Re: Well Ordering the Reals


Tony Orlow wrote:
> 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. ;)

You're starting to figure out that the reals and the naturals have
different
properties "in the middle", and the difference in these two infinite
sets
is not that one is "longer" than the other. It's hard even to say what
we would mean by the "length" of an uncountably-long set of bits.

> > 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?

Again it comes down to the indexing set.

The reason there's a natural mapping of ...1111 to N is that what we
normally mean by ...1111 is something with a bit for every element of
N, in which each of those bits is 1.

Since the natural ordering is not a well-ordering and we don't know
how to construct a well-ordering of an uncountable set, your "natural
map" won't work for *N.

I'm suggesting giving up the vague confusing inadequate ellipsis
notation
and using function notation. Let's label the bits with, oh, reals in
[0,1].
So our bit-label function has b(x) = 0 or 1 for every x in [0,1]. I
think I'm
also going to suggest that we talk about mappings from *N to P([0,1]),
the powerset of a different uncountable set. Since we know how to
biject *N and [0,1], a bijection *N->P([0,1]) exists if and only if
there's a bijection *N->P(*N).

Now suppose we want to map elements of *N to these uncountable
strings. So for every element of *N there's a different b(x) function.
We can use the notation b_y(x) indicating the bit function representing
element y of *N.

If we're using these to map from *N to some powerset, say P([0,1]),
we'll say b_y(x) = w where w is some subset, constitutes a mapping
from y to w.

Now there's a natural bit function (uncountable bit string, if you
like)
that maps to the entire set. [0,1]. Namely the one where b(x) = 1 for
all x.
But you'll have to decide what element of *N uses this bit string.

If you manage to resolve this, then I know already that the problems,
the missing elements, will be other subsets of [0,1]. When we say
no mapping is bijective, it doesn't mean all maps are failing to map
the "last element". It means all maps are failing to map SOME element.
I don't think you get this yet.

- Randy

.

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