Re: Well Ordering the Reals

Daryl McCullough said:
> Tony Orlow says...
> >
> >Daryl McCullough said:
> >> Tony Orlow says...
> >> >
> >> >imaginatorium@xxxxxxxxxxxxx said:
> >>
> >> >> Anyway, perhaps you should do this one as an exercise. About three
> >> >> posts ago I pointed out that your belief in a "natural" lexicographic
> >> >> ordering of leftward infinite digit strings is false. Remember? Because
> >> >> you can't say which of the following (for example) comes "after":
> >> >> ......33003300330033
> >> >> ......47152471524715
> >> >
> >> >Actually, using the mirror approach, so that you consider them as the real
> >> >fractions, since 0.3..... comes before 0.5....., they are quite orderable.
> >>
> >> That isn't a well-ordering, as has been pointed out many times to you.
> >> This is an infinite descending chain:
> >>
> >> .1111...
> >> .01111...
> >> .001111...
> >> .0001111...
> >> ...
> >>
> >> (assuming you are using the normal ordering).
> >That's quite an assumption, especially since I specifically said we
> >are using the mirror of the naturals in their normal ordering.
>
> You are going around in circles. Brian asked how you compare
> the infinite strings
>
> >> >> ......33003300330033
> >> >> ......47152471524715
>
> You say reverse them, and instead compare the strings
>
> .33003300330033...
> .51742517425174...
>
> Fine. How do you compare *those*? Now, you are saying you
> are comparing them by reversing them? Then you are back
> to the question of how do you compare
>
> >> >> ......33003300330033
> >> >> ......47152471524715
>
> It doesn't matter. There is no known well ordering of
> infinite strings.
Alright. It seems that you can eliminate any infinite descending chain by
reversing the bit order, but then others become possible. Part of my issue
stems from my understanding of the set of naturals, I think, which I have to
suspend if working with your system. Oh it hurts. ;)
>
> >That's actually an infinite ascending chain in this ordering,
>
> No, it's not. If you start with
>
> .1111...
> .01111...
> .001111...
> .0001111...
> ...
>
> and reverse them, you get the sequences
>
> ...111
> ...1110
> ...11100
> ...111000
> ...1110000
>
> How is that an ascending chain? To get from the first to
> the second, you subtract 1. To get from the second to the
> third, you subtract 10. To get from the third to the fourth,
> you subtract 100. How is that ascending? What is your ordering?
> I have two infinite bit strings s1 and s2. How do I know whether
> s1 is greater than or less than s2? What rule are you using?
I had gone into a whole explanation which appears to have disappeared. Did you
see where I left it? It was around here somehwere.

I had said that one can consider it an ascending sequence as each is the last
multiplied by 10, but I suppose that doesn't fly in your world, so I'll put it
back in the hangar.
>
> The usual ordering on bitstrings is lexicographic: Let n be
> the first index at which s1 is different from s2. Then if
> s1=0 and s2=1, then s1 < s2. Otherwise s1 > s2.
>
> That ordering is not a well ordering. An alternative is to
> use the addition ordering: s1 < s2 if there is a nonzero
> bit string s3 such that s1 + s3 = s2 (where addition is
> defined bitwise, plus a "carry" bit).
>
> That's not a well-ordering, either.
>
> If you aren't talking about either of those, then
> WHAT ORDERING ARE YOU TALKING ABOUT?
I suppose, at it stands, it is essentially lexicographic, since that's what the
natural binary ordering seems to be. It would appear to be impossible to order
any set of infinite bit strings well, and since the set of bit strings could be
used to represent any ordered set, it seems impossible to well order any
uncountable set.
>
> >> >So, what is wrong with saying that .....11111 corresponds to the set
> >> >of all elements which are not in the subsets to which they map, aka
> >> >the entire set?
> >>
> >> What is wrong is that it is self-contradictory.
> >> If ...111 is an element of your set, and ...111 maps to the
> >> entire set, then ...111 is an element of the subset that it
> >> maps to.
> >...true. But then the set and the element are ill formed anyway.
>
> Then your set *N is ill formed, and it is nonsensical to say
> that you have a bijection between *N and P(*N).
Well, I have been saying all along that these sets with no ends have ill-
defined sizes and bounds. That's what makes all sorts of non-equal bijections
possible. The whole point is that more care needs to be taken with bijections
in general. But I don't expect that point to be well received.
>
> >Besides this contradiction
>
> You don't get to say "besides this contradiction". If your
> bijection involves a contradiction, you don't have a bijection.
>
> >> >This answer seems to have been rejected, but why?
> >>
> >> Because it is provably an incorrect answer.
> >Then why was it given to me as a map to the set of naturals?
>
> Once again, there is nothing wrong with infinite bitstrings.
> ...111 is a perfectly good bitstring. There is nothing wrong
> with claiming to have a bijection between the infinite bitstrings
> and P(N). There is a perfectly good bijection between those sets.
> However, an infinite bitstring is *not* an element of N. So
> the fact that there is a bijection between the set B of infinite
> bitstrings and the set P(N) does not mean that there is a bijection
> between N and P(N).
>
> If you want to say that ...111 is an element of *N, that's fine.
> You want to claim that there is a bijection between *N and P(N),
> that's fine. You want to say that ...111 maps to N, that's fine.
>
> 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.
If I can say ....1111 maps to N, what can I say maps to *N? What notation do I
use?
>
> --
> Daryl McCullough
> Ithaca, NY
>
>

--
Smiles,

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

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