# nonstandard models of some of the ordinals

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### sigol...@gmail.com

Dec 18, 2004, 1:49:02 PM12/18/04
to

I have been having fun reading Robert Goldblatt's book on the hyperreal
numbers. Although I am older, and am a geneticist now, I might be
considered at the level of an advanced undergraduate. As such, I have
a question which I hope some of you can help me with.

I have been thinking of the hypernatural numbers as constructible
models for some of the smaller ordinal numbers, in a very elementary
fashion as might be explained to advanced high school students, perhaps
at the level of an article in the MAA's monthly, or an exercise.

Thus, omega = {1,2,3,...}, i.e. the sequence of natural numbers, as he
described.
Then, as a first grader looking for a bigger hypernatural number would
say
omega+1 = {2,3,4, ...}
2*omega = {2,4,6, ...}
2*omega +1 ={3,5,7, ...}
omega^2 = (1,4,9,16, ...}, omega^m = {1,2^m,3^m, ...}.

Now comes my first question. I want to get to omega^omega.

Suppose I first look at {2,4,8,16, ...,2^n, ...}.
What is this? 2^omega in the hypernaturals?

If so, then form 3^omega, 4^omega, ... .

What about {1^1, 2^2, 3^3, ...} ?
Is this omega^omega?

Then as usual {2^(2^1), 2^(2^2),2^(2^3), ...}
Is this 2^(2^omega) ?

Then {1^(1^1), 2^(2^2), 3^(3^3), ...}
Is this omega^(omega^omega)) ?

Then {1, 2^2, 3^(3^3), 4^(4^(4^4)), ...}
Is this omega^(omega^(omega^( ...))) in the hypernaturals?

Second question:
Suppose I look at {Ack(1), Ack(2), Ack(3), ...} where Ack is the
unitary Ackerman function.
What ordinal does this correspond to?
Are there any very elementary references in this regard?

Thank you,

### Lee Rudolph

Dec 18, 2004, 1:57:22 PM12/18/04
to
sigol...@gmail.com writes:

>
>I have been having fun reading Robert Goldblatt's book on the hyperreal
>numbers.

...

>Thus, omega = {1,2,3,...}, i.e. the sequence of natural numbers, as he
>described.

Okay, that's standard enough.

>Then, as a first grader looking for a bigger hypernatural number would
>say
>omega+1 = {2,3,4, ...}
>2*omega = {2,4,6, ...}
>2*omega +1 ={3,5,7, ...}
>omega^2 = (1,4,9,16, ...}, omega^m = {1,2^m,3^m, ...}.
>
>Now comes my first question.

No, before it comes *my* question: are these "first grader" encodings
of "bigger hypernatural number"s than omega being done according to
a scheme from Goldblatt's book (which I've never seen, nor even heard
of before)? If so, could you explain his scheme to those of us who
haven't seen it? If not, can you provide some meaningful explanation
of your own? Not a one of those equations makes sense to me, from
what I learned (long ago, from a book by Szierpinski, I think) about
"omega+1",..., "omega^2" are the sorts of names ordinals have, I
assume you are trying to write equations in ordinals. If not--if,
indeed, you're writing equations for some kind of "hypernatural numbers"
that don't purport to be ordinals--then please clarify further.)

Lee Rudolph

### sigol...@gmail.com

Dec 18, 2004, 2:37:28 PM12/18/04
to
Robert Goldblatt's book "Lectures on the Hyperreals" (Springer 1998),
describes aspects of the theory of nonstandard analysis. It is easy

Roughly speaking,
consider a system in which numbers are replaced by equivalence classes
of infinite sequences, two sequences being equivalent iff they differ
at only a finite number of positions.
Drastically oversimplifying, this forms a ring under pointwise addition
and multiplication of sequence elements. However, the product of two
non-zero elements can equal zero.
(e.g. {0,1,0,1, ...} * {1,0,1,0,, ...} = 0)
Ordinary numbers correspond to constant sequences.
Infinite numbers correspond to divergent sequences.
Eventually the ordinary reals embed in an extension of such a system,
as well as infinitesimals
(e.g. {0,0,0, ...} < {1,1/2, 1/3, ...} < epsilon
for any ordinary real number epsilon > 0.)
and infininite numbers.
Hope this helps

Ordinary

### David C. Ullrich

Dec 18, 2004, 3:13:33 PM12/18/04
to

Well, this makes a little more sense after your second post
on the topic. Enough sense that it's clear that your
questions about which ordinal various sequences "correspond"
to do not have answers, because you have not _specified_
what the correspondence _is_.

It's true that you've defined a few hyperreals that are
ordered in the same way as a few infinite ordinals. But
there are many other ways you could have done this -
ordinal corresponds to omega+1, so there's really no
answer to the question of which ordinals those
other hyperreals correspond to until you _define_
the correspondence.

(If that's not clear:

Even if we restrict to just sequences of positive
integers, as you seem to be doing, the hyperreal
you say "is" omega is not the smallest hyperreal
larger than every natural number. For example
(1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...) is larger than
every natural number but smaller than the thing
you're saying corresponds to omega. Why does
the sequence above correspond to omega, why
isn't it this one?)

>Thank you,

************************

David C. Ullrich

### sigol...@gmail.com

Dec 18, 2004, 3:32:35 PM12/18/04
to
Yes, you are right; I didn't think of that.
Thanks.

But suppose we do everything modulo a particular choice of a
representative sequence for omega.

Since the integer sequences are constant, the first few examples seem
to go through OK.

Then, for the, more advanced examples of larger ordinals
(i.e. each time we exponentiate) maybe we have to choose
a representative again.

Admittedly, this only gives a branching tree of models,
but still interesting, if true.

### George Cox

Dec 18, 2004, 3:34:21 PM12/18/04
to
Lee Rudolph wrote:
>
> sigol...@gmail.com writes:
>
> >
> >I have been having fun reading Robert Goldblatt's book on the hyperreal
> >numbers.
> ...
> >Thus, omega = {1,2,3,...}, i.e. the sequence of natural numbers, as he
> >described.
>
> Okay, that's standard enough.

The notation is confusing. The op means the sequence (1,2,3,...) not
the set {1,2,3,...}. So omega is not the smallest infinite ordinal.

### Ross A. Finlayson

Dec 18, 2004, 3:44:55 PM12/18/04
to
You can visualize those numbers in that way, they are names that you
assign to those constructions.

>From a set-theoretic angle, one thing that you are doing is to say {0,
1, 2, 3, ...} = omega, thus any algebaric operation on omega, the
number or ordinal, applies to each element of the set that comprises
omega.

Then, for example with the case 2 * omega, a resulting aspect of the
set, now being the even numbers, is that the asymptotic density has
been halved, it's the reciprocal and in this case inverse of the scalar
multiplier of the ordinal.

Thus, in a way you have the same number, infinity is immutable.

While that is so, infinite sets are equivalent, while that is so,
there's still the notion that you can manipulate the symbol omega,
little Greek w, just as if it were the scalar unity.

For the finite scalar multiplier the asymptotic density is only the
inverse, but with the infinite scalar multiplier, or finite
exponential, then the asymptotic density of the resulting set goes to
zero. It is yet positive: it's infinitesimal.

In another way, the density of the prime number is known, leading to an
infinite ordinal that is the ordinal formed by that set of natural
integers.

Thus, while infinity is in a way immutable, you can consider performing
operations on it as if it were a number, where it automatically
equalizes to itself with the accounting in the second column.

Don't mind me too much, I'm well-known for discussing infinity.
Basically, the notion is that infinity is in various ways usable and
conincident and coexistent with zero, one and after the negative
numbers negative one.

That's where at the deep foundation of infinity, it is the same as the
void, and the difference subsumes all inbetween.

I, and sometimes we, can consider infinity and null as the same or
indistinguishable things, because they share common properties of the
necessarily unique and existent non-entity (entities). That enables
the mathematical discussion to neatly dovetail with the logical and
even philosophical, and for some, the epistemological and existential.
Happy Holidays, and warm regards,

Ross Finlayson

### sigol...@gmail.com

Dec 18, 2004, 11:17:03 PM12/18/04
to
and I now I can say things a little differently.

A few points.

First, two apologies.
A. For now, as George Cox mentioned above,
I was unfortunately using brackets to designate sequences not sets;
I will try to make it clearer in the future and less confusing.
(I borrowed the notation from the book.)

B. A correction: Professor Goldblatt never actually wrote or described
that omega,
the smallest infinite ordinal of set theory = sequence{1,2,3, ...}.
This was an overenthusiatic error of my own,
and as pointed out by Prof. Ullrich above, cannot be correct in the
naive form in which it was written,

Second, that out of the way, what am I trying to say?

Suppose, inspired by the construction of the hyperreal numbers in non
standard analysis,
we choose a particular integer sequence as a name for the smallest
infinite ordinal of set theory,
which is usually called omega. This sequence name is not uniquely
specified,
as pointed out by Prof. Ullrich above, but my hunch is that
the structure of what follows will be invariant with respect to
different
initial choices of otherwise equally good one sided sequences as names
for omega.

In particular, suppose we choose (my hunch above is without loss of
generality)
the particular sequence (not set) {1,2,3, ...}
as the new fixed name for omega, the smallest infinite ordinal of set
theory.
Then, in the ring of hyperinteger sequence names, other names suggest
themselves for other ordinals, as indicated in my original post, above.

For example,there is of course in set theory an ordinal sucessor to
omega, often called omega+1.
The sequence (not set) {2,3,4, ...} suggests itself as a name for this
ordinal,
under this naming convention, and so on as suggested above.

What's the point? If the guess of invariance with respect to naming
convention is right,
then the set of names under fixed convention have the structure one
might have guessed when one was 9 years old and wondering about
infinity,
i.e. a commutative ring, just like the integers.
So in the age old discussion, one could have said,
"No, infinity is not a number, but its name is a hypernumber", or
something.
E.g seq{1,2,3, ...} - seq {2,4,6, ...} = - seq {1,2,3, ...}, etc.

This seems cool. The question is, is it ever at variance with the
actual structure of the ordinary ordinals,
or as I assumed, can the ordinals be embedded in the hyperintegers with
naming convention,
in the same way that the reals can be embedded in the hyperreals?

P.S. Why the guess of invariance with respect to the initial choice of
names for omega?
Because whatever name we choose, the ring structure still applies.

### G. A. Edgar

Dec 19, 2004, 6:06:12 AM12/19/04
to

> Because whatever name we choose, the ring structure still applies.

You know, addition and multiplication in the hyperreals satisfy the usual
identities, addition and multiplication in the ordinals do not...
not commutative, for example.

There is a scheme used by logicians where rapidly-growing sequences if
natural numbers correspond to ordinals; but they do not bring in nonstandard
analysis as an intermediary.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

### George Cox

Dec 19, 2004, 9:55:09 AM12/19/04
to
sigol...@gmail.com wrote:
>
>
> For example,there is of course in set theory an ordinal sucessor to
> omega, often called omega+1.
> The sequence (not set) {2,3,4, ...} suggests itself as a name for this
> ordinal,
> under this naming convention, and so on as suggested above.

No, if omega is the smallest infinite ordinal, omega + 1 does not equal
{2,3,4, ...} (sequence or set).

If omega is 1,2,3,... then omega + 1 is 1,2,3,...,1.

### KRamsay

Dec 19, 2004, 11:16:16 PM12/19/04
to

sigol...@gmail.com writes:

(I shortened the lines here.)

|In particular, suppose we choose (my hunch above is without loss
|of generality) the particular sequence (not set) {1,2,3, ...}

Probably (1,2,3,...) would be better.

|as the new fixed name for omega, the smallest infinite ordinal
|of set theory. Then, in the ring of hyperinteger sequence names,
|other names suggest themselves for other ordinals, as indicated
|in my original post, above.

Ordinals correlate to well-orderings. So if you want to relate them
to the hyperintegers, what you want to do is find a well-ordered
subset of the hyperintegers. I think this is what you want to do. :-)
The hyperintegers >0 are not well-ordered. You have for example an
infinite descending sequence (1,2,3,...)>(1,1,2,3,...)>
(1,1,1,2,3,...)>.... So if you want to have a well-ordered subset
of the hyperintegers, you have to just decide to take only some of
them.

Someone pointed out already that the operations on the hyperintegers
and the operations on the ordinals don't correspond. I'll give you
another example of that. As ordinals, 2^omega=omega. This is different
from how hyperintegers work, where 2^omega>omega, and is also different
from how operations on infinite cardinals work. So you need to be
careful. If I were you, I'd pick a different name for (1,2,3,...),
actually, than "omega" just to be sure I didn't confuse the two.
The sum, product, or exponential of two ordinals will not generally
map to the sum, product, or exponential of the hyperintegers you
mapped them to. But that's okay as long as you remember it.

Given a countable collection of hyperintegers, there exists a
hyperinteger greater than all of them. (Exercise!) So any countable
ordinal can be embedded in them, and even aleph_1. I don't know
what ordinals can be embedded in them; I don't see offhand whether
aleph_1+1 embeds in them.

Keith Ramsay

### anal...@gmail.com

Dec 20, 2004, 7:28:55 AM12/20/04
to
Isn't everything defined componentwise?

I.e. (A,B,...)^(a,b,...)=(A^a,B^b,...)

(when you wrote (A,B,...)^2 this is to be interpreted as
(A,B,...)^(2,2,...), because 2 corresponds to (2,2,...) (or the
equivalence class containing (2,2,...) anyway ) )

### George Cox

Dec 20, 2004, 1:30:16 PM12/20/04
to
anal...@gmail.com wrote:
>
> Isn't everything defined componentwise?

If you wish, or if someone wishes. But if you use "ordinal" and "omega"
near one another, people[1] will think that omega is the ordinal of that
name, and ordinal arithmetic is not defined componentwise (there are no
components).

[1: I did, sorry if I got you wrong.]

### anal...@gmail.com

Dec 20, 2004, 3:54:02 PM12/20/04
to

> If you wish, or if someone wishes. But if you use "ordinal" and
"omega"
> near one another, people[1] will think that omega is the ordinal of
that
> name, and ordinal arithmetic is not defined componentwise (there are
no
> components).
But then don't you lose the whole ultraproduct structure?

### sigol...@gmail.com

Dec 23, 2004, 11:08:37 PM12/23/04
to
Thanks, this has been a big help, a little like having a lunch
together.

Summarizing, what I have learned so far:
1. I should not have been talking about the ordinals, but simply about
the hyperintegers as algebraic representations of infinities. Any
relation to any ordinal numbers is yet to be shown. My mention of the
ordinal omega above was an error in the above context.

2. The positive hyperintegers are not well ordered ,but rather
partially ordered. There is no smallest infinite hyperinteger.

3. Athough the hyperintegers form a commutative ring, the ordinals are
not commutative under addition, multiplication, or exponentiation.
Note that some additions, multiplications, and exponentiations in
the ordinals appear consistent with what might be expected in the
positive hyperintegers, namely if a larger limit number is to the left
of a smaller number in the binary operation. The incompatibility
appears when the larger limit ordinal is to the the right of the
smaller one.

Speculation
4. Limit ordinals are limits. It is not clear how the hyperintegers I
was mentioning are limits, although it might be interesting to try to
define sucessors.

5. "Ordinals correlate to well-orderings. So if you want to relate them

to the hyperintegers, what you want to do is find a well-ordered
subset of the hyperintegers. I think this is what you want to do. :-)
The hyperintegers >0 are not well-ordered. You have for example an
infinite descending sequence (1,2,3,...)>(1,1,2,3,...)>
(1,1,1,2,3,...)>.... So if you want to have a well-ordered subset
of the hyperintegers, you have to just decide to take only some of

them." -from KRamsay (above)

Yes, I think this is precisely what I want to do. Intuitively, I feel
that if I look at the branching tree of the partially ordered
hyperintegers > (0,0,0, ...), some of the branches will contain
constructible representations of apparently very big entities, maybe
even something like Cantor's epsilon0.
Then, if we choose just the paths leading to this representation of
something apparently large, we will get an ordered seguence, which
might, somehow, be compared with the ordinals. The non commutativity
remains an issue, of course.

Thanks again,

Message has been deleted

### sigol...@gmail.com

Dec 23, 2004, 11:20:57 PM12/23/04
to
PS Note:

" 4. Limit ordinals are limits. It is not clear how the hyperintegers I
was mentioning are limits, although it might be interesting to try to
define sucessors. "
By "successors" I meant nonunique partially ordered analogs of
successors,.

### George Cox

Dec 24, 2004, 2:39:15 PM12/24/04
to
sigol...@gmail.com wrote:
>
> constructible representations of apparently very big entities, maybe
> even something like Cantor's epsilon0.

Does epsilon_0 count as very big? It's countable.

### sigol...@gmail.com

Dec 27, 2004, 10:32:11 PM12/27/04
to
That was surprising and at first shocking news to me. I'll look for
the bijection. It just goes to show that if you start out ignorant,
you can learn a relatively large amount quickly.

Two ideas occur in response.
First, there appear to be an uncountable number of hypersequences,
which as above, can be arranged in a branching partially ordered
infinite tree.
Second, the question arises, if the whole program outined above works,
can one say that no totally ordered path (chain) ever reaches the
representation of an uncountable ordinal?
This would say that however rapidly a sequence grows, considered as a
hypersquence, it always corresponds to something countable, i.e. the
length of the chain going back to the sequence (1,2,3, ...) is always
countable I am specifically thinking of sequences (perhaps like the
"busy beaver" sequence?), which I have read are not recursively
computable by a Turing machine, and grow very rapidly.
This seems like an interesting question.