+1,\frac{3a}{b^2\delta}\right\}
$$
  Since $N\geq 2a+1$, we have $N^2\geq N(2a+1)\geq 2Na+1$. And since $N>\frac{3a}{b^2\delta}$, we have $N^2b^2\delta>3Na\geq 2Na+1$, which forces
  $$
\frac{2Na+1}{N^2b^2}<\delta.
$$
  Let $l:=\frac{Na+1}{Nb}$. Then
  $$
l^2=\frac{N^2a^2+2Na+1}{N^2b^2}=k^2+\frac{2Na+1}{N^2b^2}<k^2+\delta=2;
$$
  that is, $l$ belong to $S$. We also have $l=\frac{a}{b}+\frac{1}{Nb}=k+\frac{1}{Nb}>k$. So we have $l\in S$ and $k<l$, contradicting that $k$ is an upper bound for $S$ so $k^2>2$.
  Now let $\delta:=k^2-2>0$. Fix $N\in\mathbb{N}$ such that $N>\frac{2a}{b^2\delta}$, which implies that
  $$
\frac{2a}{Nb^2}<\delta=k^2-2.
$$
  Let $l:=\frac{Na-1}{Nb}$. Then
  $$
l^2=\frac{N^2a^2-(2Na-1)}{n^2b^2}=k^2-\frac{2Na-1}{N^2b^2}>k^2-\frac{2a}{Nb^2}>k^2-\delta=2.
$$
  Hence $l>\alpha$ whenever $\alpha^2<2$. That is, $l$ is an upper bound in $\mathbb{Q}$ for $S$. Moreover,
  $$
l=\frac{Na-1}{Nb}=\frac{a}{b}-\frac{1}{Nb}<k.
$$



*

*What is the correct way to think about "For $q\in S$, we have $-2<q<2$" part? where does $-2<q<2$ come from? I think because $q^2<2$, then it follows that $-\sqrt{2}<q<\sqrt{2}$ and so $-2<q<2$; so we could also have $-3<q<3$. Am I on the right track or there is a different way to look at this?

*I do not feel comfortable with the part "To see that it has no least upper bound in $\mathbb{Q}$, we fix an upper bound $k=\frac{a}{b}\in\mathbb{Q}$ for $S$. We must show that there is an upper bound $l$ in $\mathbb{Q}$ for $S$ such that $l<k$" I think it makes more sense (at least to myself) to change it into



To see that it has no least upper bound in $\mathbb{Q}$, we proceed by contradiction. Suppose that $S$ has the least upper bound $k=\frac{a}{b}$ in $\mathbb{Q}$. Without loss of generality, we can assume that $a,b\in\mathbb{N}$.



*Again I do not like the part "We first need to establish that $k^2>2$. To do this, we suppose $k^2\leq 2$ and seek a contradiction. By Lemma 1, we can not have $k^2=2$ so we must have $k^2<2$." I think I am more comfortable with something like



Since $k$ is the least upper bound for $S$, then there are two possibilities for $k$: either $k\in S$ or $k\not\in S$. If $k\in S$, then we have $k^2<2$ and if $k\not\in S$, then we have $k^2\geq 2$. We need to eaxmine three cases: (1) $k^2<2$ (2) $k^2=2$ (3) $k^2>2$. By Lemma 1, case 2 can not happen so we are left with two cases: either $k^2>2$ or $k^2<2$.



*I can not see how the part "So for every $N\in\mathbb{N}$, we have $\frac{N^2a^2}{N^2b^2}=\frac{a^2}{b^2}<2$" is even needed!

*I think the condition $N\geq\max\left\{2a+1,\frac{3a}{b^2\delta}\right\}$ is not needed; all we need is that $N>\frac{3a}{b^2\delta}$. I can not see where and how the inequality $N\geq 2a+1$ is used in the proof.

*I am not sure how the inequality $N^2b^2\delta>3Na\geq 2Na+1$ is derived. My initial thought is that since $N>\frac{3a}{b^2\delta}$, then $Nb^2\delta>3a$ and so $N^2b^2\delta>3Na=2Na+Na$. Since $a\in\mathbb{N}$, we then have $N^2b^2\delta>3Na=2Na+Na\geq 2Na+N\geq 2Na+1$.

 A: You have made very nice observations and your questions are genuine / well framed. Such a relief to see such level of quality from a new user. +1
To be frank the proof given in your book is needlessly complicated terrible. To answer your queries first we note that $q^2<2$ so $q^2<4$ and hence $-2<q<2$. We could equally well use $q^2<9$ or $q^2<100$ and get the bounds for $q$. But there is no need to bring irrationals like $\sqrt{2}$ in the picture. Looking at the exercise it appears to make use of the properties of rationals only and does not use anything beyond what is covered in a typical seventh grade syllabus.
Next one can proceed with your way also assuming that there is a least upper bound say $k$ for $S$. And yes we do need to consider the cases when $k\in S$ and $k\notin S$ and like you have described this ultimately leads you to the cases $k^2<2,k^2>2$. In case you have not noted $k>0$. You can now give a completely new argument which is much simpler than the one in your book.
Let's begin with $k^2<2$ and then we prove that there is an $l\in\mathbb {Q} $ such that $l>k$ and $l^2<2$ so that $l\in S$ and this contradicts that $k$ is least upper bound of $S$. To find $l$ just write $l=k+h$ where $h>0$ needs to be chosen suitably. We have $d=2-k^2>0$ and we want $l^2<2$ ie $(k+h) ^2<2$ ie $$k^2+2kh+h^2<2$$ ie $$h(2k+h)<2-k^2=d$$ If we choose $h<1$ then $h(2k+h)<h(2k+1) $ and this can be made less than $d$ if we have $h<d/(2k+1)$. Thus we just need to choose $h<\min(1,d/(2k+1))$ and then $l=k+h\in S$.
Next we can handle the case when $k^2>2$. In this case we show that there is an $l$ such that $0<l<k$ and $l^2>2$. This would mean that $l$ is also an upper bound for $S$ and we get the desired contradiction. I hope you can proceed like in previous paragraph to get an $l$ in the form $k-h$ with suitable $h>0$ such that $l^2=(k-h)^2>2$.
