Re: Well Ordering the Reals

David R Tribble said:
> Tony Orlow wrote:
> >> Ugh. When you enumerate the power set according to the natural ordering of the
> >> binary representations of the subsets, the first is the null set, and the last
> >> is the set itself, represented by a string of 1's as long as the set is big.
> >
>
> Brian Chandler said:
> >> Let's take this a bit at a time. This is fine for finite sets. But for
> >> an infinite set, in normal terminology, you cannot _enumerate_ the
> >> power set, because "enumerate" means mapping to the pofnats; a list
> >> that starts with one element, followed by a second element followed by
> >> lots of elements each following the preceding elements, without
> >> necessarily ever ending. You *can* represent the power set (of the
> >> pofnats) by binary representations, but these have to be unending
> >> binary strings, just like the binary representations of reals in [0,1].
> >> Of course there is a binary string (.000000... [without ever ending])
> >> that represents 0, and another (.111111... [without ever ending]) that
> >> represents 1, but this is not an enumeration, because between any pair
> >> of reals is another one. So no real has a "next real". (E.g. there is
> >> no second binary string following .000000... .
> >
>
> Tony Orlow wrote:
> > I don't think I need to remind you that one can always reverse-rder the strings
> > as the mirror of the natural binaries, so that .100000... is the successor to
> > .000000.
> >
> > .000....
> > .100....
> > .010....
> > .011....
> > .100....
> > etc..
> > .111.... [*]
>
> But this list is only a countably infinite number (Aleph_0) of reals.
According to whom? If those bit strings are finite, then perhaps, but they're
countably infinite strings of bits, one for each n in N, and forman uncountably
infinite set of strings.
> Specifically, it lists only binary fractions with a finite number
> of nonzero digits, those that end in a sequence of all zero bits.
No, there is a bit for every natural. You could have 1's forever, 1 for each
natural.
> The list omits, in particular, any binary fraction with nonterminating
> digits, such as 1/3 = 0.01010101... .
No, that corresponds to a 1 in every bit indexed by an odd natural (assuming 0
first natural).
> This is an obvious consequence
> of the fact that there are only a countable infinitude of natural
> numbers whose bitstrings can be reversed to form the list of fractions.
The fractions in [0,1] are considered an uncountably infinite set, and the bits
are considered a countably infinite set.
>
> The last fraction in your list (marked [*]) does not actually exist in
> your list. If it did, it would have a predecessor, but no such
> predecessor exists, either. You can't get from finite-length
> bitstrings to infinite-length bitstrings. But I know you don't
> understand this.
So, you say now that one cannot have infinite length bit strings in any set?
You better discuss this with your colleagues. Funny how all these folks that
supposedly agree can't get the line-toeing answer consistent.
>
>
> Tony Orlow wrote:
> >> When the set is infinite, this is still the LAST element of the set, which you
> >> will never reach, ...
> >
>
> Brian Chandler said:
> >> Ah, there you go. Repeat after me: sets are collections of those
> >> elements that are their members. There is nothing more to it. "Never
> >> reach" does not come into it. .111111... certainly exists, and the
> >> corresponding subset exists. The end.
> >
>
> Tony Orlow wrote:
> > What is the corresponding subset to .11111....., specifically?
>
> Fraction n = 0.111... maps to subset Sn = {0,1,2,3,...}, which is
> the infinite set of all finite naturals. But your list above cannot
> include n.
>
>

--
Smiles,

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

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