Re: Well Ordering the Reals

Randy Poe said:
>
> Tony Orlow wrote:
> > Daryl McCullough said:
> > > Yes, the contradiction follows from your assumption that you
> > > have a bijection between a set and its power set. Normal people
> > > would say "My assumptions lead to a contradiction. Therefore,
> > > one of my assumptions is wrong."
> > Except that, when applied to my bijection, this element in question is
> > obviously the last of an unending set, which is the root of the contradiction.
>
> So if I construct a map between the countable bit strings and
> P(N) and claim it is a bijection, and specifically that there exists
> a bit string that maps to N, doesn't that have the same root?
> Shouldn't it give rise to the same contradiction?
You mean the bit strings which are actually allowed to have infinite length, as
in *N? Which of those maps to N?
>
> Yet for some reason it does not. There does indeed exist a
> countable bit string in my construction which maps to N, despite
> the fact that P(N) has no last element. I make no claim that N is
> the "last element" of P(N), or that my enumeration has a last element.
> Yet I can prove that every element of P(N) is mapped by a bit
> string.
Can you tell me the string that maps to N?
>
> Why is that? Why is it that only in *your* putative map from *N to
> P(*N) does requiring something to map to *N means you have to
> have a last element of P(*N) and that's contradictory, where as
> in my map to P(N), requiring something to map to N does NOT mean
> I have a last element of P(N) and there is no contradiction?
I don't know. What maps to N?
>
> You keep asking what other elements of P(*N) are not mapped besides
> *N? That's a little hard to pin down since you refuse to be pinned
> down on how many bits are in your bitstrings. However, if I
> assume that the bitstrings in your map have countably many digits,
> i.e. one bit for every finite natural, then no subset of *N containing
> an "infinite natural", whatever that looks like, is mapped.
The bitstrings in *N are infinite. You may consider them uncountably infinite,
such that there is a never-ending string of them extending to the left. The
question is, which bit does not have a corresponding natural in *N, and which
natural in *N does not index a bit? I don't see the point where this breaks, as
an infinite bijection.
>
> - Randy
>
>

--
Smiles,

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

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