Set defined for proving existence of $ r > 0 $ s.t. $ r^2 = 2 $ is problematic Theorem: There exists a positive real number $ r $ s.t. $ r^2 = 2 $
The proof in my book begins with defining a set $ L = \{ x \in \mathbb{R} | x \geq 0 , x^2 \leq 2 \} $  and then proceeds to show that $ r^2 = 2 $ is the maximum of this set. However, I don't understand the following:
Since the set L contains real numbers $ x $ such that $x \geq 0, x^2 \leq 2 $, then specifically, it contains a number $ x>0 $ s.t. $ x^2 = 2 $ . How am I supposed to show the existence of such a number ( $ r^2 = 2 $ ) if it already exists  in the set I've defined? ( In which case, the set $ L $ is problematic since it already contains the number which I need to prove exists )
 A: The point is that you know there exists some $x$ such that $x \ge 0$ and $x^2 \le 2$, (e.g., $1$) and you know that for all large enough $x$, $x^2 > 2$ (e.g., that is true for $x \ge 2$). Hence (without knowing whether there is an $x$ with $x^2=2$), you can define $L = \{x \in \Bbb{R} \mid x \ge 0, x^2 \le 2\}$ and you know that it is non-empty ($1 \in L$) and bounded above (if $x \in L$, then $x < 2$). But you also know that every non-empty bounded-above set of real numbers has a supremum $s$, say. And then you can prove that $s^2 = 2$. There is no circularity here, even though $s \in L$.
In this case (as pointed out by fleablood in a comment), if you were to change $\le$ in the definition of $L$ to $<$, you would get the same result, but the supremum $s$ would not be a member of $L$. However, there are proofs where you need to reason about a set that must contain a particular number of interest in order to establish what that number is. For example, to prove that any family $F$ of open intervals that covers the closed interval $I = [0, 1]$ has a finite subfamily that covers $I$, you look at the set $M$ of $x \in [0, 1]$ such that a finite subfamily of $F$ covers $[0, x]$ and then prove that $1 \in M$. It would be hard to recast that without allowing $1 \in M$.
A: We don't know much about real numbers at this stage.  We know that every rational number is a real.  And we know that every set of real numbers that is bounded above will have a least upper bound.  But that's about all. (Okay, we know it's an ordered fields and... more stuff)
We do not know that there actually is any number $r$ so that $r^2 = 2$?  Why should there be?  Sure,  if there were such a number than $r\in L$ but we don't know any number exists.
(Here's an analogy.  Suppose I asked you to prove that something that doesn't exist does exist.  Suppose I asked you to prove that there is a smallest positive number $s$.  That is a number $s$ so that $s > 0$ and for any $t>0$ then $s \le t$.  There is no such number.  But I'm asking you to prove it.  So you say let $R= \{x| x > 0\}$.  But then you declare that because $s> 0$ then $s \in R$ so what's left to be done.....)
(Or maybe if we defined $S = \{x| 5< x \le \infty\}$ and we argue well if $w = \infty$ then $w \in S$......  Just because the set is defined to allow $w = \infty$ doesn't mean that it does have it.)
So we don't know that there is an $r$ so that $r^2=2$.
But we do know that $s=\sup L$ exists. (That because $L$ is not empty [$1\in L$] and $L$ is bounded above [If $s\ge 2$ then $s^2 \ge 4>2$ so $s\not \in L$ so $2$ is an upper bound] and the reals have the least upper bound property.)
So how does $s^2$ compare to $2$?  Either $s^2 < 2$ or $s^2 > 2$ or $s^2 = 2$.  If we can prove $s^2 < 2$ and $s^2 > 2$ are both impossible than we'd have to conclude $s^2 \ge 2$ and $s^2 \le 2$ and we must conclude $s^2 = 2$
=======

 if $s^2 < 2$ then if we let $\epsilon = \frac {2-s^2}3$.  As $1$ and $1\in L$ we know that $\epsilon \le \frac 13$ so $\epsilon^2 < \epsilon$.  And so $(s+ \epsilon)^2 = s^2 + 2\epsilon + \epsilon^2< s^2 + 3\epsilon =s^2-(2-s^2) = 2$.  So $s < s+\epsilon \in L$ contradicting that $s$ is an upper bound of $L$.


 So $s^2 \ge 2$.


 Do something similar if we assume $s^2 > 2$.  Then let $\delta = \frac{s^2 -2}2$.  As $y \ge 1.5\implies y^2 \ge 2.25$ we know $\delta < 1$ so $\delta^2 < \delta$ and .... same stuff.   $(s-\delta)^2 = s^2 - 2\delta + \delta^2> s^2 -2\delta = s^2 -(s^2-2) = 2$.  (And if $t\ge  (s-\delta)$ then $t^2 \ge (s-\delta)^2 > 2$.) So $s-\delta$ is an upper bound contradicting that $s$ is the least upper bound of $L$.

So $s^2 \le 2$.
Well if $s$ exists, and $s^2 \ge 2$ and $s^2 \le 2$ then we must have $s^2 = 2$.
And that's that.
And, yes, $s \in L$.
(Actually the proof would have worked just as well if we had defined $L$ as $L = \{x| x\ge 0; x^2 < 2\}$.  ANd just as an aside there was no need to specify $x \ge 0$--- it doesn't matter that negative values are in $L$ because the $\sup$ would still be positive.)
