- #71
Hurkyl
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I don't feel like a very long response today, so I'll keep it short:
millions, billions, centillions, googolplexes, so what? Those are all finite. And aside from some technical details and the philosophical problem that you can't speak of a "gap" between two points without already having some a priori notion of places between them, I agree with you.
But, you haven't addressed infinitely many points at all. While any two individual points may be isolated, you've said absolutely nothing about the whole.
Sure. I'll get you started by posting the alledged proof that a line segment has infinitely many points:
notation: A*B*C means "B lies between A and C". By definition, it means that A, B, and C are distinct collinear points, and this relation satisfies the axioms of betweenness.
A point X is said to lie on the line segment YZ if and only if X=Y, X=Z, or Y*X*Z.
Lemma: Let A and B be distinct points. Then, there exists a point, C, such that A*C*B.
(I can prove this one too, if you need it)
Theorem: Let AB be any line segment. For every positive integer n, I can construct points C1, C2, ..., Cn such that
(1) they all lie on AB
(2) A*Cx*Cy if 1 <= y < x <= n.
Proof:
By the lemma, there exists a point X such that A*X*B. Choose C1 to be any such point. Condition (1) is satisfied by definition of line segment, and condition (2) is vacuously true. Thus, the theorem has been proven for n = 1.
Suppose the theorem is true for n = k. Then, by the lemma, there exists a point X such that A*X*Ck. Choose C(k+1) to be any such point.
We already know that Cx lies on AB if 1 <= x <= k. Because A*C(k+1)*Ck and A*Ck*B, we have A*C(k+1)*B (axioms of betweenness, or maybe it was a theorem), so C(k+1) lies on the line segment AB.
We alraedy know that A*Cx*Cy if 1 <= y < x <= k. Now, suppose x = k+1.
case 1: y = k. C(k+1) was constructed such that A*C(k+1)*Ck, so this case is proven.
case 2: y < k. Both A*C(k+1)*Ck and A*Ck*Cy, so we have that A*C(k+1)*Cy.
So, we see that A*Cx*Cy if 1 <= y < x <= k
So, we see that conditions (1) and (2) are both true for n = k+1. By the principle of mathematical induction, the theorem is proven for every n.
Corollary: Any line segment has infinitely many points lying on it.
Proof: suppose otherwise: that there exists a line segment AB that doesn't have infinitely many points lying on it. Let n be the number of points lying on AB.
By the theorem, we can construct points C1, C2, ..., C(n+1) such that
(1) they all lie on AB
(2) A*Cx*Cy if 1 <= y < x <= n+1.
Now, if p and q are different integers with 1 <= p, q <= n+1, we have either p < q or q < p. So, either A*Cq*Cp or A*Cp*Cq. Either way, this means Cp and Cq are distinct.
Therefore, each of the Ck are distinct, and there n+1 of them, contradicting the fact that AB has only n points lying on it.
Therefore, the initial assumption was wrong, so we conclude that every line segment has infinitely many points lying on it.
QED
Now if you start thinking about wanting to break this gap down into smaller and smaller gaps by considering more and more locations between these two original locations then you've really missed the whole concept. The whole purpose to this discussion is to simply address the fundamental nature that must exist between any two point that are not the same point. So considering millions and billions of points existing between our original two is to simply miss the point altogether actually.
millions, billions, centillions, googolplexes, so what? Those are all finite. And aside from some technical details and the philosophical problem that you can't speak of a "gap" between two points without already having some a priori notion of places between them, I agree with you.
But, you haven't addressed infinitely many points at all. While any two individual points may be isolated, you've said absolutely nothing about the whole.
Well, actually I can show that while this reasoning appears to be sound it is actually quite flawed. The bottom line is that the argument does not support the conclusion. I can actually prove this mathematically if anyone is interested in seeing the proof.
In fact, I can then go on and use calculus to prove why a finite line cannot possibly contain an infinite number of points.
Sure. I'll get you started by posting the alledged proof that a line segment has infinitely many points:
notation: A*B*C means "B lies between A and C". By definition, it means that A, B, and C are distinct collinear points, and this relation satisfies the axioms of betweenness.
A point X is said to lie on the line segment YZ if and only if X=Y, X=Z, or Y*X*Z.
Lemma: Let A and B be distinct points. Then, there exists a point, C, such that A*C*B.
(I can prove this one too, if you need it)
Theorem: Let AB be any line segment. For every positive integer n, I can construct points C1, C2, ..., Cn such that
(1) they all lie on AB
(2) A*Cx*Cy if 1 <= y < x <= n.
Proof:
By the lemma, there exists a point X such that A*X*B. Choose C1 to be any such point. Condition (1) is satisfied by definition of line segment, and condition (2) is vacuously true. Thus, the theorem has been proven for n = 1.
Suppose the theorem is true for n = k. Then, by the lemma, there exists a point X such that A*X*Ck. Choose C(k+1) to be any such point.
We already know that Cx lies on AB if 1 <= x <= k. Because A*C(k+1)*Ck and A*Ck*B, we have A*C(k+1)*B (axioms of betweenness, or maybe it was a theorem), so C(k+1) lies on the line segment AB.
We alraedy know that A*Cx*Cy if 1 <= y < x <= k. Now, suppose x = k+1.
case 1: y = k. C(k+1) was constructed such that A*C(k+1)*Ck, so this case is proven.
case 2: y < k. Both A*C(k+1)*Ck and A*Ck*Cy, so we have that A*C(k+1)*Cy.
So, we see that A*Cx*Cy if 1 <= y < x <= k
So, we see that conditions (1) and (2) are both true for n = k+1. By the principle of mathematical induction, the theorem is proven for every n.
Corollary: Any line segment has infinitely many points lying on it.
Proof: suppose otherwise: that there exists a line segment AB that doesn't have infinitely many points lying on it. Let n be the number of points lying on AB.
By the theorem, we can construct points C1, C2, ..., C(n+1) such that
(1) they all lie on AB
(2) A*Cx*Cy if 1 <= y < x <= n+1.
Now, if p and q are different integers with 1 <= p, q <= n+1, we have either p < q or q < p. So, either A*Cq*Cp or A*Cp*Cq. Either way, this means Cp and Cq are distinct.
Therefore, each of the Ck are distinct, and there n+1 of them, contradicting the fact that AB has only n points lying on it.
Therefore, the initial assumption was wrong, so we conclude that every line segment has infinitely many points lying on it.
QED
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