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Jul 20, 2009, 4:44:51 PM7/20/09

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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_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."

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

to

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

news:5714cbb4-0e18-4d3e...@t13g2000yqt.googlegroups.com...

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

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

> "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:

>

> > "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.

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.

>

>> 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.

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

to

Yes, that's what I mean.

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

to

1. Potential

2. Actualized

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

to

Huh?

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.

> 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

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

to

news:a4b0f09c-ed28-4697...@s31g2000yqs.googlegroups.com...

Hi.

I stand corrected! You are right.

Karl-Olav Nyberg

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 > 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

>

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.

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:

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).

>Because I read this:

>

>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."

(John Jones, "My talk about Godel to the post-grads."

in sci.logic.)

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...

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

Jul 21, 2009, 12:06:42 PM7/21/09

to

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, 8:32=A0pm, Don Stockbauer <donstockba...@hotmail.com> wrote:

>> On Jul 20, 8:00=A0pm, mike3 <mike4...@yahoo.com> 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!

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.

>

> On Mon, 20 Jul 2009 13:44:51 -0700 (PDT), mike3 <mike4...@yahoo.com>

> >"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?

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