Re: Well Ordering the Reals
- From: Virgil <ITSnetNOTcom#virgil@xxxxxxxxxxx>
- Date: Thu, 10 Nov 2005 17:22:10 -0700
In article <1131667512.624822.245800@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"David R Tribble" <david@xxxxxxxxxxx> wrote:
> Daryl McCullough said:
> >> But you were claiming that you have a mapping between
> >> *N and P(*N). That's not fine. As a matter of fact, it is provably
> >> false.
> >
>
> Tony Orlow wrote:
> > If I can say ....1111 maps to N, what can I say maps to *N? What notation
> > do I
> > use?
>
> That's a very good question. You see that there is a problem
> here, right?
>
> n = ...111 maps to set N. n has a countably infinite number of 1 bits,
> each of which designates a finite natural of N. So even though n is
> an infinite value, it only has enough bits to enumerate the finite
> naturals of N (or *N).
>
> All of this is just one more way of saying that *N can biject P(N),
> but *N is not large enough to biject P(*N).
>
> To map numbers to subsets of *N, you would have to invent an
> even longer type of infinite number, with more than a countably
> infinite number of bits, to use it to map to any subset containing
> infinite naturals from *N. How you do that is going to be very
> interesting.
>
> But even if you did pull that off, those new numbers would not be
> members of *N, so you still could not biject *N and P(*N).
The way that I understand TO's *N set, every element of *N is a
countably infinite sequence of binary digits. Thus, in effect, every
member of *N is of the form of a function f:N -> {0,1}.
And similarly TO's P(*N) corresponds to the set of all functions
g:*N -> {0,1}, where g(*n) = 1 or 0 according to whether *n is in the
set represented by g or not.
That creates a problem for TO, though, as any bijection from the set of
all functions f:N -> {0,1} to the set of all functions g:*N -> {0,1}
requires generates a bijection from *N to N, according to category
theory.
Thus any mapping *f: *N -> P(*N) can only be a bijection if there is a
bijection f: N -> *N.
Well, TO? Is there any bijection from N to *N?
.
- References:
- Re: Well Ordering the Reals
- From: Robert Low
- Re: Well Ordering the Reals
- From: David Kastrup
- Re: Well Ordering the Reals
- From: Tony Orlow
- Re: Well Ordering the Reals
- From: Daryl McCullough
- Re: Well Ordering the Reals
- From: Daryl McCullough
- Re: Well Ordering the Reals
- From: Tony Orlow
- Re: Well Ordering the Reals
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