LudovicoVan
Posts:
3,792
From:
London
Registered:
2/8/08


Can we count N ?!
Posted:
Jun 8, 2013 9:55 AM


It seems easy enough to count the subsets of N, i.e. P(N), by traversing the complete binary tree breathfirst and a mapping between nodes and subsets of N. Something like (details omitted):
"" "*" "0*", "1*" "00*", "01*", "10*", "11*" "000*", "001*", "010*", "011*", "100*", "101*", "110*", "111*" ...
"" "1" "01", "11" "001", "011", "101", "111" "0001", "0011", "0101", "0111", "1001", "1011", "1101", "1111" ...
{} {0} {1}, {0,1} {2}, {1,2}, {0,2}, {0,1,2} {3}, {2,3}, {1,3}, {1,2,3}, {0,3}, {0,2,3}, {0,1,3}, {0,1,2,3} ...
Now, the immediate objection to any such attempts is that: infinite subsets are not captured.
To which we reply, informally (as we cannot be more formal than the objection is!), that, despite we cannot but write down few entries of any infinite sequence, the sequence itself cannot be more "incomplete" than N={0,1,2,3,...} itself is, i.e. that infinite subsets are surely captured as long as N itself is fully captured. Namely, the objection is illogical, as an attempt at its formalisation would possibly show.
But they might retort with the question: what is the index of the set of even numbers?
To which we'd have to get into nonstandard integers (the ordinally transfinite), while contending that there can be no such thing as "unfinished sets", that the standard N and similar sets are incongruent within a *set theory of infinite sets*, and that all infinite sets are rather (extended)countable, i.e. there is a bijection with the set N*, the compactification of N, and that in omega we already have two directions of counting, forwards from zero and backwards from w (actual infinity as a doublypotential infinity), etc. etc.
Anyway, the point of contention is that we do have (can produce) an answer to the question of what is the index of the set of even numbers, it's the standard that is inadequate to answer and even just ask that question.
Julio

