Download American Mathematical Monthly, volume 117, May 2010 by Daniel J. Velleman PDF

By Daniel J. Velleman

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Extra resources for American Mathematical Monthly, volume 117, May 2010

Example text

Lk } in R2 passing through the origin of R2 and define L = {l ⊂ R2 : l is a line parallel to some li , where i ∈ [1, k]}. A straightforward verification shows that L is the required family. , it does not appeal to the axiom of choice. , is carried out within ZF set theory (see [6], [8]). In this context, the following example seems to be of some interest. Example 1. Consider the question of whether there exists a family of great circles on the sphere S2 such that every point of S2 belongs to exactly 2 circles of the family.

The existence of y is obvious. Define Cm as the circle with center y and radius 1/2. Case 2. The point xm belongs to fewer than k circles from the family {Cn : n < m}. In this case, let y be a point of A2 such that y = yn (n = 0, 1, . . , m − 1), ||y − z|| = 1/2 (z ∈ Z \ {xm }), ||y − xm || = 1/2. The existence of y follows from the facts that the circle in A2 centered at xm with radius 1/2 is countably infinite and the first two requirements concerning y rule out only finitely many points on this circle, so there are infinitely many y satisfying all three requirements.

To finish our paper, let us sketch the proof of Theorem 1. , [6], [8]), and on the method of transfinite induction. More precisely, in his argument Mazurkiewicz exploits the fact that any set of cardinality continuum (= c) can be well-ordered. He starts with an α-sequence {lξ : ξ < α} which consists of all distinct lines in R2 , where α denotes the smallest ordinal number of cardinality continuum (this simply means that card(α) = c and card(ξ ) < c for each ordinal ξ < α). , no three of them belong to a line); card(X ξ ∩ lξ ) = 2.

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