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From: Transfer Principle on 26 Jun 2010 00:01
On Jun 24, 10:59 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-06-24, Transfer Principle <lwal...(a)lausd.net> wrote:
> > So tav appears to contain all the pofnats. But it also contains
> > elements which aren't pofnats.
> And so we have a contradiction.  Gee, that didn't take long.

But now TO is posting again, so we can fix this error based
on TO's latest comments.

Let's summarize what TO has stated.

1. The set omega is exactly what we expect it to be.

2. N+ is the set of all pofnats, i.e., omega\{0}. It is a
countably infinite set.

3. *N contains the nonstandard naturals (which Chandler calls
"tnats," but TO calls T-riffics).

Notice that sometimes the notation *N is used for Robinson's
hyperreals as well, with *R being the set of hyperreals and
*N the set of hypernaturals. Although TO has stated that his
T-riffics are _not_ the hypernaturals, the use of similar
notation implies that the concepts are analogous.

How can we fix the axioms? At this point, it appears that we
can just go back to ZF and assign Bigulosities to the sets
of ZF, as I originally intended to do. This means that we can
go back to determining which bijections are "strong."

TO states that "tav" isn't a set, at least in the way that I
have been using it. But in ZF, _everything_ is a set (i.e.,
there are no urelements), so tav must be a set.

A similar dilemma occurs when we wish to extend the set R of
standard reals to the extended reals. We wish to append two
new elements, oo and -oo, to R, but what sort of objects are
oo and -oo, anyway? They must be sets since every object in
ZF is a set. If one were to use Dedekind cuts to construct
the reals, then one might define oo to be Q and -oo to be the
empty set, but what if we wanted to find a definition that
wasn't depended on the D-cut construction of R? It doesn't
matter what sort of objects we use for oo and -oo as long as
they differ from each other and from every element of R. (For
example, the Metamath website defines oo to be P(U(C)) and
-oo to be P(P(U(C))).)

And likewise with tav. All we demand of tav is it to be
different from any other object assigned to be the Bigulosity
of some set. It doesn't matter whether tav is a finite or an
infinite set -- what matters is that the object assigned the
Bigulosity tav (such as N+) be infinite. This is trickier
than the extended reals since we only needed two new objects,
oo and -oo, while we need infinitely many new objects to be
Bigulosities (since tav-n is a separate Bigulosity for each
natural number n).

After doing so, then we simply treat tav and the other
Bigulosities as symbols, just as TO demands.
From: Transfer Principle on 26 Jun 2010 00:37
On Jun 25, 8:57 am, Tony Orlow <t...(a)lightlink.com> wrote:
> On Jun 24, 10:01 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> > In Walker's defense, he said that the formula phi should not contain the
> > symbol tav.  
> Good point. Maybe phi should not include "finite" either, but at
> least, specifying an inequality without a limit of 0 is sufficient to
> avoid supposed contradictions.

This was something that I was wondering earlier as well. We
could add a new primitive "finite" and declare that phi
cannot contain the "finite" primitive. Then we can add some
axioms stating which objects are finite, such as:

finite(0)
(Ax (finite(x) -> Ay (finite(xu{y}))
~finite(tav)

I wasn't sure whether to accept this though, since at the
time I was wondering whether a set of pofnats could be
infinite without containing an infinite element. Although
phi in this version of the ICI schema can't contain the
symbol "finite," recall that we are considering the other
axioms of ZF-Infinity, including the Separation Schema,
which has no such restriction regarding which symbols can
appear in its wff phi. So we could have:

w = {ne(tav) | finite(n)}

and w would apparently be an infinite set, all of whose
elements are finite, hence a contradiction.

But of course, all of this was based on treating tav as a
set in the way that finite naturals are sets. But TO tells
us that tav should _not_ be treated this way, thus making
this version of the schema incorrect.

Of course, tav is still a set, since all objects in ZF must
be sets. But tav can be _any_ set whose existence is provable
in ZF, including a _finite_ set, as long as it doesn't equal
any other set assigned to be a Bigulosity (including the
finite Bigulosities).

For example, we could define tav to be {{0}}. Sure enough,
this would mean that Bigulosity(tav) = 1 -- but tav is to be
treated as a _symbol_. We couldn't care less what Bigulosity
tav has -- we only care about what the Bigulosity of sets
such as N+. It's not Bigulosity(tav), but Bigulosity(N+) that
matters, and as long as Bigulosity(N+) = tav, then Bigulosity
is working as it's supposed to. Of course, this also means
that Bigulosity(Bigulosity(N+)) = 1, but no one required
Bigulosity to be idempotent.

Similarly, in standard ZF, complex numbers such as 3, -1, or
i each have cardinality 2, if we define a complex number to
be an ordered pair of real numbers. And if the complex number
has equal real and imaginary parts, such as 1+i, then the
(Kuratowski) ordered pair reduces to a singleton, so its
cardinality would be 1. But statements such as "card(i) = 2"
and "card(1+i) = 1" have no practical value, and likewise
"Bigulosity(tav) = 1" has no practical value.

So the focus should be on assigning Bigulosities to sets, not
worrying about what Bigulosities actually are.
From: Transfer Principle on 26 Jun 2010 00:46
On Jun 25, 11:30 am, Brian Chandler <imaginator...(a)despammed.com>
wrote:
> > When it comes to set C, it *looks* like set A, however, if there are
> > tav+1 naturals (starting at 0) in decimal, there are log_10(tav+1) bit
> > positions. Given N=S^L, with S=2 and L=log_10(tav+1), we have
> > 2^log_10(tav+1)=log_5(tav+1) elements.
> This doesn't seem to be right. Just suppose tav=999. Then 2^(log10(tav
> +1)) = 2^3 = 8. And 5^8 does not equal 1000. Not that details like
> this seem very important, somehow.

Let's see whether we can simplify this correctly. If
t+1 is a finite number, then:

2^log_10(t+1) = (t+1)^log_10(2)

Checking this with t = 999 gives:

(999+1)^log(2) = 1000^log(2) = 8

as expected.

So via ICI, we conclude that the Bigulosity of set C
must be (tav+1)^log(2).
From: Virgil on 26 Jun 2010 01:32
In article
<600d5e30-3ee9-4163-a862-7fa20893c0e6(a)j4g2000yqh.googlegroups.com>,
Tony Orlow <tony(a)lightlink.com> wrote:

> On Jun 25, 5:51�pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote:
> > Tony Orlow <t...(a)lightlink.com> writes:
> > > Where you have a minimum positive difference between successive
> > > elements (like 1), then any infinite number of them means an infinite
> > > difference between the first and last.
> >
> > Is it your opinion that N+ has a last element?
>
> No, and that's beside the point. If there are an infinite number of
> unit increments between any two values then there is an infinite
> difference between them. If not, then it's a countable sequence. N+
> is countable. Some subsets of R *might* be measurable even though not
> continuous (perhaps the Cantor set?), but most are countable, and
> therefore without real measure.

Since R is not countable, MOST of its subsets are also not countable,
including a variety of sets related to the Cantor 'ternary' set.
Measurability is quite a dfferent property.
From: Virgil on 26 Jun 2010 01:36
In article
<f213f380-e97c-4b76-9fc0-30c694abfc69(a)a30g2000yqn.googlegroups.com>,
Tony Orlow <tony(a)lightlink.com> wrote:

> > Tony Orlow wrote:
> > > How far is H-riffic 3 from 2 (in base 2 H-riffic) ?
> >
> > That fails to answer my questions. I'll ask again.
>
> I have already answered this question to a large extent with the H-
> riffics. There is some proof left to do, but the fact is that taking
> log(|x|) repeatedly tends around 0, and so the probability of the root
> of any number becoming 0 through this recursive process eventually
> approaches 1.

Nonsense. since the set of numbers that can be generated by TO's process
is at most countable, the probability that a real number can be so
generated is necessarily zero, even should it turn out that the set of
TO's numbers in dense in R.
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