Difference between finding $\sup\{x\in\mathbb{Q}: x>0,x^2<2\}$ and when $x\in \mathbb{R}$ What is a difference between proofs of next two statements?
$$\sup{E}=\sup\{x\in\mathbb{Q}: x>0,x^2<2\}=\sqrt{2}$$ and $$\sup{F}=\sup\{x\in\mathbb{R}: x>0,x^2<2\}=\sqrt{2}.$$
For second one I can do the next thing. At first we show that this set is not empty and bounded above. Then:
$$\sup{F}=\sup\{x\in\mathbb{R}: x>0,x^2<2\}=\sup\{0<x<\sqrt{2}\}.$$
Then it is obvious that $\sqrt{2}$ is the upper bound for $F$ because for $\forall x\in F$ we have $x\leq \sqrt{2}$. Then we take any other upper bound $M$. And we know that it must be greater than $0$. If we suppose that $M<\sqrt{2}$ then we can pick a real number $x_0$ such that $0<M<x_0<\sqrt{2}$. Then we see that $x_0$ must be belong to the set F. But this contradicts our assumption that $M$ is an upper bound. So for any upper bound $M$ we have $M\geq \sqrt{2}.$ And it is the end of our proof.
QUESTION 1: Is it correct? Is it rigorous?
QUESTIIN 2: What the difference will be between this proof and proof for set $E$? I seek this difference during about two weeks and can not found it. (It would be nice if this proof is given in the same fashion.)
QUESTION 3: Such a question is rised by the following solution from Kaczor's "Problems in mathematical analysis 1". What is the idea behind this solution? What does he want to do? Can someone say the idea behind this solution? I went through dozens of books on real analysis searching this and did not find any promt.

 A: It would be a good idea for you to post the actual question.
The problem is that this sort of problem is risen to prove that there is a real number whose square is 2. If you look at the picture never have they tried to use $\sqrt{2}$, this is not because they want to make it particularly complicated but because the very objective of the problem is to prove that $\sqrt{2}$ exists.
For anyone familiar with the reals both of your claims $\text{sup} F=\text{sup}E=\sqrt{2}$ are completely obvious statements. However the idea of the exercise is that starting from the knowledge of the rational numbers and the construction of the real numbers as Dedekind cuts there exist a square root of 2. In particular note that $E$ is a Dedekind cut that you can make sense of only knowing about the rational numbers. However you can not really make sense $F$ unless you know stuff about the real numbers.
To answer your questions.

*

*your argument is completely rigorous provided you can use your knowledge of the real numbers (which, as I said before, makes the question completely trivial).

*Your statement that you can find an $x_0$ such that... relies on delicate properties of the real numbers that are the very thing you are trying to construct, so this should be systematically avoided. That is why in the picture they make a big deal of constructing explicitly the number $w$ that is actually a rational.

*The picture proof basically follows the same train of thought as your argument. The only thing they are trying to avoid is to always stay in the solid ground of the rational numbers.

A: The second case is straightforward since F=(0,$\sqrt{2}$).In the first case it is clear that $\sqrt{2}$ is upper bound of E. We will show that this is supE. ie the least upper bound of E. Assume on the contrary that the least upper bound is $\sqrt{2}$-$\epsilon$ . Then q<$\sqrt{2}$-$\epsilon$ for all q in E. Also,$\sqrt{2}-\epsilon < \sqrt{2}-\epsilon ^{2}$. There is certainly a rational number r in ($\sqrt{2}-\epsilon ^{2},\sqrt{2}$). This r belongs to E and it is greater than supE=$\sqrt{2}-\epsilon$. This is an obvious contradiction which proves the result. (There are many ways for defining irrational numbers. The Dedekind cuts (eg supE is defined as  $\sqrt{2}$). We can also define real irrational numbers as equivalent classes of Cauchy sequences where equivalence of $x_{n}\,$ and, $y_{n}\,$ is defined as $ x_{n}-y_{n}\to 0$.
