Re: Well Ordering the Reals
- From: "David R Tribble" <david@xxxxxxxxxxx>
- Date: 22 Nov 2005 12:06:28 -0800
Tony Orlow wrote:
>> 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.
>
David R Tribble wrote:
>> 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).
>>
>> But this omits a vast (uncountably infinite) number of reals, all of
>> which are not some power of 2.
>
Tony Orlow wrote:
> It's countably infinite if you only allow finite numbers of successions. With
> infinite successions, all reals can be specified exactly.
>
If we allow only a finite number of successions, it will be only a
countably finite list. Since we are, in fact, allowing an infinite
number of successions, it is a countably infinite list.
But this is still much too short to be an uncountably infinite list,
and therefore it cannot possibly denumerate all the reals.
You really ought to learn the terminology, lest some will think you
are talking out of your hat. Start with "countably infinite".
>> 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.
It's your math, Tony. How can you draw any conclusions from it if it's
not established yet?
.
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