
Re: A intuitive notion of set size.
Posted:
Jan 13, 2013 7:13 AM


On Jan 13, 4:20 am, Zuhair <zaljo...@gmail.com> wrote: > For any two sets A,B: > [1] A is bigger than B iff A > B Or A,B are sets of naturals and > there exist a non empty set C of naturals such that for every element > n of C: > A(n) > B(n) and A(n*) / B(n*) >= A(n) / B(n). > where X(n) = {y y in X & y <' n}; > <' stands for natural strict smaller than relation; >   stands for cardinality defined after Cantor's. > n* stands for the immediate successor of n in C with respect to > natural succession. > [2] A is smaller than B iff B is bigger than A. > [3] A is equinumerous to B iff ~ A bigger than B & ~ A smaller than > B. > /
Example:
For natural k, let n_k = 3*2^{k1} + 1; so n_1 = 4, n_2 = 7, n_3 = 13, n_4 = 25, and so on.
Partition N into two seta A and B as follows:
A = {1} u [n_1, n_2) u [n_3, n_4) u [n_5, n_6) u ... = {1, 4, 5, 6, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 49, 50, ...};
B = {2, 3} u [n_2, n_3) u [n_4, n_5) u [n_6, n_7) u ... = {2, 3, 7, 8, 9, 10, 11, 12, 25, 26, 27, 28, 29, 31, 32, ...}.
In other words: the first natural, 1, is in A: the next 2 naturals are in B; the next 3 are in A, the next 6 in B, the next 12 in A, the next 24 in B, and so on.
By your definition, A is bigger than B. Hint: let C = {n_2, n_4, n_6, n_8, n_10, ...}.
By your definition, B is bigger than A. Hint: let C = {n_1, n_3, n_5, n_7, n_9, ...}.
So A is both bigger and smaller than B. B is both bigger and smaller than A. Since A is equinumerous to itself, the relations "bigger than" and "smaller than" fail to be transitive.
This does not seem very intuitive to me.

