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Topic: Simple analytical properties of n/d
Replies: 20   Last Post: Mar 11, 2013 11:01 PM

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ross.finlayson@gmail.com

Posts: 1,199
Registered: 2/15/09
Re: Simple analytical properties of n/d
Posted: Mar 9, 2013 11:11 PM
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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



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