Abstract
We conclude the third part of our excursion with a presentation of Hausdorff’s paradox, conjecturing on its origin in BDT.
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Notes
- 1.
We quote from Hausdorff 1914a. There the theorem appears in the appendix (p 469), in an endnote to a passage (p 402) in §1, titled “the problem of definition of content” (Inhaltsbestimmung), of Chap. 10, titled “Contents [Inahlte] of point sets”. The theorem, with some of its background from section X.1, also appeared in Mathematische Annalen, perhaps a couple of months before the book (Hausdorff 1914b).
- 2.
See below.
- 3.
We maintain Hausdorff’s terminology though currently the problem is referred to as the ‘measure problem’.
- 4.
Hausdorff relates the formulation in this presentation to Lebesgue.
- 5.
It is tacitly assumed that A, B are disjoint; this point is noted on p 400. For (δ) it surely must be assumed that the sets are pairwise disjoint.
- 6.
- 7.
- 8.
Hausdorff was directed towards rotations as the form of congruence by his use of rotations of a circle in his counterexample to the possibility of ω-additive content. He turned to spatial figure no doubt because he needed independent rotations.
- 9.
Moore (1982 p 187) remarks that Hausdorff, in his proof of this theorem, “extended earlier arguments yielding a non-measurable set”, but he does not give any references.
- 10.
The number of these differences is equal to the number of pairs of elements from Q, which is denumerable (see Sect. 1.2.). As the number of axes passing through the center of the sphere is not denumerable, and likewise is the number of rotation angles, the choice of an axis and rotation angle as required is possible.
- 11.
Namely, f(K)/n.
- 12.
In the proof that f(Q) = 0, a chain of Q is generated, proof-processing from Zahlen.
- 13.
These are reduced forms, after omitting ϕ 2 or ψ 3 that may occur when words of different types are combined.
- 14.
Here Hausdorff reduces the case of γ to the not yet handled case δ.
- 15.
Hausdorff does not explain what α′ stands for. If for the m 1, m n of δ, m 1 + m n ≠ 3, then α′ is α; otherwise, α′ is γ. This is transformed by ϕγϕ to a δ with n-2 ϕ and n-1 ψ, if originally δ had n ψ and n-1 ϕ. Repeating this procedure up to n times, we must encounter an α variant for otherwise we reduce the formula to ϕ which is clearly not 1.
- 16.
To obtain the transformation matrix for (ϕψ) we have to multiply the transformation matrix of ψ on the left of the transformation matrix of ϕ, namely, we first apply ϕ and then ψ; Hausdorff uses here the same convention as did Bernstein, namely, that ϕψ means that first ϕ is executed then ψ; this convention applies to all members of G.
- 17.
In both (ψ) and (ϕψ); this is because sin4/3 π = -sin2/3 π.
- 18.
This is not the α′ mentioned above.
- 19.
The claim is laid for all x, y, z that are obtained from 0, 0, 1 by α-type members of G.
- 20.
To obtain these equations put in place of x, y, z of (ϕψ) (or (ϕψ 2)) their expansion under the induction hypothesis. For z′ we obtain the factor sinθ2 which should be replaced by 1-cosθ 2.
- 21.
There are only n values for cosθ for which the equation (3/2)n-1cosθ n + … -1 = 0 is fulfilled, and the totality of these values (for every n) is denumerable. It is thus possible to take a θ such that cosθ is not in that totality.
- 22.
The partitioning of G given here is different from the one we used in Sect. 1.1 because here 1 is included and put in A.
- 23.
Every member ρ of G can be represented by one rotation, the matrix of which is the multiplication of all the matrices of the components of ρ. All these rotations are different. The poles of the axis of this rotation are its fixed points. Since G is denumerable so is Q.
- 24.
Note Hausdorff’s convention xϕ (and below Mϕ) instead of the standard today ϕ(x). It explains why in composite mappings the order of execution is from the left. Bernstein did not use this convention.
- 25.
- 26.
Each P x is similarly partitioned into three partitions with the same relations. But a P x is not a counterexample to Lebesgue’s problem because its content is 0. Taking a x′ ≠ x in P x will generate a different partitioning of P x and so to obtain the desired partitioning a specific member from each network must be chosen by AC. Changing the choice-set M will generate different partitioning of P.
- 27.
We can assume that θ is acute for otherwise we get a symmetrical case with the rotation axis on the other side of Z. Thus we can use in the proof the familiar identities cos(π/2-θ) = sinθ, sin(π/2-θ) = cosθ.
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Hinkis, A. (2013). The Origin of Hausdorff Paradox in BDT. In: Proofs of the Cantor-Bernstein Theorem. Science Networks. Historical Studies, vol 45. Birkhäuser, Basel. https://doi.org/10.1007/978-3-0348-0224-6_27
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