Re: Well Ordering the Reals

David R Tribble said:
> Tony Orlow wrote:
> > Now, as I said below, you can avoid the ends of the line and mention of
> > infinity, if you wish, although I think that's a mistake. Still, for the faint
> > of infinite, I offer the Peano-esque definition:
> >
> > 1) 0 is a real number
> > 2) If x is a real number 2^x and -2^x are real numbers.
> >
> > There, now you don't have to touch that dirty oo.
>
> So that gives us 0 as a real,
> and 2^0 = +1 and -2^0 = -1 as reals,
> and 2^1 = +2, -2^1 = -2, 2^-1 = +1/2, and -2^-1 = -1/2 as reals,
> and continuing, we get:
> +4, -4, +1/4, -1/4, +sqrt(2), -sqrt(2), +1/sqrt(2), -1/sqrt(2),
> 16, 1/16, root4(2), 1/root4(2), 2^sqrt(2), 1/2^sqrt(2), and their
> negatives,
> etc.
>
> This defines a countably infinite list of reals of the form
> r(n) = 2^p or -2^p
> where
> p = 0 or 2^r(n-1) or -2^r(n-1).
It's countably infinite if you only allow finite numbers of successions. With
infinite successions, all reals can be specified exactly.
>
> But this omits a vast (uncountably infinite) number of reals, all of
> which are not some power of 2. For example, where is 3 or 1/5 in
> the list? This is equivalent to asking, what is p for 2^p = 3 or
> 2^p = 1/5?
I can't answer that at this time. I am not familiar enough with this tetration-
like math which has not been established yet.
>
> If your list were exhaustive, then you should be able to determine
> at what step n any given real x = r(n) produced. So what is n
> for 3 or 1/5?
When using a base of 1/2, both of those would almost surely require infinite
stings of bits, like they do in regular binary.
>
>

--
Smiles,

Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.

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