R = {x: x is not a member of itself},and so Russell's Paradox cannot be carried out. In NBG set theory, on the other hand, we can form R; it is a class. In that setting, Russell's argument does not yield a contradiction; it merely yields a proof that the class R is a proper class.
Indeed, in most cases the only picture drawn is a periodic tiling of the plane (or some portion of it) by congruent triangles. This picture suggests that, in a similar fashion, 3-dimensional space can be tiled with congruent tetrahedrons. It can't! Try it (with cardboard models); you'll find that you need to alternate tetrahedrons and octahedrons. This sort of thing is not easy to visualize, and so pictures can be misleading.
There seem to be two common ways to get around the difficulty, although neither method completely eliminates the difficulty:
(i) Instead of a periodic tiling, use an aperiodic one. Any polygon in the plane can be subdivided into triangles (generally not congruent); indeed, any n-gon can be subdivided into n-2 triangles. Similarly, any polyhedron in 3-space (e.g., any octahedron) can be subdivided into tetrahedrons, etc. A mathematician with a reliable imagination can then convince himself or herself of the validity of various assertions about the number of simplexes that any facet is contained in. However, the relevant pictures are hard to visualize and impossible to draw in dimensions higher than 2, and so it is not clear just how "reliable" are the imaginations of the mathematicians involved. I would not want to rely on mine for this purpose. -- The subject of triangulations is not a simple one. There are entire books devoted to this subject, and they're not simple books. For instance, see
C. Y. Dang, Triangulations and simplicial methods, Lecture Notes in Economics and Mathematical Systems 421, Springer-Verlag, Berlin, 1995. (Math Reviews 96f:90079.)
(ii) Instead of a geometric approach, we could take an algebraic approach. Describe the relevant simplexes and facets entirely in terms of their barycentric coordinates, without trying to visualize the picture for any dimensions higher than 2. This approach lends itself more readily to being made into a rigorous proof, but the proof includes some rather long and tedious steps, verifying in higher dimensions what was obvious in two dimensions. Among the fully detailed expositions that I've looked at thus far, the most readable is the one in
R. Webster, Convexity, Oxford University Press, 1994.However, because the proof of Sperner's Lemma in that book is complete, it is also quite long. There are two parts to the proof:
card{Borel subsets of R} = card(R).For a proof, continue the iteration indicated above, but don't just use the positive integers for the subscripts; use the ordinals. Define H(alpha) = (union of H(beta)'s for beta less than alpha)*. Let W be the first uncountable ordinal. The union of countably many predecessors of W is a predecessor of W; from that it follows that the sigma-algebra generated by G is equal to H(W). This result can be found in Section 5, Exercise (9) in Halmos's book MEASURE THEORY. -- A related result: it is easy to show that
card{Lebesgue measurable subsets of R} is greater than card(R).Indeed, define C as in 24.39, and then consider the collection of all subsets of C.
{0, {0}, {{0}} } is transitive (easy to verify). But it is not an ordinal. Perhaps the easiest (i.e., laziest) way to prove that it is not an ordinal is as follows: By 5.46.c, all the members of an ordinal are ordinals. But {{0}}={1} is not transitive, as we noted in 5.43.a, so it's not an ordinal.
Let T be any ordinal greater than 1, and let X be the powerset of T -- that is, X=P(T). I claim that X is transitive but not an ordinal. To see that X is transitive, suppose A is a member of S and S is a member of X; we want to show A is a member of X. Since S is a member of X=P(T), S is a subset of T. Since A is a member of S, A is a member of T. Since T is transitive, A is a subset of T. Thus A is a member of P(T)=X, and so X is transitive. To show X is not an ordinal, note that 1 is a member of T, so {1} is a subset of T, so {1} is a member of X. But {1} is not an ordinal, as we've already noted above. So by 5.46.c, X is not an ordinal.
I. Fleischer, Order-convergence in posets, Math. Nachr 142 (1989), 215-218.