
Re: Simple analytical properties of n/d
Posted:
Mar 9, 2013 11:11 PM


On Mar 9, 6:28 pm, William Elliot <ma...@panix.com> wrote: > On Sat, 9 Mar 2013, Ross A. Finlayson wrote: > > On Mar 8, 9:31 pm, William Elliot <ma...@panix.com> wrote: > > > What's R_[0,1]? So now your considering any function f:N > [0,1] > > > and want to discuss rang f. > > > > > Obviously and directly as N is countable, ran(f) > > > > is countable, > > > > Is ran(f) = [0,1]? > > > > Of course not, you showed that range f is countable and > > > unable to be any uncountable set. > > > Apply the antidiagonal argument to ran(f). > > Why? > > > the only item different from each is and, ran(f) includes 1.0. > > Huh? > > > Apply the nested intervals argument to ran(f). > > What's that? > > > The interval is [.0, .1], there's no missing element from ran(f)'s > > [0,1]. > > > The antidiagonal argument and nested intervals argument don't support > > that ran(f) =/= [0,1]. > > > In fact, remarkable among functions N > R, is that the antidiagonal > > argument and nested intervals argument, DON'T apply to f. > > It doesn't? What if f:N > [0,1] is the constant function f(N) = {0}?
It is that: f is remarkable (or unique up to translation, mirroring, or finite scaling) among functions, with range everywhere dense in [0,1] as an onto mapping would be (N onto> [0,1]), the antidiagonal and nested intervals arguments: don't apply. It is remarkable that ran(f) is everywhere dense in [0,1], and the otherwise proofs of uncountability of [0,1], and that ran(f) =/= [0,1]: don't apply.
Then, the general notion for progress, is then to work up what it means that d > oo and d = oo, and again what features of the numbers exist that were ignored because there exists an exception to modern set theory's result.
Regards,
Ross Finlayson

