Re: Well Ordering the Reals

David R Tribble wrote:
>> If your binary numbers contain more than a countable infinitude of
>> bits, then they are not members of *N. Also, once you have lost
>> the ability to denumerate the bits, you have lost the ability to derive
>> the value (the sum of the digits) of such a number, and consequently
>> such numbers cannot be compared nor ordered.
>

Tony Orlow wrote:
> So, you have no way to represent a member of P(*N).

Sure I do. A member of P(*N) is a subset of *N, containing zero or
members of *N. But each member of *N is either a finite natural, or
an infinite nautral composed of a countably infinit number of bits.

Some examples:
{2}
{0, 1, 2}
{4, 6, 18, 218, ...}
{...806, ...807, ...808}
{0, 1, 2, 3, ..., ...001, ...002, ...003, ...}

The last three subsets contain infinite numbers from *N, using your
notation. The last set is *N itself.

But if you start using numbers with an uncountable infinitude of
bits, you can't use such simple notation for them.

Huyah Huyah Huyah Ommmmmmm.

.

Quantcast