Let me start by stating that I do not have a strong math background (I'm doing Software Engineering) and I'm studying this by myself, so I might get something wrong or make a mistake.
I want some help understanding the proof of theorem 140 from Landau's Foundations of Analysis book (pages 51-53). I already kind of understood almost all of the proof. The only problem I'm having is understanding why did he choose Y as an upper, not lower, number for $\eta$.
i saw this post, and it makes sense that you should pick an upper since it is the negative of a lower number from $-\eta$, but the set $-\eta$ hasn't been introduced yet in this part, so I want to know how can I understand it only from the knowledge built up to this part on the book, how do you know you should pick an upper number to form the set.
I want first to thank @n901 for the comment. I understood intuitively that you should use an upper number from $\eta$. The problem I had is that by reading the book, it seemed that he just chose that from intuition, and not from actual proof. What I mean is, what would make sense to me would be to prove that all rational numbers that are solutions to the equation by choosing rational numbers from the sets $\xi$ and $\eta$ would form a set that is, by itself, a cut. Hence, you would choose lower numbers from both cuts (since you would choose numbers from inside the sets).
I tried to make a proof from that, and would like to share it to clarify my idea, although I'm not sure about its validity:
UPDATE:
I got it now, after reading again. What he did was to choose that cut arbitrarily (in a intuitive way), prove that it is indeed a cut and then prove that that cut was exactly the one that solved the equation. Sorry for not understanding that simple part. When I tried proving it, I just thought about doing it from the ground up instead of finding the cut intuitively and then proving that it was the right one.
Theorem 140: If
$$ \xi > \eta $$
then
$$ \eta + \nu = \xi $$
Proof: I) (Same as Landau's proof)
There exists at most one solution; for if
$$ \nu_1 \neq \nu_2 $$
then, by Theorem 135,
$$ \eta + \nu_1 \neq \eta + \nu_2 $$.
II)
What we want to prove is that if we choose $X$ and $Y$ from the sets $\eta$ and $\xi$ respectively, then the set $\nu$ of all numbers $Z$ that solve the equation:
$$ X + Z = Y $$
constitutes a cut.
1)
From Theorem 94, we have that:
$$ X + Z = Y > X $$
Then, the equation will only have a solution for chosen $X$ and $Y$ when $ Y > X $. In that case, from Theorem 101, we have that there will always be a solution, which is:
$$ Z = Y - X $$
Then, the set $\nu$ has at least one number.
By choosing $ Z = Y $ and by using Theorem 94, we have:
$$ X + Z = X + Y = Y + X > Y $$
$$ X + Z \neq Y $$
Then, there is a number which is not in the set $\nu$.
2)
Here, we need to prove that for any number in the set, all the numbers below that number are also in the set.
Suppose we have a number $Z$ which is a solution to our equation. We know that there is an $X$ from $\eta$ and a $Y$ from $\xi$ where $ X + Z = Y $ and $ Y > X $. Let's choose a rational $Z_2$ such that $ Z_2 < Z $. From Theorem 101, we have that there is an $U$ that solves $ Z_2 + U = Z $. By choosing $ Y_2 = Y - U $, we have:
$$ \begin{align*} X &+ Z &=\;& Y \\ X &+ (Z_2 + U) &=\;& Y \\ (X &+ Z_2) + U &=\;& (Y - U) + U \\ X &+ Z_2 &=\;& Y_2 \\ \end{align*} $$
Since $ Y = Y_2 + U $, we also have that $ Y_2 < Y $, hence $Y_2$ is also in $\xi$. $Z_2$ is, then, a solution to the equation by choosing $X$ from $\eta$ and $Y_2$ from $\xi$, meaning that $Z_2$ is also in $\nu$.
Then, for a given $Z$ in $\nu$, any $ Z_2 < Z $ will also be inside $\nu$.
3)
Suppose we choose $X$ and $Y$ from the sets $\eta$ and $\xi$ respectively, such that $Y > X$. We know that there is a solution $Z = Y - X$. Now, lets choose an $Y_2$ from $\xi$ such that $Y_2 > Y$. From $ Y_2 > Y > X $, we get that $ Y_2 > X $. The equation has, then, a solution $Z_2$ for $X$ and $Y_2$:
$$ X + Z_2 = Y_2 $$
From Theorem 101, there is a $U$ which is solution for $ Y + U = Y_2$, which gives:
$$ \begin{align*} X + Z_2 &= Y_2 \\ X + Z_2 &= Y + U \\ Z_2 + X &= U + Y \\ Z_2 + X &= U + (X + Z) \\ Z_2 + X &= (U + Z) + X \\ Z_2 &= U + Z \\ Z_2 &= Z + U \\ Z_2 &> Z \end{align*} $$
Then, for any given $Z$ that solves the equation, there is a solution $Z_2$ which is bigger than $Z$.