# Complex multi-valued functions taking on uncountably infinitely many values?

4 views

### mike3

Jul 20, 2009, 4:44:51 PM7/20/09
to
Hi.

Is it possible for a complex, analytic "multi-valued function" to take
on an _un_countably infinite number of values? If so, is there a good
example? If not, why not? Note that "log" takes on countably
infinitely many, not uncountably infinitely many.

http://www.tetration.org/tetration_net/tetration_net_Tetration_Dynamics.htm
"My position is that any complete extension to the domain of tetration
must account for the fact that ^b a may take an uncountable infinity
of values."

### Karl-Olav Nyberg

Jul 20, 2009, 5:21:38 PM7/20/09
to

"mike3" <mike...@yahoo.com> skrev i melding

Hi.
The first "functions" that strikes my mind are the inverse trigonometric.
Given sin, cos, tan or cotan of an angle, there ar an_un_countably infinite
number of values for the angle (I guess?).

Karl-Olav Nyberg

### mike3

Jul 20, 2009, 6:56:37 PM7/20/09
to
On Jul 20, 3:21 pm, "Karl-Olav Nyberg" <konyb...@online.no> wrote:
> "mike3" <mike4...@yahoo.com> skrev i meldingnews:5714cbb4-0e18-4d3e...@t13g2000yqt.googlegroups.com...

>
> > Hi.
>
> > Is it possible for a complex, analytic "multi-valued function" to take
> > on an _un_countably infinite number of values? If so, is there a good
> > example? If not, why not? Note that "log" takes on countably
> > infinitely many, not uncountably infinitely many.
>
> > Because I read this:
>
> >http://www.tetration.org/tetration_net/tetration_net_Tetration_Dynami...

> > "My position is that any complete extension to the domain of tetration
> > must account for the fact that ^b a may take an uncountable infinity
> > of values."
>
> Hi.
> The first "functions" that strikes my mind are the inverse trigonometric.
> Given sin, cos, tan or cotan of an angle, there ar an_un_countably infinite
> number of values for the angle (I guess?).
>
> Karl-Olav Nyberg

Nope, its only countable infinity, like "log". If sin(x) = y then sin
(x + 2npi) = x
for integers, and only integers n. This gives only a countable set of
values for
the inverse sin^-(y) = x + 2npi, as the integers are countable.

### David R Tribble

Jul 20, 2009, 8:39:14 PM7/20/09
to
mike3 wrote:
>> Is it possible for a complex, analytic "multi-valued function" to take
>> on an _un_countably infinite number of values? If so, is there a good
>> example? If not, why not? Note that "log" takes on countably
>> infinitely many, not uncountably infinitely many.
>

You mean for any given single value of x.

### mike3

Jul 20, 2009, 9:00:27 PM7/20/09
to

Yes, that's what I mean.

### Don Stockbauer

Jul 20, 2009, 10:32:09 PM7/20/09
to

1. Potential

2. Actualized

### mike3

Jul 21, 2009, 12:56:21 AM7/21/09
to

Huh?

### Mariano Suárez-Alvarez

Jul 21, 2009, 3:34:53 AM7/21/09
to
On Jul 20, 5:44 pm, mike3 <mike4...@yahoo.com> wrote:
> Hi.
>
> Is it possible for a complex, analytic "multi-valued function" to take
> on an _un_countably infinite number of values? If so, is there a good
> example? If not, why not? Note that "log" takes on countably
> infinitely many, not uncountably infinitely many.

No 'multivalued function' takes uncountably many
values at a point.

One way to make sense of 'multivalued' functions is to
consider Riemann surfaces S and pairs of analytic maps

p f
C <----- S -----> C

(where C is the complex numbers) with p an open map.
The idea is that the 'domain' D of the 'function' is the
image of p, that the values of the 'function' at a point
z in D are the values that f takes on the preimage p^{-1}(z),
and the surface S encodes the way analytic continuation
works for the 'function'.

Indeed, this is the kind of object one obtains by
doing analytic continuation in the usual way.

For such a function to take uncountably many values
at a point z of its domain, the preimage p^{-1}(z)
would have to be uncountable. But this preimage
would then accumulate on some point w of S, for S is
a separable space. Now the map p is constant on
a sequence of points of S converging to w, so the map
p is constant.

This is the argument, up to details...

-- m

### N�ste Nyberg

Jul 21, 2009, 5:41:15 AM7/21/09
to

"mike3" <mike...@yahoo.com> skrev i melding

Hi.
I stand corrected! You are right.

Karl-Olav Nyberg

### Bacle

Jul 21, 2009, 6:37:55 AM7/21/09
to
>>
> For such a function to take uncountably many values
> at a point z of its domain, the preimage p^{-1}(z)
> would have to be uncountable. But this preimage
> would then accumulate on some point w of S, for S is
> a separable space. Now the map p is constant on
> a sequence of points of S converging to w, so the map
> p is constant.
>
> This is the argument, up to details...
>
> -- m
>
Mariano: is every surface or at least Riemann surface
separable?. (and do you mean by separable having a countable dense subset?), and if this is the def. of separability, why it implies the accumulation issue?.

There is the same result for R^n and C^2n of every
uncountable subset in each having an accumulation point,
but the argument for this that I have seen is that
each is Lindeloff -- my salute to the Swedes here :)--
meaning every cover of R^n (C^2n) by open sets has
a countable subcover. Then a cover by, e.g., open balls
of finite radius has a countable subcover. Then
one of the balls in this countable subcover must
contain infinitely many points. Then we use a thm.
by Weirstrass that every bounded infinite subset has
a limit point. Are you using something similar?

> Still, we may not be able to use Lindeloff for
a surface as a subspace, since Lindeloff is not
hereditary, i.e., not every subspace of a Lindeloff
space is Lindeloff.

### David C. Ullrich

Jul 21, 2009, 8:42:12 AM7/21/09
to
On Mon, 20 Jul 2009 13:44:51 -0700 (PDT), mike3 <mike...@yahoo.com>
wrote:

>Hi.
>
>Is it possible for a complex, analytic "multi-valued function" to take
>on an _un_countably infinite number of values? If so, is there a good
>example? If not, why not? Note that "log" takes on countably
>infinitely many, not uncountably infinitely many.

First some definitions, or hints at definitions, to clarify things:

A "function element" is a (single-valued) analytic function in some
disk. A function element can (sometimes) be "analytically continued"
along a curvc. The usual notion of "multivalued function" is the
set of all function elements that can be obtained by starting
with one fixed function element and continuing along some
curve.

If that's what we mean by "multivalued function" then the set
of possible values of f(x) is countable. There are uncountably
many curves from x to x. But given a curve c_1 from x to x
there exists delta > 0 such that if c_2 is another curve from
x to x and |c_2(t) - c_1(t)| < delta for all t then the values
of f(x) you get by continuation along c_1 and c_2 are
the same. Say c_2 is "close enough" to c_1 when this happens.

Now, although there are uncountably many curves from x to x,
there exists a countable set S of such curves such that any curve
from x to x is close enough to some curve in S. Hence there
are only countably many values of f(x).

>
>http://www.tetration.org/tetration_net/tetration_net_Tetration_Dynamics.htm
>"My position is that any complete extension to the domain of tetration
>must account for the fact that ^b a may take an uncountable infinity
>of values."

You can easily find web pages where the author's position is that the
Earth is flat, the Holocaust never happened, the Apollo moon landings
were fakes, and all sorts of other things, including bizarre
mathematics.

David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
in sci.logic.)

### David C. Ullrich

Jul 21, 2009, 8:48:30 AM7/21/09
to
On Tue, 21 Jul 2009 06:37:55 EDT, Bacle <ba...@yahoo.com> wrote:

>>>
>> For such a function to take uncountably many values
>> at a point z of its domain, the preimage p^{-1}(z)
>> would have to be uncountable. But this preimage
>> would then accumulate on some point w of S, for S is
>> a separable space. Now the map p is constant on
>> a sequence of points of S converging to w, so the map
>> p is constant.
>>
>> This is the argument, up to details...
>>
>> -- m
>>
> Mariano: is every surface or at least Riemann surface
> separable?.

Assuming a "surface" is _connected_ (as the surfaces that arise
from analytic continuation are) then yes.

>(and do you mean by separable having a countable dense subset?),

Yes.

>and if this is the def. of separability, why it implies the accumulation issue?.

Say D is a countable dense set and S is uncountable. Then there
exists x_0 in D such that the ball about x_0 of radius 1 contains
uncountably many elements of S.

Now the ball about x_0 of radius 1 is contained in the union
of finitely many balls of radius 1/2...

> There is the same result for R^n and C^2n of every
> uncountable subset in each having an accumulation point,
> but the argument for this that I have seen is that
> each is Lindeloff -- my salute to the Swedes here :)--
> meaning every cover of R^n (C^2n) by open sets has
> a countable subcover. Then a cover by, e.g., open balls
> of finite radius has a countable subcover. Then
> one of the balls in this countable subcover must
> contain infinitely many points. Then we use a thm.
> by Weirstrass that every bounded infinite subset has
> a limit point. Are you using something similar?
>
>> Still, we may not be able to use Lindeloff for
> a surface as a subspace, since Lindeloff is not
> hereditary, i.e., not every subspace of a Lindeloff
> space is Lindeloff.

But any (connected) Riemann surface is Lindelof.
For example, any locally compact metric space
is Lindelof...

### I.N. Galidakis

Jul 21, 2009, 9:55:53 AM7/21/09
to

I think that his "uncountable" simply refers to the range of possible tetration
extensions a^^b, for fixed a, in contrast to the countable range of normal
tetration a^^n, for fixed a.
--
Ioannis

### Bacle

Jul 21, 2009, 12:06:42 PM7/21/09
to
Excellent, very cleat. Thank You.

### Michael Stemper

Jul 21, 2009, 12:55:43 PM7/21/09
to
In article <7a6d9415-f58d-42e2...@n11g2000yqb.googlegroups.com>, mike3 <mike...@yahoo.com> writes:
>On Jul 20, 8:32=A0pm, Don Stockbauer <donstockba...@hotmail.com> wrote:
>> On Jul 20, 8:00=A0pm, mike3 <mike4...@yahoo.com> wrote:

>> > On Jul 20, 6:39=A0pm, David R Tribble <da...@tribble.com> wrote:
>> > > mike3 wrote:

>> > > >> Is it possible for a complex, analytic "multi-valued function" to =
>take
>> > > >> on an _un_countably infinite number of values? If so, is there a g=

>ood
>> > > >> example? If not, why not? Note that "log" takes on countably
>> > > >> infinitely many, not uncountably infinitely many.
>>
>> > > You mean for any given single value of x.
>>
>> > Yes, that's what I mean.
>>
>> 1. Potential
>> 2. Actualized
>
>Huh?

Oh, it doesn't mean anything; it's just something that he says.

--
Michael F. Stemper
#include <Standard_Disclaimer>
Visualize whirled peas!

### mike3

Jul 21, 2009, 4:42:48 PM7/21/09
to
On Jul 21, 6:42 am, David C. Ullrich <dullr...@sprynet.com> wrote:
> On Mon, 20 Jul 2009 13:44:51 -0700 (PDT), mike3 <mike4...@yahoo.com>
> >http://www.tetration.org/tetration_net/tetration_net_Tetration_Dynami...

> >"My position is that any complete extension to the domain of tetration
> >must account for the fact that ^b a may take an uncountable infinity
> >of values."
>
> You can easily find web pages where the author's position is that the
> Earth is flat, the Holocaust never happened, the Apollo moon landings
> were fakes, and all sorts of other things, including bizarre
> mathematics.
>

I know. Which is why I was skeptical of it.

So I guess this is another piece of stupid whack-pot mathematics that
doesn't make any sense.

But now that my suspicion that an analytic multi-valued function can
have only countably many values has been confirmed, then the
"argument"
on the page must contain a fallacy or other type of error in it.

So what is that?

The idea, it seems, is this. Suppose I had an analytic tetration
function,
call it tet_b(z). Such a function will have branch points at _at
least_
z = -2, -3, -4, -5. Furthermore, we have log(tet_b(z+1)) = tet_b(z).
Yet log is multi-valued. It appears what is being said is that at
some point "z", you an get "extra" values for tetration by taking
log(log(log(...(b^...b^z)...))) and using different branches for the
logs.
I.e. we have ^1 e = e. ^2 e = e^e. Now log(e^e) = e + (2npi i).
^3 e = e^e^e so log(log(e^e^e)) = log(e^e + (2npi i)) = Log(e^e +
(2npi i)) + 2mpi i
for integers n and m, and so on and so on, i.e.

^1 e = 2s_0 pi i + Log(2s_1 pi i + Log(2s_2 pi i + ... Log(2s_(k-1) pi
i + exp^k(1))...))

where exp^k(1) = e^e^e...^e (k high), single-valued, and the s_j are
any integers. (Note that if s_j = 0 for all j this just returns the
principal value "e".) As these strings are finite length, there's
still
only countably many. But I guess he thinks then that if we take a
limit
as k->oo, we can set up a value for an infinite sequence of integers,
and so achieve _un_countably many values. Yet, as has been shown here,
a complex-analytic multi-valued function cannot take on an uncountable
number of values. (Note that on his page he numbers these in the
opposite order, i.e. what I call s_(k-1) he calls s_0, etc. but it's
still the same idea.)

Thus arguing this way must fail somehow. This would imply possibly one
or more of the following: not all values constructible this way are
actually valid Riemann branches of the tetration function, that the
limit as k->oo is not a valid Riemann branch, and there are many
sequences of s_j that yield the same values, enough to reduce the
amount
of _distinct_ values obtainable via this method to a countably
infinite set.

What do you think?