Re: Well Ordering the Reals

Randy Poe said:
>
> Tony Orlow wrote:
> > Randy Poe said:
> > >
> > > Tony Orlow wrote:
> > > > David R Tribble said:
> > > > > David R Tribble wrote:
> > > > > >> All members of *N will map to subsets of *N. That's not the problem.
> > > > > >> There are subsets of *N that are not mapped by any member of *N.
> > > > > >> That's the problem.
> > > > > Tony Orlow wrote:
> > > > > > Oh. Can you name one, please?
> > > > >
> > > > > I already have, several times. You just keep ignoring them.
> > > > >
> > > > No you haven't. You have spooken in vague terms, claiming that N indexes the
> > > > bits of each element in *N
> > >
> > > What is vague about "there is one bit for every n in N"?
> > >
> > > > when in fact you need bits with actually infinite
> > > > bit positions to achieve the infinite values in *N.
> > >
> > > This claim is either (a) wrong or (b) unjustified, since you haven't
> > > defined
> > > what you mean by this set *N. The only definition I've seen so far
> > > is that "*N is the set of bit strings whose bits are labelled with
> > > elements
> > > of *N". From that I conclude that *N is the empty set, with no bits,
> > > and that there is no need for infinite labels.
> >
> > But even with no bits you index one element,
>
> You're confusing *N and P(*N). The empty set does not have one
> element to enumerate. The powerset of the empty set does.
Right. I am saying that if you have a base set which requires no bits to
specify an element in the set, then you can still have one element there, as
there is nothing needed to specify the sole element, in which case the power
set will have two elements, the null sets and the singleton set, and require a
bit to distinguish between them.

I suppose if you have the null set, the power set of the null set is
essentially itself, because the null set and the base set, which are always the
first and last subsets in the power set, are the same, so the power set would
have 1 element, and also require 0 bits to specify that element. So, if *N
weren't specified as the naturals with no restriction of finiteness, then the
null set would satisfy the description of a set where it takes the same number
of bits to specify an element in the set as to specify an element in the power
set. Of course, we are talking about infinitely long strings, not null strings.
>
> > so *N would be a singleton,
>
> Eh? What is this single element of the empty set?
I am suggesting that requiring no bits to specify an element doesn't make the
set empty, but I suppose that is a little irrelevant.
>
> > with a
> > two-member power set, requiring a bit.
>
> > *N is the set of naturals including infinite naturals, which necessarily
> > contain bits in infinite positions.
>
> Too vague. You haven't given me a separate meaning of
> "infinite positions" by telling me a set other than *N itself which I
> can use for indexing.
If *N is the set including al naturals finite and infinite, then it is the set
of choice for indexing a set which requires infinite indexes into the set.
>
> > Unfortunately, you are not going to agree
> > with this, since it is the standard position that there are an infinite number
> > of finite naturals,
>
> No, I'm willing to discuss a set with infinite "positions" in the sense
> that
> there is a bit-function which assigns 0 or 1 to every element of an
> uncountable set. I don't agree that N can serve as that set. So how
> about [0,1]?
Well, I thought about that. N can't serve as the index because no n in N is
infinite. The reals in [0,1] really aren't discrete in their natural order in
that there is no discrete successor, but I could apply my ordering scheme for
the reals to create a succession of sorts. Still, that doesn't seem likely to
work. Since we are "counting" an infinite number of elements, it is natural to
use counting numbers, but they muct include infinite values. It seems strange,
but *N seems to be the natural index for the bits in *N. The only way to
resolve this, as far as I see, is to develop notation that distinguishes
between |N| bits, |*N| bits, and |P(*N)| bits.
>
> My objection to saying that "*N is the set of numbers whose binary
> representations have one bit for every element of *N" is meaningless
> nonsense.

Your objection is meaningless nonsense? Oh don't be so hard on yourself. I
mean, it's meaningless, and nonsense, just don't be so hard on yourself. ;)

> The way definitions work is that they are defined in terms of
> other terms on which you and your audience agree. You can't define
> *N in terms of *N.
Why not? The set of binary numbers with uncountable bits to the left of the
point and all zeros to the right, such that all elements are whole numbers,
itself indexes the uncountably infinite set of bits used by each of its
elements. If you can formulate an actual contradiction that this causes, that
would be interesting to see.
>
> The empty set has no elements. No bits are required to represent its
> elements. Therefore { } meets this definition, your only definition of
> *N so far.
No, I specified that it is the set of naturals, finite and infinite, such that
it requires an uncountably infinite number of bits to specify an element in the
set. Maybe you think this set doesn't exist, but it's certainly conceivable to
me.
>
> Saying "*N is the set of naturals including infinite naturals" doesn't
> work either. What's an infinite natural? Is it the set of things which
> one
> bit for every finite n in N? That will do for me, but I think not for
> you.
No, according to my understanding, any set of finite naturals or of finite
strings on a finite alphabet may have no finite upper bound, but is also not
actually infinite. It is potentially infinite but limited.

> If you want to say "the set of infinite naturals is the set of numbers
> with one bit for every infinite natural" well then we're back in the
> circle again aren't we?
Yes, kind of. We will say it is the set of binary numbers with uncountably many
bits for each element. Does that seem like it leads you anywhere comfortable?
(probably not)
>
> Are you incapable of defining the representation of *N in terms of some
> set that isn't *N?
I am capable in my system of specifying the number of bits formulaically, to
finer distinctions than the power set distinction here, but whatever. If I were
you, I would probably say, "Okay. You want aleph_1 bits for elements in *N?
Fine, you have aleph_2 bits for elements in P(*N)". Personally, I prefer
leaving hebrew letters out of the discussion if they are unnecessary, and
saying that if set S has n elements, then P(S) has 2^n elements, and that this
IS using the regular arithmetic exponentiation operator in the infinite case.
As you count through S, adding elements, each time you double the size of the
power set, which is simply adding another bit to the strings in the power set.
That doesn't seem like a concept that breaks down for infinite n.
>
> - Randy
>
>

--
Smiles,

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

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