Re: Well Ordering the Reals

David R Tribble said:
> Tony Orlow wrote:
> >> You know, I am starting to get confused when asked about bit string length,
> >
>
> Randy Poe said:
> >> Well then it's a good thing nobody but you is talking about bit
> >> string "length" for infinite collections of bits, isn't it?
> >
>
> Tony Orlow wrote:
> > No? How many of you said that there aren't enough bits in the elements of *N to
> > correspond to the number of elements in P(*N)? If the string never ends, how
> > can it be too short for anything? And you're not talking about length.....
>
> I said that. I was referring to the fact that a simple binary
> bitstring mapping of the finite and infinite naturals in *N to
> the subsets of *N is not sufficient to cover all the subsets, and
> hence is not a bijection.
>
> If you choose a particular infinite natural m in *N to map to the
> entire [sub]set *N, then since it maps to all the members of *N
> it must be larger than any other member in *N (because it has
> all its bits set to 1). But m must have a 1 bit that corresponds
> to m itself in the subset. But since that 1 bit is in position log2(m)
> of m, this implies that there is another member of *N that is larger
> than m (mapped by the next higher bit position in m, which is also
> a 1, because m > log2(m)), which is a contradiction.
>
> Hence my comment that no member of *N has "enough bits" to map to *N.
Well, good. That comment is actually correct. I wish that these considerations
of value range were kept in mind in general. Indeed that's what this is. You
have noted that the number that could map to the last element of the set is
outside the range of the set. This is also true when mapping the evens. What
even is twice the size of ....11111? Is it .....111110? Oh wait, that's one
LESS than ....1111, or is it?
>
>

--
Smiles,

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

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