
Re: Can we count N ?!
Posted:
Jun 9, 2013 9:37 PM


On Saturday, June 8, 2013 6:55:51 AM UTC7, Julio Di Egidio wrote: > 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
I'll agree that a breadthfirst traversal, of the last row, which for the infinite or unbounded tree is the same as the depthfirst traversal of the infinitelength paths, sees the order being the sweep from the path with all 0branches through the path through all 1branches. Here particularly the antidiagonal argument of the tree doesn't apply to the mathematics of the infinite graph of elements of the Cantor space 2^w (i.e. {0,1} X {0,1} x ...), and the range of the sweep is thus not uncountable. Visavis, the Cartesian product and ordinals, rays through countably many ordinal lattice points are dense in the paths, and: ordinals are wellordered, that the range of the sweep is thus countable.
It's still quite particular that where the range of the sweep is the Cantor sweep, any other function still sees the uncountability results about the range apply, then as regards to ordinal points dense in the paths having the same order type, but being countable.
Similarly in the unit interval of reals, the sweep (or "natural/unit equivalency function") f is not shown uncountable, and indeed ran(f) is R_[0,1]. As well similarly to the countable rays through ordinal points dense in the paths, the rationals are dense in the irrationals, with that the reals are trichotomous and elements of 2^w lexicographically ordered.
Then the conscientious mathematician would wonder: what are the properties of the sweep besides that unique among functions it is onto, and 11, the Cantor space and R_[0,1]? How are the unit's reals _both_ uniformly composed unit, and, unit of the complete ordered field?
Because they are.
Regards,
Ross Finlayson

