In the book "The Moment of Proof" by Donald C. Benson, Chapter 4 page 51, he gave a geometric application of infinite descent.
proposition 4.7:
Let $S$ be a finite set of points. Suppose that a straight line connecting any two points of $S$ always contains a third point of $S$. Then all of the points of $S$ lie in a straight line.
This is the proof he provided. I rephrased it a little:
Proof. Suppose that the points in $S$ do not belong to the same straight line. Then at least one of the points in $S$, we call $P$, is at a positive distance from a line determined by two other points in S, $A \ and\ B$. Let that positive distance be $d_0$ and the line $AB$ be $l_0$. Assume that there is a third point $C$ that belongs to $S$ that lies on $l_0$.
Define $Q$ to be a point on $l_0$ such that $PQ \perp l_0$ and $\bar{PQ} = d_0$. The right or left line emanating from Q must contain two of three points $A,B,C$. Let the line emanating from Q such that that line contains two of the three points be $h_1$. Out of that two points, one point is farthest from Q and one point is nearer to Q. Call it $F$ (far) and $N$ (near) respectively. $N$ lies on segment $QF$. Define $l_1$ to be line $PF$ and $d_1$ the be perpendicular distance from $N$ to $l_1$. Let point $R$ to be on a point $l_1$ such that $RN \perp l_1$ and $\bar{RN} = d_1$
Since $\triangle FRN \sim \triangle FQP$, and $FR<FP$, we can say that $RN<PQ$. Therefore, $d_1<d_0$. That means that $d_0$ isn't the shortest distance. $\blacksquare$
I have two questions regarding this proof.
We have taken $d_0$ to be the smallest distance and shown that $d_1<d_0$. But if we chose $d_1$ to be the smallest distance, there will not be any smaller distance, so how would the proof continue?
Even if we have shown that $d_0$ isn't the smallest distance, how does that prove or disprove the statement?
Thank you!
Edit: For question 1, I believe that the proof will continue if I add a new point in $S$ such that it is between $R$ and $F$. From there, we can use the same logic (have a line emanate out of $R$... so on and so forth). This shows that there is indeed no such thing as "shortest distance". But I still cannot figure out question 2.
Since there are only finitely many points, some triple $A, B$ and $C$ must realize the minimum. By continuing to find triples with a smaller $d$ we must eventually arrive at this minimum triple or to a triple that we've already seen. Both lead to contradiction. In fact, we could assume that the $A, B, C$ we pick in the first place is the minimum. I mean, just pick it!