I am trying to understand a proof of Pappus' Theorem. This is taken from the book Geometry and Topology by Miles Reid and Balázs Szendröi.
Definition 1 Let $PQ$ be the line through two points $P=(x_0 : \cdots : x_n)$ and $Q=(y_0: \cdots : y_n)$ of $\mathbb{P^{n}}$. First lift to $\mathbb{R^{n+1}}$ by setting $\widetilde P = (x_0, \ldots, x_n)$ and $\widetilde Q = (y_0, \ldots, y_n)$ (that is, pick the values $x_i$ and $y_i$ in the given ratio), then set
$$PQ = \langle P,Q\rangle = \{ \text{ratios}\ (\lambda x_0 + \mu y_0): \cdots : (\lambda x_n + \mu y_n)\ \text{for all}\ (\lambda, \mu) \neq (0,0) \}.$$
Definition 2 The standard frame of reference is
$$P_i = (0: \cdots :1: \cdots : 0),$$ where $1$ is in the ith place and $$P_{n+1}= (1: \cdots :1: \cdots : 1).$$
Theorem 1 Let $\{P_0, \ldots, P_{n+1}\}$ be the standard frame of reference. Then there is a one-to-one correspondence between projective transformations and frames of reference, defined by $T \mapsto T(P_0), \ldots,T(P_{n+1})$.
Theorem 2 (Pappus' Theorem) Let $L, L' \subset \mathbb{P^2}$ be two lines and $$P, Q, R \subset L \quad \textit{and} \quad P', Q', R' \subset L'$$ two triples of distinct points on $L$ and $L'$ (not equal to $L \cap L')$. Then the three points $$QR' \cap Q'R = A, \quad PR' \cap P'R = B \quad \text{and} \quad PQ' \cap P'Q = C$$ are collinear.
Proof This can also be proved via a lifting to $\mathbb{P^3}$, but this requires a bit more information about $\mathbb{P^3}$ (specifically, quadric surfaces in $\mathbb{P^3}$ and properties of lines on them). I sketch the easy proof in coordinates.
By Theorem 2, I can choose homogeneous coordinates $(x : y : z)$ such that $$P = (1 : 0 : 0),\ Q = (0 : 1 : 0),\ P' = (0 : 0 : 1) \quad \text{and} \quad Q'= (1 : 1 : 1).$$
Why are these coordinates chosen for $P, Q, P'$ and $Q'$? They correspond to the standard frame of reference, but could an arbitrary set of points have been chosen instead?
Then $$L = PQ : \{z = 0\},\ P'Q : \{x = 0\},\ L' = P'Q' : \{x = y\} \quad \text{and} \quad PQ' : \{y = z\}.$$
How do I determine that $L = PQ : \{z=0\}?$ I know from Definition 1 that $$PQ = \langle P,Q\rangle = \{ \text{ratios}\ (\lambda \cdot 1 + \mu \cdot 0: \lambda \cdot 0 + \mu \cdot 1 : \lambda \cdot 0 + \mu \cdot 0) = (1:1:0) \ \text{for all}\ \ (\lambda, \mu) \neq (0,0) \}$$ which corresponds to $z=0$, but why can't I put $x=y$ instead (since $x=y=1$)?
Therefore$$C = P'Q \cap PQ' = (0 : 1 : 1).$$
Now let $R = (1 : \beta : 0)$ and $R' = (1 : 1 : \gamma)$. Then we have $$PR' : \{z = \gamma y\} \quad \text{and} \quad P'R : \{y = \beta x\}$$ so that $B = (1 : \beta : (γ))$ and $$QR' : \{z = \gamma x\} \quad \text{and} \quad Q'R : \{y − z = \beta(x − z)\}$$ so that $A = (1 : (\beta + \gamma − \beta \gamma) : γ).$
Finally, $A, B, C$ are all on the line $${y − z = \beta(1 − \gamma)x}.$$ Q.E.D
How does this conclusion follow?
I appreciate your help. Thank you.
Due to Theorem 1, changing the frame of reference is the same as applying a projective transformation. So given four points, no three of them are collinear, you can always find a projective transformation which maps these to the standard frame of reference. And since Pappos' theorem is an incidence theorem, and incidences are preserved under projective transformations, you can choose any four points in general position (i.e. no three of them collinear) as the frame of reference, without loss of generality.
You misread that formula. Actually what you get is
$$ L=\{(1\lambda+0\mu:0\lambda+1\mu:0\lambda+0\mu)\mid(\lambda,\mu)\neq(0,0)\}=\{(\lambda:\mu:0)\mid(\lambda,\mu)\neq(0,0)\} $$
which is the set of all three-element vectors which have a zero in their last coordinate, with the exception of the null vector. Assuming $\lambda=\mu=1$ or $\lambda=\mu$ is not valid here.
Any line is described by a linear equation $ax+by+cz=0$ where the $(a,b,c)$ are unique up to scalar multiples (i.e. $(a:b:c)$ is unique). So if there is one equation describing a line, there can be no fundamentally different equation describing the same line. If you know $z=0$ describes this line (since both the points defining it have zero $z$ coordinate), you can be sure that $x=y$ does not describe the same line (and here indeed describes a different line that you'll need, namely the one where both defining points have equal $x$ and $y$ coordinates, i.e. $P'Q'$).
$C$ is fixed by the frame of reference. $R$ and $R'$ are semi-free: they are constrained to lie on the line $PQ$ resp. $P'Q'$ but can move freely on that line, amounting to one real degree of freedom. That's expressed by choosing two real parameters $\beta$ and $\gamma$ to describe their position. The choice of the ones in their coordinate vectors is just a choice of representative, and doesn't affect the ratios.
Once you choose coordinates for $R$ and $R'$, you can compute coordinates for $A$ and $B$ from these. So at that point, you hold coordinates for $A,B,C$ and can verify that there exists a single line (still parametrized by $\beta$ and $\gamma$) incident to all three of them.
As an alternative, instead of computing the equation of that line, one could also compute the determinant of the homogeneous coordinates of the points, which will be zero if they are collinear. Perhaps your book will cover that at some later point?