Spacy Colors II (Posted on 2014-10-26)

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The negation of the theorem is Rn is partitioned into n subsets S1, S2, ... , Sn and ∀ i∈In ( |Si| ≠ R+ ). Since ∀ i∈In ( |Si| ⊆ R+ ) our negation becomes ∀ i∈In ( ∃ ai∈R+ ( ai∉|Si| ) ). WLOG we can assume that a1 ≥ a2 ≥ ... ≥ an > 0. The " > 0" follows from S≠Φ ⇒ 0∈|S|. Definition of spheres in Rn and their intersections: Let [P,a] denote a sphere with center P and radius a.    x12 + x22 + ... + xk2 = r2 is a k-sphere of radius r with center located at the origin of Rn with k∈In. Topologists would call it a (k-1)-sphere where geometers would call it a k-sphere. For this theorem we will go with the geometer's definition. x12 + x22 + ... + xn-12 + xn2 = r12 (1) x12 + x22 + ... + xn-12 + (xn-r1)2 = a22 (2) The n-spheres (1) and (2) have radii r1 and a2 respectively. The center of (1) is located at the origin of Rn and the center of (2) is located on (1). Combining (1) and (2) we get x12 + x22 + ... + xn-12 = r12 - xn2 = a22 - (xn-r1)2 or xn = r1 - (a22/2r1) (3) Plugging (3) into (2) gives, x12 + x22 + ... + xn-12 = a22*[1 - (a2/2r1)2] = { a2√[1 - (a2/2r1)2] }2 = r22 (4) Thus, the intersection is a (n-1)-sphere with radius r2 and center located at the origin of Rn after being translated a distance (a22/2r1) - r1 in the xn dimension. If we intersect this (n-1)-sphere with another n-sphere of radius a3 and center located on the (n-1)-sphere we get a (n-2)-sphere with radius r3 = a3√[1 - (a3/2r2)2]. If we let r1 = a1, then the Spacy Colors Lemma shows that all of these k-spheres ( k = n, n-1, n-2, ... ) are nonempty and contain infinite number of points. This would be true if we did not run out of dimensions - the 1-sphere only has two points and the 0-sphere has none. This is OK because we only have n subsets and they run out at the 1-sphere. Proof: We will now show that the negation of the theorem leads to an empty k-sphere ( k∈In ). Consider the following algorithm which I will simulate with n = 9 and one of the 2n-1 possible scenarios. Algorithm: i = 1 k = n+1 Kk = Rn while ( i ≤ n ) { if ( Kk ∩ Si ≠ Φ ) { pick Pi ∈ Kk ∩ Si k = k-1 Kk = Kk+1 ∩ [Pi,ai] print " i Kk = [P1,a1] ∩ ... ∩ [Pi,ai] " } else { print " i Kk ∩ Si = Φ " } i = i+1 } Where i = index of subsets Si and numbers ai k = index of k-spheres Kk Scenario: 1 K9 = [P1,a1] 2 K8 = [P1,a1] ∩ [P2,a2] 3 K8 ∩ S3 = Φ 4 K8 ∩ S4 = Φ 5 K8 ∩ S5 = Φ 6 K7 = [P1,a1] ∩ [P2,a2] ∩ [P6,a6] 7 K6 = [P1,a1] ∩ [P2,a2] ∩ [P6,a6] ∩ [P7,a7] 8 K5 = [P1,a1] ∩ [P2,a2] ∩ [P6,a6] ∩ [P7,a7] ∩ [P8,a8] 9 K5 ∩ S9 = Φ Analysis:    Rule 0: Find " i Kk =" line with lowest k integer    Rule 1: Kk ⊆ [Pi,ai] ⇒ Kk ∩ Si    Rule 2: k ≤ j and Kj ∩ Si = Φ ⇒ Kk ∩ Si = Φ    Rule 3: Kk ∩ Si = Φ and KK ∩ Sj = Φ ⇒ Kk ∩ (Si ∪ Si) = Φ    Rule 0: K5 @ line i=8    Rule 1, Rule 3, and line i=8 ⇒ K5 ∩ (S1 ∪ S2 ∪ S6 ∪ S7 ∪ S8) = Φ    Rule 2, Rule 3, and lines i=3, 4, 5, and 9 ⇒ K5 ∩ (S3 ∪ S4 ∪ S5 ∪ S9) = Φ    Therefore, K5 ∩ (S1 ∪ ... ∪ S9) = K5 ∩ Rn = K5 = Φ The Rules 0-3 will apply to all n ≥ 1. Therefore, the negation of the theorem implies an empty k-sphere with k∈In. Therefore, The theorem is true. QED I would not like to give a detailed explanation for n = 101 and its 2100 scenarios.

	Subject	Author	Date
	Nice work	JLo	2020-01-22 10:39:30
	Corrections to The Solution	Bractals	2015-06-07 10:25:31