fixed point set theory
amy666
you know or you should know that ' fixed points ' are fundamental in many branches of math.
however standard set theory with its large cardinals is inconsistant with the concept of ' fixed points '.
PART I
1 2 3 4 ... 'oo'.
'oo' is a fixpoint of the successor function :
x + 1 = S(x) = x
has only this 'oo' as a solution.
one can argue
'oo' = aleph_0 ( ordinary integers )
'oo' = aleph_1 ( AP integers and the alike )
but that has no influence on the rest of this thread , same conclusions follow.
remember , only this 'oo' and not
e.g. aleph_'oo' , aleph_34 , oo + 2 , 2 oo - 1 , w^w^...^w
since 'oo' is a fixpoint , and btw also otherwise the set of integers 1 2 3 ... oo is not well defined.
-------------
PART II
http://upload.wikimedia.org/wikipedia/commons/8/83/Omega_squared.png
this w^2 can clearly be mapped onto w.
thus w^2 = w*w = w
apart from the way we count it.
basicly w^2 is just the ways we count subsets of w.
w^n = w
all ordinals belong to a cardinal.
it is thus clear that we cannot increase the cardinality of an ordinal by operations that do would not change the cardinality if applied to cardinality.
( symbolic w^2 = w <=> R^2 = R )
----------------
PART III
since PART I and PART II have shown that applying the successor function and using ordinals cannot increase cardinality beyond the way cardinal functions do , we are only left with cardinal functions.
the simplest function that increases cardinality is 2^x.
in fact , up to isomorphism , its the only one together with its repetitions.
2^ aleph_x = aleph_(x + 1)
for finite n this is a simple structure.
but now follows the intresting part :
conjecture :
aleph_'oo' is the largest cardinal.
PROOF : ( with the above in mind )
2^aleph_x = aleph_x+1
2^aleph_x = aleph_S(x)
where S(x) is again the successor function
since 'oo' was the fixed point of S(x) = x we get
2^aleph_'oo' = aleph_'oo'
thus aleph_'oo' is also a fixed point , the fixed point of 2^x = x.
thus aleph_'oo' is the largest cardinal.
QED
note 'oo' is at most aleph_1 ( AP - adic or cardinal equivalent )
thus largest ( existing ! ) cardinal <= aleph_aleph_1
----------------
PART IV
conclusion :
you know or you should know that ' fixed points ' are fundamental in many branches of math.
( e.g. algebra , chaos , dynamics , calculus , number theory )
however standard set theory with its large cardinals is inconsistant with the concept of ' fixed points '.
thats a big thing.
standard set theory with large cardinals not so solid afterall.
even stronger , ANY set theory that can model the integers but has large cardinals is thus not so solid.
furthermore large cardinals should have been stated as a conjecture , not axioms.
since it is now disproved and axioms cannot be disproved.
copyright
tommy1729
Dave L. Renfro
> where S(x) is again the successor function
>
> since 'oo' was the fixed point of S(x) = x we get
The successor function is a map from a well ordered
set to a well ordered set (ordinals to ordinals,
a collection of cardinals to a collection of
cardinals, etc.) that has no fixed points. The
cardinal operation "add 1" is not the same as
the successor operation.
Dave L. Renfro
amy666
successor =/= +1 ?
see :
http://en.wikipedia.org/wiki/Successor_function#Definition
Successor function: The 1-ary successor function S, which returns the successor of its argument (see Peano postulates), is primitive recursive. That is, S(k) = k + 1.
( end quote )
what is the successor of 1 ? 2
what is the successor of x ? x + 1.
so at least there is one successor function which matches.
furthermore
2^aleph_x = aleph_x+1
there , without the term ' successor '
also x + 1 = x
without S(x) = x.
so my arguments still holds , even without " successor "
dont underestimate me
tommy1729
Dave L. Renfro
> 2^aleph_x = aleph_x+1
>
> there , without the term ' successor '
>
> also x + 1 = x
Assuming GCH, the x+1 in the first of your lines above
is ordinal addition, not cardinal addition. For ordinal
addition, x + 1 is always strictly greater than x.
Dave L. Renfro
amy666
sigh ...
fixed point fixed point fixed point
x = x + 1
fixed point !
David Belanger
> basicly w^2 is just the ways we count subsets of w.
what
Aatu Koskensilta
Don't you recall amy's helpful explanation: "fixed point! fixed point!
fixed point!"?
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Owen Jacobson
That's just it. In ordinal arithmetic, the expression 'x = x + 1' is
inherently false. You can't simply assert that there is an ordinal x
such that x = x + 1, and therefore you can't assert that there is an
ordinal x such that S(x) = x under the usual definition of the
successor function.
Sorry,
-o
MoeBlee
> 1 2 3 4 ... 'oo'.
>
> 'oo' is a fixpoint of the successor function :
>
> x + 1 = S(x) = x
>
> has only this 'oo' as a solution.
I take that by 'oo' you mean w (omega), the set of natural numbers?
Dave Renfro mentioned this, but I'd like to mention it also:
w (omega) is NOT a fixed point for the ORDINAL successor operation.
Where 'x+1' stands for xu{x}, it is not the case that w+1 = w.
But where '+' stands for CARDINAL addition, yes, it is the case that w
+1 = w, though this is more usually notated as aleph_0 + 1 = aleph_0.
And in that context it is NOT the case that w is the only x such that
x = x+1, contrary to your claim "only this 'oo' as a solution".
When you see the symbol '+' you have to be careful to distinguish what
it stands for in a given context. Ideally, in different contexts you
should regard '+' as 'ordinal_+' and in other contexts as 'cardinal_
+', and yet other operations in other contexts, as if they are
different symbols even.
> http://upload.wikimedia.org/wikipedia/commons/8/83/Omega_squared.png
>
> this w^2 can clearly be mapped onto w.
You linked to a graphic of some kind.
If by 'w^2' by mean wXw (the Cartesian product), then, yes, w^2 is 1-1
with w.
> thus w^2 = w*w = w
Okay, if by '*' you mean cardinal multiplication, the correct
formulation is:
card(w^2) = w*w = w
or, more usually notated:
card(w^2) = aleph_0*aleph_0 = aleph_0 (I'll not mention the notational
matter of 'w' and 'aleph_0' from now on.)
> basicly w^2 is just the ways we count subsets of w.
No, it's not. It's the cardinality of the set of ordered pairs of
members of w. That is different from the cardinality of the set of
subsets of w.
> w^n = w
No, it's this:
card(w^n) = w (where n is a natural number greater than 0)
> all ordinals belong to a cardinal.
Okay, for every ordinal b there exists a cardinal k such bek.
Do we need choice for that? Hmm?
> it is thus clear that we cannot increase the cardinality of an ordinal by operations that do would not change the cardinality if applied to cardinality.
I don't know what you mean by that.
> ( symbolic w^2 = w <=> R^2 = R )
Maybe you mean card(w^2) = w <-> card(R^2) = card(R) (where 'R' stands
for the set of real numbers).
Anyway the above is a theorem of ZFC, since both sides of the
biconditional are themselves theorems.
> the simplest function that increases cardinality is 2^x.
>
> in fact , up to isomorphism , its the only one together with its repetitions.
I don't know about that. Would you mention the particular theorem you
have in mind?
> 2^ aleph_x = aleph_(x + 1)
That is the generalized continuum hypothesis.
> for finite n this is a simple structure.
I don't know what you mean by that.
> but now follows the intresting part :
>
> conjecture :
>
> aleph_'oo' is the largest cardinal.
>
> PROOF : ( with the above in mind )
>
> 2^aleph_x = aleph_x+1
If you take the continuum hypothesis as an axiom.
> 2^aleph_x = aleph_S(x)
>
> where S(x) is again the successor function
In this case, it would be the ORDINAL successor.
> since 'oo' was the fixed point of S(x) = x we get
No, 'w' is NOT a fixed point for ORDINAL sucessor.
> 2^aleph_'oo' = aleph_'oo'
No. That doesn't follow from ANYTHING here, because it is NOT the case
that S(x) = x where 'S' stands for ORDINAL successor.
There's more in your post, but I'm going to stop at this point.
MoeBlee
David C. Ullrich
wrote:
>since only set theory can hold the attention of the majority at sci.math ill post about it again , despite being more intrested in other subjects.
>
>you know or you should know that ' fixed points ' are fundamental in many branches of math.
>
>however standard set theory with its large cardinals is inconsistant with the concept of ' fixed points '.
What a strange thing to say.
>PART I
>
>1 2 3 4 ... 'oo'.
>
>'oo' is a fixpoint of the successor function :
>
>x + 1 = S(x) = x
If you say so. Exactly what _set_ are you talking
about when you say oo, and exactly what is the
_definition_ of S(x) for any set x?
I don't expect you'll actually provide us with those
definitions, but until you do you haven't said anything
meaningful.
>has only this 'oo' as a solution.
>
>one can argue
>
>'oo' = aleph_0 ( ordinary integers )
>
>'oo' = aleph_1 ( AP integers and the alike )
>
>but that has no influence on the rest of this thread , same conclusions follow.
>
>remember , only this 'oo' and not
>
>e.g. aleph_'oo' , aleph_34 , oo + 2 , 2 oo - 1 , w^w^...^w
>
>since 'oo' is a fixpoint , and btw also otherwise the set of integers 1 2 3 ... oo is not well defined.
>
>-------------
>
>PART II
>
>http://upload.wikimedia.org/wikipedia/commons/8/83/Omega_squared.png
>
>this w^2 can clearly be mapped onto w.
>
>thus w^2 = w*w = w
>
>apart from the way we count it.
>
>basicly w^2 is just the ways we count subsets of w.
Now you don't have to give your definitions, because
the things you're talking about are standard gizmos.
No, w^2 does _not_ count the subsets of w.
>w^n = w
>
>all ordinals belong to a cardinal.
>
>it is thus clear that we cannot increase the cardinality of an ordinal by operations that do would not change the cardinality if applied to cardinality.
>
>( symbolic w^2 = w <=> R^2 = R )
>
>----------------
>
>PART III
>
>since PART I and PART II have shown that applying the successor function and using ordinals cannot increase cardinality beyond the way cardinal functions do , we are only left with cardinal functions.
>
>the simplest function that increases cardinality is 2^x.
>
>in fact , up to isomorphism , its the only one together with its repetitions.
You just love to babble nonsense.
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.)
David C. Ullrich
wrote:
>Dave L Renfro wrote :
>
>> amy666 wrote (in part):
>>
>> > where S(x) is again the successor function
>> >
>> > since 'oo' was the fixed point of S(x) = x we get
>>
>> The successor function is a map from a well ordered
>> set to a well ordered set (ordinals to ordinals,
>> a collection of cardinals to a collection of
>> cardinals, etc.) that has no fixed points. The
>> cardinal operation "add 1" is not the same as
>> the successor operation.
>>
>> Dave L. Renfro
>
>successor =/= +1 ?
>
>see :
>
>http://en.wikipedia.org/wiki/Successor_function#Definition
>
>Successor function: The 1-ary successor function S, which returns the successor of its argument (see Peano postulates), is primitive recursive. That is, S(k) = k + 1.
>
>( end quote )
That's the definition for integers. Not for infinite ordinals.
>what is the successor of 1 ? 2
>
>what is the successor of x ? x + 1.
>
>so at least there is one successor function which matches.
>
>furthermore
>
>2^aleph_x = aleph_x+1
>
>there , without the term ' successor '
>
>also x + 1 = x
>
>without S(x) = x.
>
>so my arguments still holds , even without " successor "
>
>
>dont underestimate me
>
>
>tommy1729
David C. Ullrich
David C. Ullrich
<aatu.kos...@uta.fi> wrote:
>David Belanger <dbel...@csclub.uwaterloo.ca> writes:
>
>> On Wed, 25 Feb 2009, amy666 wrote:
>>
>>> basicly w^2 is just the ways we count subsets of w.
>>
>> what
>
>Don't you recall amy's helpful explanation: "fixed point! fixed point!
>fixed point!"?
I noted that, but it wasn't very clear to me. Oh - I just got it.
He meant to say "fixed point! fixed point! fixed point!
ficed point!". Now it all makes sense.
Jesse F. Hughes
> dont underestimate me
.
--
Scissors and string, scissors and string,
When a man's single, he lives like a king.
Needles and pins, needles and pins,
When a man marries, his trouble begins. --- Mother Goose
amy666
> On Wed, 25 Feb 2009 11:58:50 EST, amy666
> <tomm...@hotmail.com>
> wrote:
>
> >since only set theory can hold the attention of the
> majority at sci.math ill post about it again ,
> despite being more intrested in other subjects.
> >
> >you know or you should know that ' fixed points '
> are fundamental in many branches of math.
> >
> >however standard set theory with its large cardinals
> is inconsistant with the concept of ' fixed points '.
>
> What a strange thing to say.
no , as the OP explained ' fixed points ' are fundamental to math.
it is thus logical , despite rarely considered.
>
> >PART I
> >
> >1 2 3 4 ... 'oo'.
> >
> >'oo' is a fixpoint of the successor function :
> >
> >x + 1 = S(x) = x
>
> If you say so. Exactly what _set_ are you talking
> about when you say oo, and exactly what is the
> _definition_ of S(x) for any set x?
i already explained in the OP.
1 2 3 4 ... oo
can only mean the set of the integers or the set of the " AP - integers ".
no other interpretation is meaningfull.
i didnt say that , i said w^2 is just the way we count subsets of w.
i gave a link to a picture of w^2.
it clearly counts till w , w times.
as you can see on the picture ( = proof )
since card(w^2) = card(w)
w^2 is thus simply a way of putting w into subsets.
>
> >w^n = w
> >
> >all ordinals belong to a cardinal.
> >
> >it is thus clear that we cannot increase the
> cardinality of an ordinal by operations that do would
> not change the cardinality if applied to cardinality.
> >
> >( symbolic w^2 = w <=> R^2 = R )
> >
> >----------------
> >
> >PART III
> >
> >since PART I and PART II have shown that applying
> the successor function and using ordinals cannot
> increase cardinality beyond the way cardinal
> functions do , we are only left with cardinal
> functions.
> >
> >the simplest function that increases cardinality is
> 2^x.
> >
> >in fact , up to isomorphism , its the only one
> together with its repetitions.
>
> You just love to babble nonsense.
and here you just love to get personal again , instead of rational.
nonsense ? ok then , lets turn the tables again !
i said :
> >the simplest function that increases cardinality is
> 2^x.
> >
> >in fact , up to isomorphism , its the only one
> together with its repetitions.
give a counterexample !!
with proof !!
>
> >2^ aleph_x = aleph_(x + 1)
> >
> >for finite n this is a simple structure.
if card(R) = card(C) => GCH
"Understanding FLT isn't about following the formal
proof.
That would make a mockery of everything Fermat was up
to."
(Jon C. James, "My talk about FLT to the sci.math-grads."
in sci.math.)
( just to show how silly it is not to value proof enough , i modified ullrich's quote )
regards
tommy1729
amy666
i remembered because he usually defends my ideas but was on the other side that time.
but he didnt find a disproof , not even after many months.
there is no disproof !
Mariano Suárez-Alvarez
> David wrote :
>
> > On Wed, 25 Feb 2009 11:58:50 EST, amy666
But that is not how it works...
If you claim something, it is up to *you* to
provide a proof of that claim. That others
cannot (and, in many cases, have no wish to)
come up with a counterexample to your claims
does most certainly not count as support for
the veracity of those claims, let alone it
is a proof of that veracity.
-- m
amy666
yes , that is exactly how it works !!
if you call a claim nonsense
then you should at least be capable of disproving it.
otherwise it might be true and is thus FAR FROM NONSENSE.
should be easy too disproof , since its so clearly 'nonsense' ?
HA !
>
> If you claim something, it is up to *you* to
> provide a proof of that claim. That others
> cannot (and, in many cases, have no wish to)
> come up with a counterexample to your claims
> does most certainly not count as support for
> the veracity of those claims, let alone it
> is a proof of that veracity.
>
> -- m
>
>
the excuses are pathetic !
you cannot call someone saying something nonsense and then claim you cant and wont prove its wrong !!
its your job to show its nonsense , if you said so , but you guys ( david , mariano ) cant !
every time i ' turn the tables ' , your real faces appear !
MoeBlee
> it clearly counts till w , w times.
>
> as you can see on the picture ( = proof )
A picture here is not a proof. But anyway, okay, we do understand the
ORDINAL w^2 = w*w where '^' and '*' are for ordinal operations.
(So I'd revise my own previous comments to reflect that.)
> since card(w^2) = card(w)
>
> w^2 is thus simply a way of putting w into subsets.
Whatever "a way of putting w into subsets means".
It is not the case that w is PARTITIONED in some way you've given.
Anyway, you haven't proven
2^aleph_w = aleph_w where '^' stands for CARDINAL exponentiation, if
that is what you mean by your claim:
> 2^aleph_'oo' = aleph_'oo'
MoeBlee
Mariano Suárez-Alvarez
Nonsense is not easy or hard to disprove: it is simply
nonsense. What you wrote is very well in the "not even
wrong camp" and the particular claim which I quoted
is so vague, has so many undefined terms, and it is
so poorly expressed that, to be honest, I do not really
know what it is supposed to be claiming.
While it may be that you have a clear and precise
idea in your mind about what the claim claims,
you certainly did not manage to communicate it.
As far as I can see, what you wrote is nonsense.
Since there are many, many things out there which
are not nonsense, I simply cannot find any justification
in trying to figure out what you mean and on top of
that, provide a counterexample to whatever it is
that you meant.
-- m
amy666
> On Feb 26, 11:42 am, amy666 <tommy1...@hotmail.com>
> wrote:
>
> > it clearly counts till w , w times.
> >
> > as you can see on the picture ( = proof )
>
> A picture here is not a proof. But anyway, okay, we
> do understand the
> ORDINAL w^2 = w*w where '^' and '*' are for ordinal
> operations.
>
> (So I'd revise my own previous comments to reflect
> that.)
>
> > since card(w^2) = card(w)
> >
> > w^2 is thus simply a way of putting w into subsets.
>
> Whatever "a way of putting w into subsets means".
>
> It is not the case that w is PARTITIONED in some way
> you've given.
yes it is , just look at the pic again.
>
> Anyway, you haven't proven
>
> 2^aleph_w = aleph_w where '^' stands for CARDINAL
> exponentiation, if
> that is what you mean by your claim:
>
> > 2^aleph_'oo' = aleph_'oo'
i already explained what i meant in the OP and yes i did prove it.
snipping a long OP to a few lines doesnt mean i only said what you quoted.
in genereral , all is relevant in my post(s).
>
> MoeBlee
>
>
>
tommy1729
MoeBlee
> Jack wrote :
> > On Feb 26, 11:42 am, amy666 <tommy1...@hotmail.com>
> > wrote:
>
> > > it clearly counts till w , w times.
>
> > > as you can see on the picture ( = proof )
>
> > A picture here is not a proof. But anyway, okay, we
> > do understand the
> > ORDINAL w^2 = w*w where '^' and '*' are for ordinal
> > operations.
>
> > (So I'd revise my own previous comments to reflect
> > that.)
>
> > > since card(w^2) = card(w)
>
> > > w^2 is thus simply a way of putting w into subsets.
>
> > Whatever "a way of putting w into subsets means".
>
> > It is not the case that w is PARTITIONED in some way
> > you've given.
>
> yes it is , just look at the pic again.
No, you don't know what a partition IS.
And there is no PICTURE of an infinitistic object such as in this
instance that is a PROOF of anything.
> > Anyway, you haven't proven
>
> > 2^aleph_w = aleph_w where '^' stands for CARDINAL
> > exponentiation, if
> > that is what you mean by your claim:
>
> > > 2^aleph_'oo' = aleph_'oo'
>
> i already explained what i meant in the OP and yes i did prove it.
What you said depends on what 'oo' stands for. If 'oo' stands for
'w' (omega, the set of natural numbers), then I already explained why
you've not proven what you claim; as did several other people.
As best that can be discerned from your notation, you are conflating
ordinal and cardinal operations. So you're coming up with incorrect
results. I'm not the only person who has pointed this out to you. Your
refusal to recognize such matters and to LISTEN to other people who
are explaining for you the mistake you've made is part of the reason
you are a crank and not taken seriously by a good number of
knowledgable people. If you wish to discuss mathematics seriously with
such people as Mr. Renfro then you need to LISTEN to and learn to
UNDERSTAND the gratis instruction he has given you; while, of course,
it is still crucial that you get a good book on this subject so that
you have some organized context for learning it. Of course, though,
you won't sit down to study a book on this subject, given that you are
still literally a child without the patience and discipline for such
things or you are an adult acting the age and mentality of such a
child.
> snipping a long OP to a few lines doesnt mean i only said what you quoted.
And I don't claim that all of your remarks are contained only in the
portions I quoted.
MoeBlee
MoeBlee
> > > It is not the case that w is PARTITIONED in some way
> > > you've given.
>
> > yes it is , just look at the pic again.
>
> No, you don't know what a partition IS.
P.S. I don't mean to say that w is not a denumerable union of
denumerable non-empty pairwise disjoint sets. Indeed, it is. But
rather, that simply showing a picture of something is not in itself a
proof in such an infinitistic case as this.
MoeBlee
David C. Ullrich
wrote:
>[...]
>
>dont underestimate me
I didn't notice this until I saw a few people reply
to it. Here's what they _should_ have said:
Relax. I doubt that it's _possible_ to underestimate you.
>
>tommy1729
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
David C. Ullrich
wrote:
>David wrote :
>
>> On Wed, 25 Feb 2009 11:58:50 EST, amy666
>> <tomm...@hotmail.com>
>> wrote:
>>
>> >since only set theory can hold the attention of the
>> majority at sci.math ill post about it again ,
>> despite being more intrested in other subjects.
>> >
>> >you know or you should know that ' fixed points '
>> are fundamental in many branches of math.
>> >
>> >however standard set theory with its large cardinals
>> is inconsistant with the concept of ' fixed points '.
>>
>> What a strange thing to say.
>
>no , as the OP explained ' fixed points ' are fundamental to math.
Of course they are. What's strange is saying that large cardinals
are inconsistent with the concept.
In you "arguments" you seem to be assuming that
"fixed point theory" says that every function
has a fixed point. That's not so.
>it is thus logical , despite rarely considered.
>
>>
>> >PART I
>> >
>> >1 2 3 4 ... 'oo'.
>> >
>> >'oo' is a fixpoint of the successor function :
>> >
>> >x + 1 = S(x) = x
>>
>> If you say so. Exactly what _set_ are you talking
>> about when you say oo, and exactly what is the
>> _definition_ of S(x) for any set x?
>
>i already explained in the OP.
>
>1 2 3 4 ... oo
>
>can only mean the set of the integers or the set of the " AP - integers ".
>
>no other interpretation is meaningfull.
That doesn't come close to answering my question.
What set is the notation oo referring to, and
what is the definition of S(x) for sets x?
Yes it does. Nobody said that was not so. What's
hilariously false is your assertion that it counts
the subsets of w.
How in the world do you get from "counts w,
w times" to "counts the subsets of w"?
>as you can see on the picture ( = proof )
>
>since card(w^2) = card(w)
>
>w^2 is thus simply a way of putting w into subsets.
Oh. You didn't mean it counts _all_ the subsets of w,
just some collection of them.
That's deep. You're saying that there exists a countable
collection of subsets of w. So what?
>
>>
>> >w^n = w
>> >
>> >all ordinals belong to a cardinal.
>> >
>> >it is thus clear that we cannot increase the
>> cardinality of an ordinal by operations that do would
>> not change the cardinality if applied to cardinality.
>> >
>> >( symbolic w^2 = w <=> R^2 = R )
>> >
>> >----------------
>> >
>> >PART III
>> >
>> >since PART I and PART II have shown that applying
>> the successor function and using ordinals cannot
>> increase cardinality beyond the way cardinal
>> functions do , we are only left with cardinal
>> functions.
>> >
>> >the simplest function that increases cardinality is
>> 2^x.
>> >
>> >in fact , up to isomorphism , its the only one
>> together with its repetitions.
>>
>> You just love to babble nonsense.
>
>and here you just love to get personal again , instead of rational.
You _do_ babble nonsense, really a lot, including here.
If it's not true that you love to do so it's hard to see
why you _do_.
In other words, for some reason you decided to lie about what I said.
>
>regards
>
>tommy1729
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
David C. Ullrich
wrote:
Huh?
You haven't given anything to disprove. There is
no "conflict" between set theory and the concept
of fixed points. Right inside set theory we define
x to be a fixed point of f if f(x) = x. There you
are, a proof that there's no "conflict".
amy666
they havent told me anything i didnt know.
i know that for ' traditional ' ordinals
1 + w =/= w
1 + w =/= w + 1
w + 1 =/= w
maybe it wasnt clear but
i used ' PART I ' , ' PART II ' , ' PART III ' and the conclusion part IV.
none of the OP is standard , no standard ordinals for example.
its important to realise
Part II depends upon Part I
part II depends on part I and part II
and none of it is standard.
so no point in trying to explain me ' standard ordinals ' , im talking about new viewpoints , that of
' fixed point set theory '
you call me childish , but its childish to correct me on things i am not even talking about ...
even mariano had to admit " not even wrong "
regards
tommy1729
amy666
i meant the pic is sufficient to ' see ' how to construct a proof and to see what it ' really is '.
amy666
this is an intresting comment by david.
although i did not say that every function has a fixed point , my ideas are indeed pretty close to that.
david claims that not every function has a fixed point.
thats why i changed the title , this is a bit off topic but intresting.
ill start :
conjecture :
every analytic function has a fixed point in C*
Toni...@yahoo.com
As usual you write nonsenses, and you even boast about them: 1 + w =
w, and I think w is a very "traditional" ordinal
Don't you get tired of making an ass of yourself ALL the time and in
ALL the categories?
Tonio
MoeBlee
> maybe it wasnt clear but
Your stuff on such topics is virtually never clear, and the reason for
that is just what I mentioned: your childish refusal to learn the
basic principles, terminology, and notation of the subject matter.
> its important to realise
that if you want to be clear and to communicate with people about this
subject then you need to learn its basic principles, terminology, and
notation.
MoeBlee
MoeBlee
> i meant the pic is sufficient to ' see ' how to construct a proof and to see what it ' really is '.
You haven't even clearly said a proof of WHAT, let alone that I should
rely on your claim that a picture is sufficient to see a proof.
We do know a proof that w can be partitioned into denumerably many
denumerable sets. But if that is what you mean, then it is still
obscure as to how your picture "proves" that.
In any case, you have not proven your later conclusion that 2^aleph_w
= aleph_w.
MoeBlee
amy666
yes i did.
and its simply the OP.
not a part that people will snip , but the entire OP.
>
> MoeBlee
>
regards
tommy1729
amy666
but i dont want to communicate about the standards !
im talking about new stuff and thus i have my own terminology.
im not here to learn ' old math ' , im here to bring you ' new math '.
amy666
wrong , with standard ordinals 1 + w =/= w.
HA !
! this shows your ignorance !
>
> Don't you get tired of making an ass of yourself ALL
> the time and in
> ALL the categories?
>
> Tonio
since you just showed your ignorance by writing 1 + w = w ( standard ordinals )
you have to start realizing you are the one making an ass out of yourself.
at least people like ullrich and feldmann understand that 1 + w = w is FALSE ( for standard ordinals )
you , however only copy some insults , but cant copy/understand the math ( nor mine nor david nor dennis or anyone elses math ) and you are thus my weakest opponent.
tommy1729
Dave L. Renfro
> wrong , with standard ordinals 1 + w =/= w.
O-K, so what's wrong with the following?
Let A = {1} and B = {2, 3, 4, ...}.
Then, using the usual real number ordering,
(A,<) is a representative of the order type '1',
(B,<) is a representative of the order type 'w',
and A is disjoint from B.
Then one can obtain from this the set {1, 2, 3, ...},
with the usual real number ordering, as a representative
of the order type '1 + w'. Since {1, 2, 3, ...} with
the usual real number ordering is also a representative
of the order type 'w', it follows that '1 + w' = 'w'.
Dave L. Renfro
MoeBlee
> > In any case, you have not proven your later
> > conclusion that 2^aleph_w
> > = aleph_w.
>
> yes i did.
>
> and its simply the OP.
Okay, then you should go right along believing that. You stick to your
convictions. Don't let anyone talk you out of them!
MoeBlee
MoeBlee
> > On Feb 27, 4:07 am, amy666 <tommy1...@hotmail.com>
> > wrote:
>
> > > maybe it wasnt clear but
>
> > Your stuff on such topics is virtually never clear,
> > and the reason for
> > that is just what I mentioned: your childish refusal
> > to learn the
> > basic principles, terminology, and notation of the
> > subject matter.
>
> > > its important to realise
>
> > that if you want to be clear and to communicate with
> > people about this
> > subject then you need to learn its basic principles,
> > terminology, and
> > notation.
> but i dont want to communicate about the standards !
>
> im talking about new stuff and thus i have my own terminology.
>
> im not here to learn ' old math ' , im here to bring you ' new math '.
But you're using old vocabulary (and in certain other postings) mixed
with UNDEFINED new notation. So either you need to establish a new
vocabulary of concepts and terminology by a method of primitives and
definitions and/or establish a glossary that translates your use of
old vocabulary to the new sense that uniquely intend.
Otherwise, as your postings continue to read as gobbleldygook.
MoeBlee
MoeBlee
> Antonio wrote :
> > As usual you write nonsenses, and you even boast
> > about them: 1 + w =
> > w, and I think w is a very "traditional" ordinal
>
> wrong , with standard ordinals 1 + w =/= w.
>
> HA !
>
> ! this shows your ignorance !
Wrong, you puerile ignoramus, where '+' stands for ordinal addition,
we have
1+w = w.
> > Don't you get tired of making an ass of yourself ALL
> > the time and in
> > ALL the categories?
> since you just showed your ignorance by writing 1 + w = w ( standard ordinals )
>
> you have to start realizing you are the one making an ass out of yourself.
>
> at least people like ullrich and feldmann understand that 1 + w = w is FALSE ( for standard ordinals )
They do? Really?
MoeBlee
Toni...@yahoo.com
Well, be my guest: continue with your idiocies. Apparently you're
unable to feel ashamed anymore.
*Sigh*
Tonio
amy666
your elements overlap.
only card(1 + w) = card(w) follows.
Dave L. Renfro
>>> wrong , with standard ordinals 1 + w =/= w.
Dave L. Renfro wrote:
>> O-K, so what's wrong with the following?
>>
>> Let A = {1} and B = {2, 3, 4, ...}.
>>
>> Then, using the usual real number ordering,
>> (A,<) is a representative of the order type '1',
>> (B,<) is a representative of the order type 'w',
>> and A is disjoint from B.
>>
>> Then one can obtain from this the set {1, 2, 3, ...},
>> with the usual real number ordering, as a
>> representative
>> of the order type '1 + w'. Since {1, 2, 3, ...} with
>> the usual real number ordering is also a
>> representative
>> of the order type 'w', it follows that '1 + w' = 'w'.
amy666 wrote:
> your elements overlap.
>
> only card(1 + w) = card(w) follows.
This makes no sense on several levels. For example,
what does it mean for elements to overlap? The term
"overlap" uaually means that the interiors of the
sets have a nonempty intersection, but the present
context doesn't allow for this interpretation.
Anyway, maybe you can tell the people at these
web pages that they've got it wrong:
See equation (1) for 1 + w = w at
http://mathworld.wolfram.com/OrdinalAddition.html
See "addition" for 3 + w = w at
http://en.wikipedia.org/wiki/Ordinal_arithmetic
See examples 5.4 (on p. 12) for 1 + w = w at
http://www.peter-dixon.staff.shef.ac.uk/teaching/STDN.PDF
Dave L. Renfro
amy666
..
seems i wrote 1 + w =/= w
im sorry i mixed up.
my apologies , im tired ...
Dave L. Renfro
> seems i wrote 1 + w =/= w
>
> im sorry i mixed up.
>
> my apologies , im tired ...
<speechless>
<sci.math takes a collective pause and gasps>
Dave L. Renfro
Denis Feldmann
Not exactly : yes, apologies are good. But there is still a slight
difference between writing 1+w=/=w when one wanted to write w+1=/=w and
writing in *answer* to
> As usual you write nonsenses, and you even boast about them: 1 + w
=w, and I think w is a very "traditional" ordinal
<quote>
wrong , with standard ordinals 1 + w =/= w.
HA ! ! this shows your ignorance !
</quote>
It can hardly be couted as a mistype due to fatigue...
Denis Feldmann
Is the function f(z) =exp(z)+z analytic ?
How seriously do you test your conjectures before postiong them?
MoeBlee
> w + 1 =/= w
>
> seems i wrote 1 + w =/= w
>
> im sorry i mixed up.
>
> my apologies , im tired ...
Come on, why apologize? Stand your ground! You're a courageous pioneer
of a new mathematics. Surely, you can devise an explanation of the NEW
standard ordinal operations by which we have 1+w ~= w.
MoeBlee
Mariano Suárez-Alvarez
> David wrote :
>
>
>
> > On Thu, 26 Feb 2009 14:42:02 EST, amy666
> > wrote:
>
> > >David wrote :
>
> > >> On Wed, 25 Feb 2009 11:58:50 EST, amy666
Mariano Suárez-Alvarez
wrote:
> amy666 a écrit :
>
>
>
> > David wrote :
>
> >> On Thu, 26 Feb 2009 14:42:02 EST, amy666
> >> wrote:
>
> >>> David wrote :
>
> >>>> On Wed, 25 Feb 2009 11:58:50 EST, amy666
Are you seriouesly asking that? I would say the answer is quite
obvious :-)
-- m
Toni...@yahoo.com
****************************************************************
No, no "apologies", kid: you did write
"i know that for ' traditional ' ordinals
1 + w =/= w
1 + w =/= w + 1
w + 1 =/= w "
So you wrote 1 + w =/= w several times, ALSO when you did tried to
trash me for pointing the equality, and now you're trying to make it
better my writing "I'm tired"...and that's disingenous from you,
because you REALLY meant what you wrote.
You should write: "Sorry, but once again I write stupidities about
things I know nothing or almost nothing, and then I claim nonsenses
and boast and cry and write "HA" a lot and..... In short, I make a
complete fool of myself, and I'm going to begin checking more
carefully stuff I write about BEFORE I write about it and not after I
am said so".
Write the above, THEN apologize with everybody, and then PERHAPS we
will begin to believe you're beginning to change.
Tonio
Aatu Koskensilta
> wrong , with standard ordinals 1 + w =/= w.
I am afraid you're mistaken, Amy.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Ioannis
[snip]
> You should write: "Sorry, but once again I write stupidities about
> things I know nothing or almost nothing, and then I claim nonsenses
> and boast and cry and write "HA" a lot and..... In short, I make a
> complete fool of myself, and I'm going to begin checking more
> carefully stuff I write about BEFORE I write about it and not after I
> am said so".
>
> Write the above, THEN apologize with everybody, and then PERHAPS we
> will begin to believe you're beginning to change.
How about we give him a huge joint to smoke, instead?
Perhaps it will make him write more clearly :-)
> Tonio
--
Ioannis --- "There's _always_ a mistake, somewhere".
David C. Ullrich
wrote:
If you realize that not every function has a fixed point then
how is it you're saying that the fact that the successor
function for ordinals has no fixed point is some sort
of problem?
>david claims that not every function has a fixed point.
Yes, I "claim" that. Just like I "claim" that 2 + 2 = 4.
>thats why i changed the title , this is a bit off topic but intresting.
>
>ill start :
>
>conjecture :
>
>every analytic function has a fixed point in C*
What an immensely stupid conjecture. It's obviously
equivalent to the conjecture that every analytic
function has a zero in C*, which is obviously false.
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.)
David C. Ullrich
wrote:
First, if true that's no defense - in fact if you're using your
own terminology it's vital that you _define_ the terms
you're using, which you never do, no matter how many
times I ask (there's an example in this thread).
Second, either you're lying or you have an incredibly
bad memory even for things you wrote yourself.
Here's a quote from the OP, where you explained
the point to the thread:
"however standard set theory with its large cardinals is inconsistant
with the concept of ' fixed points '."
Note the phrase "standard set theory".
>im not here to learn ' old math ' , im here to bring you ' new math '.
Your "new math" is going to be nonsense until you learn some
"old math".
Arnold
>
> fixed point fixed point fixed point
"What I tell you three times is true": Lewis Carroll.
>
David Bernier
> amy666 <tomm...@hotmail.com> writes:
>
>> wrong , with standard ordinals 1 + w =/= w.
>
> I am afraid you're mistaken, Amy.
>
Amy: Imagine an infinite queue of people P1, P2, P3, P4 ...
If X goes in from of the queue, we get:
X, P1, P2, P3, P4 ... (ordering: 1 + omega )
and if X goes at the end:
P1, P2, P3, P4 ... X ( ordering: omega + 1)
David Bernier
Ben Standeven
>
> > In you "arguments" you seem to be assuming that
> > "fixed point theory" says that every function
> > has a fixed point. That's not so.
> >
>
> this is an intresting comment by david.
>
> although i did not say that every function has a
> fixed point , my ideas are indeed pretty close to
> that.
>
Let f(1) = 2, f(n) = 1 for all n >= 2. What is the fixed point of f?
amy666
yes.
so ?
fixpoint at oo.
oo is an element of C*
>
> How seriously do you test your conjectures before
> postiong them?
>
you really believe you have just given a counterexample ?
regards
tommy1729
Robert Israel
> > > conjecture :
> > >
> > > every analytic function has a fixed point in C*
> >
> > Is the function f(z) =exp(z)+z analytic ?
>
> yes.
>
> so ?
>
> fixpoint at oo.
>
> oo is an element of C*
But exp(z) + z is not defined there. It has an essential singularity at
infinity. In order to say f(infty) = infty, your analytic function f must
have a pole at infinity. The only entire functions with a pole at infinity
are nonconstant polynomials.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
amy666
>
> > > > conjecture :
> > > >
> > > > every analytic function has a fixed point in C*
> > >
> > > Is the function f(z) =exp(z)+z analytic ?
> >
> > yes.
> >
> > so ?
> >
> > fixpoint at oo.
> >
> > oo is an element of C*
>
> But exp(z) + z is not defined there. It has an
> essential singularity at
> infinity. In order to say f(infty) = infty, your
> analytic function f must
> have a pole at infinity. The only entire functions
> with a pole at infinity
> are nonconstant polynomials.
> --
hmm whats wrong with having an essential singularity as fixpoint ?
amy666
ok ok
i was wrong ...
darn alcohol ...
Mariano Suárez-Alvarez
> > amy666 <tommy1...@hotmail.com> writes:
>
> > > > > conjecture :
>
> > > > > every analytic function has a fixed point in C*
>
> > > > Is the function f(z) =exp(z)+z analytic ?
>
> > > yes.
>
> > > so ?
>
> > > fixpoint at oo.
>
> > > oo is an element of C*
>
> > But exp(z) + z is not defined there. It has an
> > essential singularity at
> > infinity. In order to say f(infty) = infty, your
> > analytic function f must
> > have a pole at infinity. The only entire functions
> > with a pole at infinity
> > are nonconstant polynomials.
> > --
>
> hmm whats wrong with having an essential singularity as fixpoint ?
That it simply does not make any sense?
-- m
Denis Feldmann
Interesting. I) In my notation, C* is the complex plane without 0, not
teh Riemann sphere C u{oo}
>
2) What makes you believe there exist analytic functions on the whole
Riemann spheze which are not polynomials ? What ar your definitons for
continuity at OO ?
>> How seriously do you test your conjectures before
>> postiong them?
>>
>
> you really believe you have just given a counterexample ?
>
>
Yes, I did :-)
>
> regards
>
> tommy1729
Robert Israel
> > amy666 <tomm...@hotmail.com> writes:
> >
> > > > > conjecture :
> > > > >
> > > > > every analytic function has a fixed point in C*
> > > >
> > > > Is the function f(z) =exp(z)+z analytic ?
> > >
> > > yes.
> > >
> > > so ?
> > >
> > > fixpoint at oo.
> > >
> > > oo is an element of C*
> >
> > But exp(z) + z is not defined there. It has an
> > essential singularity at
> > infinity. In order to say f(infty) = infty, your
> > analytic function f must
> > have a pole at infinity. The only entire functions
> > with a pole at infinity
> > are nonconstant polynomials.
> > --
>
> hmm whats wrong with having an essential singularity as fixpoint ?
Do you know the Weierstrass-Casorati theorem (or the Great Picard Theorem)?
--
amy666
> amy666 <tomm...@hotmail.com> writes:
>
> > > amy666 <tomm...@hotmail.com> writes:
> > >
> > > > > > conjecture :
> > > > > >
> > > > > > every analytic function has a fixed point
> in C*
> > > > >
> > > > > Is the function f(z) =exp(z)+z analytic ?
> > > >
> > > > yes.
> > > >
> > > > so ?
> > > >
> > > > fixpoint at oo.
> > > >
> > > > oo is an element of C*
> > >
> > > But exp(z) + z is not defined there. It has an
> > > essential singularity at
> > > infinity. In order to say f(infty) = infty, your
> > > analytic function f must
> > > have a pole at infinity. The only entire
> functions
> > > with a pole at infinity
> > > are nonconstant polynomials.
> > > --
> >
> > hmm whats wrong with having an essential
> singularity as fixpoint ?
>
> Do you know the Weierstrass-Casorati theorem (or the
> Great Picard Theorem)?
yes.
but if a function is entire such as exp(z) + z and its essential singularity lies at oo
i dont see why the big picard matters ?
small picard seems sufficient.
still , what is the problem with having an essential singularity as a fixpoint ?
i guess it cant be an attractive fixpoint then , is that what you mean ?
> --
> Robert Israel
> isr...@math.MyUniversitysInitials.ca
> Department of Mathematics
> http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver,
> BC, Canada
regards
tommy1729
Mariano Suárez-Alvarez
> Robert wrote :
>
>
>
> > amy666 <tommy1...@hotmail.com> writes:
Why do you think that defining exp(z) + z
to be equal to infinity when z is oo makes *any* sense?
-- m
amy666
really ?
well i mean the Riemann sphere , and i believe ive seen C* been used like that before ...
some authors do use you their own notation of course ...
> 2) What makes you believe there exist analytic
> functions on the whole
> Riemann spheze which are not polynomials ? What ar
> your definitons for
> continuity at OO ?
analytic on C , not oo.
> >> How seriously do you test your conjectures before
> >> postiong them?
> >>
> >
> > you really believe you have just given a
> counterexample ?
> >
> >
>
> Yes, I did :-)
> >
> > regards
> >
> > tommy1729
regards
tommy1729
David C. Ullrich
wrote:
>> amy666 <tomm...@hotmail.com> writes:
>>
>> > > > conjecture :
>> > > >
>> > > > every analytic function has a fixed point in C*
>> > >
>> > > Is the function f(z) =exp(z)+z analytic ?
>> >
>> > yes.
>> >
>> > so ?
>> >
>> > fixpoint at oo.
>> >
>> > oo is an element of C*
>>
>> But exp(z) + z is not defined there. It has an
>> essential singularity at
>> infinity. In order to say f(infty) = infty, your
>> analytic function f must
>> have a pole at infinity. The only entire functions
>> with a pole at infinity
>> are nonconstant polynomials.
>> --
>
>hmm whats wrong with having an essential singularity as fixpoint ?
It doesn't make any sense. You don't have any
idea what "essentual singularity" means, do you?
Look. We're already cutting you miles of slack.
For example, the function z + 1 has no fixed
point in C, and it also has no fixed point at oo,
because it's not defined there.
BUT in fact z + 1 tends to oo as z -> oo;
if f(z) = z+1 and you say f(oo) = oo then
while that's not quite right, we all know
what you mean - for that f, if you define
f(oo) = oo then f is nice and continuous
on C*, and it has a fixed point at oo, fine.
That's exactly why Denis gave the example
e^z + z instead of a simpler example like
z + 1. If we say f(z) = e^z + z then it does
_not_ make sense to say f(oo) = oo,
because f(z) does _not_ tend to oo as
z tends to oo. This function has no limit
at infinity whatever, no matter how you try
to stretch things - it makes no sense to say
it has a fixed point at infinity because it
simply doesn't have a value at infinity.
It doesn't even have-a-value-even-though-
it-really-doesn't-HAVE-a-value, like the
function z+1 does. Saying f(oo) = oo for
the function e^z + z simply makes no sense
at all.
You really _would_ be better off if you didn't
talk about things you didn't understand.
Really.
>
>> Robert Israel
>> isr...@math.MyUniversitysInitials.ca
>> Department of Mathematics
>> http://www.math.ubc.ca/~israel
>> University of British Columbia Vancouver,
>> BC, Canada
David C. Ullrich
Dave L. Renfro
> but if a function is entire such as exp(z) + z and its
> essential singularity lies at oo
>
> i dont see why the big picard matters ?
Well, for one thing, it implies that given any complex number w,
there exists a sequence {z_1, z_2, z_3, ...} such that z_n --> oo
and exp(z_n) + z_n --> w. You can define exp(oo) + oo to be oo,
-1, 3 + 2i, etc. but no matter how you define exp(z) + z at z = oo,
the function you've defined on the Riemann sphere will fail (rather
badly) to be continuous at z = oo. It's even worse than trying to
define the value of sin(1/x) (as a real-valued function of a real
variable) at x = 0.
Dave L. Renfro
David Bernier
Analytic over C \union { oo } implies continuous at oo.
In any open neighborhood of oo, e^z + z will take all values in C except
at most one. So, do you think e^z + z is continuous at oo ?
David Bernier
Toni...@yahoo.com
> amy666 wrote (in part):
>
> > but if a function is entire such as exp(z) + z and its
> > essential singularity lies at oo
>
> > i dont see why the big picard matters ?
>
> Well, for one thing, it implies that given any complex number w,
-- except one, at most --
David C. Ullrich
wrote:
The "problem" is that oo is simply _not_ a fixed point for e^z + z.
Because that function simply does not tend to oo at oo. This
follows from, for example, the little Picard theorem.
>i guess it cant be an attractive fixpoint then , is that what you mean ?
>
>
>> --
>> Robert Israel
>> isr...@math.MyUniversitysInitials.ca
>> Department of Mathematics
>> http://www.math.ubc.ca/~israel
>> University of British Columbia Vancouver,
>> BC, Canada
>
>regards
>
>tommy1729
David C. Ullrich
David C. Ullrich
wrote:
Huh?????????????
>
>> >> How seriously do you test your conjectures before
>> >> postiong them?
>> >>
>> >
>> > you really believe you have just given a
>> counterexample ?
>> >
>> >
>>
>> Yes, I did :-)
>> >
>> > regards
>> >
>> > tommy1729
>
>regards
>
>tommy1729
David C. Ullrich
amy666
again , why does that matter ?
what is wrong with an essential singularity as fixpoint ?
amy666
ah , thats why you object.
well it makes sense and it doesnt.
kind a like this
gamma ( - oo ) = 0
- oo is here a specific way of taking a limit.
thus in general - oo doesnt make sense here unless you mean that specific limit ...
sigh , this is an old debate again ...
running in circles ...
regards
tommy1729
amy666
e^(+oo) + (+oo) = + oo
e^(-oo) + (-oo) = - oo
=> two fixpoints at oo ?
>
> >i guess it cant be an attractive fixpoint then , is
> that what you mean ?
> >
> >
> >> --
> >> Robert Israel
> >> isr...@math.MyUniversitysInitials.ca
> >> Department of Mathematics
> >> http://www.math.ubc.ca/~israel
> >> University of British Columbia
> Vancouver,
> >> BC, Canada
> >
> >regards
> >
> >tommy1729
>
> 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.)
regards
tommy1729
Toni...@yahoo.com
> > On Mon, 02 Mar 2009 16:56:40 EST, amy666
> > wrote:
>
> > >Robert wrote :
>
> > >> amy666 <tommy1...@hotmail.com> writes:
>
well-known theorems, you insist in the same stupidities as before...
Unbelievable...but I think that's just natural. After all, one
participant here already wrote clearly that she admires you, so you
think (figure of speech) that you can jump up with all kinds of
nonsenses...and indeed, you do.
**sigh**
Tonio
Tonio
amy666
you didnt answer me.
you got personal instead.
your feelings about me dont matter.
and yes , galathaea admires me and loves me.
but that is off topic.
tommy1729
Dave L. Renfro
>> Well, for one thing, it implies that given any complex number w,
tonic...@yahoo.com wrote:
> -- except one, at most --
Dave L. Renfro wrote:
>> there exists a sequence {z_1, z_2, z_3, ...} such that z_n --> oo
>> and exp(z_n) + z_n --> w. You can define exp(oo) + oo to be oo,
>> -1, 3 + 2i, etc. but no matter how you define exp(z) + z at z = oo,
>> the function you've defined on the Riemann sphere will fail (rather
>> badly) to be continuous at z = oo. It's even worse than trying to
>> define the value of sin(1/x) (as a real-valued function of a real
>> variable) at x = 0.
The "except at most one" restriction occurs in a much
stronger version that involves the _values_ exp(z) + z takes
in every neighborhood of oo. In fact, just knowing that
exp(z) + z can take any complex number from a specified
set dense in the complex plane (which allows for co-countably
many exceptions, and thus is quite a bit more generous than
allowing for at most one exception) is enough to deduce
the result I stated. Moreover, I should have said "given
any w in the Riemann sphere", since that's true also.
Dave L. Renfro
David C. Ullrich
wrote:
In particular, _the_ limit of gamma(z) as z tends to infinity does not
exist. The gamma function is _not_ continuous on C*.
>sigh , this is an old debate again ...
>
>running in circles ...
Only because of your ignorance and/or stubbornness -
the actual math doesn't lead to any "debate".
You're not free to make words or phrases mean whatever
you want. You can redefine "fixed point" if you wish,
but it doesn't follow that e^z + z actually has a fixed
point at infinity, just that it has a Timmy-fixed-point.
David C. Ullrich
amy666
> but it doesn't follow that e^z + z actually has a
> fixed
> point at infinity, just that it has a
> Timmy-fixed-point.
lol.
timmy-fixed-point
however , the definition of a fixed point is that it is mapped to itself.
it doesnt mention convergeance or singularities.
>
> >
> >regards
> >
> >tommy1729
>
> 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.)
regards
tommy1729
Dave L. Renfro
> however , the definition of a fixed point is that
> it is mapped to itself.
>
> it doesnt mention convergeance or singularities.
I'm wondering if you realize that any function can be
extended so as to have a fixed point, which is a fairly
trivial observation, and people in this thread are
basically wondering what it is you're trying to do
if not this.
For example, consider f(x) = sin(1/x) for x a real number
not equal to 0. Then by putting f(0) = 0, we obtain
(an extension of f that has) a fixed point.
Another example: Let g be the function defined by the
set of ordered pairs {(5,-17), (amy666,Renfro), (oo,oo)}.
Then by defining g(Renfro) = Renfro, we obtain
(an extension of g that has) a fixed point.
Still another example: Let h be the function defined by
h(z) = z^5 if z is a complex number not equal to 3 + 2i.
Then by defining h(3 + 2i) = 3 + 2i, we obtain
(an extension of h that has) a fixed point.
Dave L. Renfro
MoeBlee
> and yes , galathaea admires me and loves me.
>
> but that is off topic.
Yes, off topic and in ane.
MoeBlee
David C. Ullrich
wrote:
>You can redefine "fixed point" if you wish,
>> but it doesn't follow that e^z + z actually has a
>> fixed
>> point at infinity, just that it has a
>> Timmy-fixed-point.
>
>lol.
>
>timmy-fixed-point
>
>however , the definition of a fixed point is that it is mapped to itself.
>
>it doesnt mention convergeance or singularities.
That's right. That's exactly why the function e^z + z has
no fixed point. There is no point z such that e^z + z = z.
The fact that the definition doesn't mention convergence
is exactly why your comments about how the limit
as z tends to infinity _in certain directions_ is infinite
are irrelevant.
amy666
> amy666 wrote (in part):
>
> > however , the definition of a fixed point is that
> > it is mapped to itself.
> >
> > it doesnt mention convergeance or singularities.
>
> I'm wondering if you realize that any function can be
> extended so as to have a fixed point, which is a
> fairly
> trivial observation, and people in this thread are
> basically wondering what it is you're trying to do
> if not this.
>
> For example, consider f(x) = sin(1/x) for x a real
> number
> not equal to 0. Then by putting f(0) = 0, we obtain
> (an extension of f that has) a fixed point.
and that is quite often done.
>
> Another example: Let g be the function defined by the
> set of ordered pairs {(5,-17), (amy666,Renfro),
> (oo,oo)}.
> Then by defining g(Renfro) = Renfro, we obtain
> (an extension of g that has) a fixed point.
i dont know if renfro is analytic :)
>
> Still another example: Let h be the function defined
> by
> h(z) = z^5 if z is a complex number not equal to 3 +
> 2i.
> Then by defining h(3 + 2i) = 3 + 2i, we obtain
> (an extension of h that has) a fixed point.
certainly not analytic.
>
> Dave L. Renfro
regards
tommy1729
amy666
> On Wed, 04 Mar 2009 11:15:57 EST, amy666
> <tomm...@hotmail.com>
> wrote:
>
> >You can redefine "fixed point" if you wish,
> >> but it doesn't follow that e^z + z actually has a
> >> fixed
> >> point at infinity, just that it has a
> >> Timmy-fixed-point.
> >
> >lol.
> >
> >timmy-fixed-point
> >
> >however , the definition of a fixed point is that it
> is mapped to itself.
> >
> >it doesnt mention convergeance or singularities.
>
> That's right. That's exactly why the function e^z + z
> has
> no fixed point.
???
> There is no point z such that e^z + z
> = z.
e^z + z = z
=> e^z = z - z
=> e^z = 0
e^z = 0 = e^(z - 1) = e^(z - 2) = ...
thus z = -oo
thus we have point z = -oo
you might not like ' signed oo ' but even for both affine and projective oo we get a fixed point :
exp(oo) + oo = oo
regards
tommy1729
Richard Tobin
amy666 <tomm...@hotmail.com> wrote:
>> There is no point z such that e^z + z
>> = z.
>e^z + z = z
>
>=> e^z = z - z
>
>=> e^z = 0
>
>e^z = 0 = e^(z - 1) = e^(z - 2) = ...
>
>thus z = -oo
>
>thus we have point z = -oo
Aren't you talking about C* ?
-- Richard
--
Please remember to mention me / in tapes you leave behind.
David C. Ullrich
wrote:
>Dave L Renfro wrote :
>
>> amy666 wrote (in part):
>>
>> > however , the definition of a fixed point is that
>> > it is mapped to itself.
>> >
>> > it doesnt mention convergeance or singularities.
>>
>> I'm wondering if you realize that any function can be
>> extended so as to have a fixed point, which is a
>> fairly
>> trivial observation, and people in this thread are
>> basically wondering what it is you're trying to do
>> if not this.
>>
>> For example, consider f(x) = sin(1/x) for x a real
>> number
>> not equal to 0. Then by putting f(0) = 0, we obtain
>> (an extension of f that has) a fixed point.
>
>and that is quite often done.
That's quite often done for _that_ function?
I don't believe it - give a reference where
someone does that.
_If_ f(0) is undefined but f(t) has a limit
as t -> 0 _then_ the value of f(0) is
quite often defined to be equal to this
limit. But that function does not _have_
a limit at 0 - that was the point to the
example.
>
>>
>> Another example: Let g be the function defined by the
>> set of ordered pairs {(5,-17), (amy666,Renfro),
>> (oo,oo)}.
>> Then by defining g(Renfro) = Renfro, we obtain
>> (an extension of g that has) a fixed point.
>
>i dont know if renfro is analytic :)
>
>
>>
>> Still another example: Let h be the function defined
>> by
>> h(z) = z^5 if z is a complex number not equal to 3 +
>> 2i.
>> Then by defining h(3 + 2i) = 3 + 2i, we obtain
>> (an extension of h that has) a fixed point.
>
>certainly not analytic.
>
>
>>
>> Dave L. Renfro
>
>regards
>
>tommy1729
David C. Ullrich
David C. Ullrich
wrote:
What I like or not is iirrelevant. You started this nonsense
by saying that that function tends to infinity _at infinity_.
It doesn't.
You seem to have forgotten what you said the other
day about how the definition of "fixed point" doesn't
mention anything about convergence, btw. That
was an excellent point - and it shows that everything
you've said here is silly.
And you're missing the point to Renfro's post.
_If_ you decide that you're allowed to define
"fixed point" any way you like then yes,
everything has a fixed point. That's what you're
doing here - instead of showing the function
_actually_ has a fixed point you're coming
up with a definition that suits you.
amy666
'oo' is no extention i made it already existed.
unlike ' renfro '.
see the huge difference ?
David C. Ullrich
wrote:
If f is an entire function and you claim that f has a fixed
point at oo you are _extending_ f. Saying it's an
entire function means that it is a map from C to C,
so there is no such thing as f(oo) until you _extend_
f to be a map from C* to C*.
amy666
map from C to C = entire
map from C* to C* = ?? name ??
Dave L. Renfro
> map from C to C = entire
>
> map from C* to C* = ?? name ??
From <http://en.wikipedia.org/wiki/Meromorphic_function>:
* When D is the entire Riemann sphere, the field of meromorphic
* functions is simply the field of rational functions in one
* variable over the complex field, since one can prove that
* any meromorphic function on the sphere is rational. (This is
* a special case of the so-called GAGA principle.)
From <http://en.wikipedia.org/wiki/Entire_function>:
* As a consequence, a (complex-valued) function which is entire
* on the whole Riemann sphere (complex plane and the point at
* infinity) is constant. Thus a (non-constant) entire function
* must have a singularity at the complex point at infinity,
* either a pole or an essential singularity (see Liouville's
* theorem below). In the latter case, it is called a transcendental
* entire function, otherwise it is a polynomial.
This last result has already been pointed out by Robert Israel
elsewhere in this thread (see [1]).
[1] http://groups.google.com/group/sci.math/msg/f72c034e1ef14c5c
Dave L. Renfro
lwa...@lausd.net
> On Fri, 27 Feb 2009 07:14:26 EST, amy666 <tommy1...@hotmail.com>
> wrote:
> >although i did not say that every function has a fixed point , my ideas are indeed pretty close to that.
> If you realize that not every function has a fixed point then
> how is it you're saying that the fact that the successor
> function for ordinals has no fixed point is some sort
> of problem?
OK, I've been away from sci.math for a few weeks, so that I've
just discovered this thread. And although recently the
discussion has changed to fixed points in complex analysis, I
will steer it back to its original topic of set theory.
All of this discussion about fixed points reminds me of a
concept from set theory, the Fixed-Point Reflection Principle
(but note that I've only seen informal discussions of it).
But I know that essentially, it states that for certain
one-place functions phi, if phi(On) = On, i.e., if the (proper)
class of all ordinals On is a fixed point of the function phi,
then there exists a (set) ordinal alpha s.t. phi(alpha) = alpha,
so that alpha is also a fixed point of phi. (If my formulation
of the Fixed-Point Reflection Principle is incorrect, then if
someone would please post the correct formulation.)
Notice that although the Fixed-Point Reflection Principle is
not a schema of ZFC (even with proper classes), certain
instances of the schema are accepted as axioms -- and these
are the large cardinal axioms (postulating the existence of,
say, inaccessible or Mahlo cardinals). Notice that tommy1729,
even in his original post, mentioned the large cardinal axioms,
which is why I feel that I'm on the right track in realizing
that this is what he currently has in mind.
Therefore, what tommy1729 seeks is a set theory in which a
form of the Fixed-Point Reflection Principle is actually a
_schema_ of the theory, something like:
"If phi(x) has the property phi(On) = On (and satisfies certain
other properties), then there exists an ordinal alpha such
that phi(alpha) = alpha."
Or since tommy1729 rejects ordinals, perhaps we can write this
without ordinals, instead using the class V of all sets:
"If phi(x) has the property phi(V) = V (and satisfies certain
other properties), then there exists set a s.t. phi(a) = a."
And of course the goal would be to determine what those certain
other properties must be to avoid inconsistency, yet still
have at least one instance of the schema whose negation is a
theorem of ZFC (so that the theory would therefore differ
essentially from ZFC).
Now tommy1729 first mentions successors -- and I'm not even
sure what it means to say S(On) = On or S(V) = V, anyway. I
suspect that classes such as On or V can't have successors --
so in particular they aren't fixed points of successor.
But all of tommy1729's comments about successors was really
intended to lead to a discussion of powersets. In particular,
we know that P(V) can't be anything other than V. Thus V must
be a fixed point of P -- so that there would exist, by
Reflection, a set a such that a = P(a). This set a would
definitely be a non-Cantorian set, and so the Reflection
Schema would prove the existence of non-Cantorian sets (so
that the theory would definitely differ from ZFC, where one
can prove that all sets are Cantorian).
> >david claims that not every function has a fixed point.
> Yes, I "claim" that. Just like I "claim" that 2 + 2 = 4.
And here we go again, where Ullrich compares the consideration
of nonstandard set theories in which functions such as the
powerset function can have a fixed point, to a theory in which
2 + 2 is not 4.
To Ullrich, "the powerset function has no fixed point" is not
a "claim," but something that is absolutely true in the same
way that 2 + 2 = 4. And so, the thinking goes, just as no
sane person would ever consider a theory in 2 + 2 is not 4, no
sane person should ever consider a theory in which the powerset
function has a fixed point. Despite the existence of established
theories such as NFU, Ullrich still compares denying Cantor's
theorem to denying 2 + 2 = 4.
Just because someone rejects one theorem of ZFC, it doesn't mean
that they reject _every_ theorem of ZFC. Someone can want the
powerset to have a fixed point and still keep 2 + 2 = 4. No
matter how many times I point this out, someone like Ullrich
continues to make this sort of remark and compare the rejection
of _any_ theorem of ZFC to the rejection of the well-known
arithmetic of small finite numbers.
Most of the so-called "cranks" here don't reject the arithmetic
of small finite numbers -- and the ones who do, I intentionally
avoid posting in such threads. Most of the so-called "cranks"
reject either _large_ finite naturals (with "large" usually
being on the order of the number of particles in the universe)
or, more commonly, _infinite_ sets.
I write this, but I know that someone will later on make yet
another comparison of rejecting ZFC to rejecting 2 + 2 = 4.
> >thats why i changed the title , this is a bit off topic but intresting.
> >ill start :
> >conjecture :
> >every analytic function has a fixed point in C*
> What an immensely stupid conjecture. It's obviously
> equivalent to the conjecture that every analytic
> function has a zero in C*, which is obviously false.
Which is why I avoid making any comments on fixed
points in complex analysis and will focus only on
fixed points in set theory.
Denis Feldmann
And here we go again, where David compares two very close meanings of
the word "claim", and lwalker deliberately misinteprets him
Troll
>
> To Ullrich, "the powerset function has no fixed point" is not
> a "claim," but something that is absolutely true in the same
> way that 2 + 2 = 4. And so, the thinking goes, just as no
> sane person would ever consider a theory in 2 + 2 is not 4, no
> sane person should ever consider a theory in which the powerset
> function has a fixed point. Despite the existence of established
> theories such as NFU, Ullrich still compares denying Cantor's
> theorem to denying 2 + 2 = 4.
>
> Just because someone rejects one theorem of ZFC, it doesn't mean
> that they reject _every_ theorem of ZFC. Someone can want the
> powerset to have a fixed point and still keep 2 + 2 = 4. No
> matter how many times I point this out, someone like Ullrich
> continues to make this sort of remark and compare the rejection
> of _any_ theorem of ZFC to the rejection of the well-known
> arithmetic of small finite numbers.
>
> Most of the so-called "cranks" here don't reject the arithmetic
> of small finite numbers -- and the ones who do, I intentionally
> avoid posting in such threads. Most of the so-called "cranks"
> reject either _large_ finite naturals (with "large" usually
> being on the order of the number of particles in the universe)
> or, more commonly, _infinite_ sets.
>
> I write this, but I know that someone will later on make yet
> another comparison of rejecting ZFC to rejecting 2 + 2 = 4.
>
Probably you will construe what they say like that, but they will not
>>> thats why i changed the title , this is a bit off topic but intresting.
>>> ill start :
>>> conjecture :
>>> every analytic function has a fixed point in C*
>> What an immensely stupid conjecture. It's obviously
>> equivalent to the conjecture that every analytic
>> function has a zero in C*, which is obviously false.
>
> Which is why I avoid making any comments on fixed
> points in complex analysis
Because you realize that there is not much room there for trolling?
David C. Ullrich
wrote:
An _analytic_ map from C* to C* turns out to be
a rational function, ie the quotient of two polynomials
(with one exception, that being f(z) = infinity for all z).
David C. Ullrich
>On Feb 28, 6:26 am, David C. Ullrich <dullr...@sprynet.com> wrote:
>> [...]
>
>> >david claims that not every function has a fixed point.
>> Yes, I "claim" that. Just like I "claim" that 2 + 2 = 4.
>
>And here we go again, where Ullrich compares the consideration
>of nonstandard set theories in which functions such as the
>powerset function can have a fixed point, to a theory in which
>2 + 2 is not 4.
Huh? It's incredibly hard to tell whether you're stupid,
or can't read, or lying about things that are staring
everyone in the face, or what. Look:
timmy said that I claim that not every function
has a fixed point. Saying I "claim" that is silly,
because it is awesomely obvious that not every
function has a fixed point. Hence my sarcastic
reply.
Now how in the world do you get from my
implicit "that's true, not every function has a
fixed point" to the idea that I'm claiming that
there cannot be a "nonstandard set theory"
in which the powerset has a fixed point?
>To Ullrich, "the powerset function has no fixed point" is not
>a "claim," but something that is absolutely true in the same
>way that 2 + 2 = 4. And so, the thinking goes, just as no
>sane person would ever consider a theory in 2 + 2 is not 4, no
>sane person should ever consider a theory in which the powerset
>function has a fixed point. Despite the existence of established
>theories such as NFU, Ullrich still compares denying Cantor's
>theorem to denying 2 + 2 = 4.
The way you put words in my mouth is simply incredible.
Especially when you do it immediately after quoting what
I actually said.
>Just because someone rejects one theorem of ZFC, it doesn't mean
>that they reject _every_ theorem of ZFC. Someone can want the
>powerset to have a fixed point and still keep 2 + 2 = 4. No
>matter how many times I point this out, someone like Ullrich
>continues to make this sort of remark
What fucking remark are you referring to? What planet are you
on, where "yes, it's true that not every function has a fixed
point" means "if someone rejects one theorem of ZFC they
must reject them all"?
No matter how many times people point out that you
never seem to get a single thing straight regarding
what people have actually _said_, you continue to
just make stuff up.
Maybe it's psychological. Some people hear voices
that aren't there - I suppose some people could see
words on their monitors that aren't there.
> and compare the rejection
>of _any_ theorem of ZFC to the rejection of the well-known
>arithmetic of small finite numbers.
In a previous post you said that you advocate overthrowing
the government of the United States by force. My advice
is to be careful about saying that sort of thing in public;
I believe that it's explicitly _not_ included in the
free speech stuff in the constitution.
_Why_ do you say that it's warmer at the North Pole
than in Texas, by the way?
amy666
so ?
Jesse F. Hughes
> But all of tommy1729's comments about successors was really
> intended to lead to a discussion of powersets. In particular,
> we know that P(V) can't be anything other than V. Thus V must
> be a fixed point of P -- so that there would exist, by
> Reflection, a set a such that a = P(a). This set a would
> definitely be a non-Cantorian set, and so the Reflection
> Schema would prove the existence of non-Cantorian sets (so
> that the theory would definitely differ from ZFC, where one
> can prove that all sets are Cantorian).
There is a natural extension of the powerset functor[1] to the category
of sets and classes, where P(X) is the class of all *subsets* of X.
As you suggest, the universe V is a fixed point for this functor.
If V is the least fixed point, then V is the category of all well-founded
sets.
If V is the greatest fixed point, then V is the category NWF (the
non-well-founded sets, the theory of which is usually denoted ZFA).
This is interesting stuff. Pretending that there is a *set* a such
that a = P(a) is rather less interesting, at least until you tell us
how to amend the axioms of ZF for consistency's sake.
> > >david claims that not every function has a fixed point.
> > Yes, I "claim" that. Just like I "claim" that 2 + 2 = 4.
> And here we go again, where Ullrich compares the consideration
> of nonstandard set theories in which functions such as the
> powerset function can have a fixed point, to a theory in which
> 2 + 2 is not 4.
And here we go again, where lwalker pretends that Tommy was speaking
about some well-defined, consistent theory (other than ZF) in which
his ridiculous axiom is true.
Footnotes:
[1] Note: You *have* to discuss this extension explicitly to make any
sense out of your claim that P(V) = V. The operator P is normally
taken to be an operator (or functor) V -> V (or Set -> Set). Thus,
P(V) is usually undefined.
--
"But remember, as long as one human being follows the rules of
mathematics, then mathematics as a human discipline survives.
Right now I'm that one human being, so mathematics survives."
-- James S. Harris
lwa...@lausd.net
> On Wed, 11 Mar 2009 22:48:10 -0700 (PDT), lwal...@lausd.net wrote:
> >And here we go again, where Ullrich compares the consideration
> >of nonstandard set theories in which functions such as the
> >powerset function can have a fixed point, to a theory in which
> >2 + 2 is not 4.
> Huh? It's incredibly hard to tell whether you're stupid,
> or can't read, or lying about things that are staring
> everyone in the face, or what. Look:
> timmy said that I claim that not every function
> has a fixed point. Saying I "claim" that is silly,
> because it is awesomely obvious that not every
> function has a fixed point. Hence my sarcastic
> reply.
Speaking of sarcastic replies and joke answers, I just
noticed another thread from last week, which is titled
"bottles vs. cartons." Lest I be accused of yet again
misinterpreting a post, let me give the quotes verbatim.
The OP wrote:
"Why are soft drinks packaged in cylindrical containers,
while milk is packaged in square containers?"
and Ullrich's reply was:
"Well, the milk inside the container is square (just look!),
so if the container were circular either the milk wouldn't fit
or there would be some wasted space inside the container.
"Similarly for the soft drinks. Next question, please."
Forget about 2 + 2 = 4 and elementary school arithmetic, but
let's consider elementary school science, in which one learns
that a liquid is a fluid that takes the shape of its container,
so that Ullrich's response doesn't answer the OP's question.
Of course, Ullrich already knows this. His response was almost
surely (and by "almost surely," I mean with probability 1, of
course) intended as a _joke_.
Now it's understandable that Ullrich would give joke responses
when dealing with so-called "cranks" such as tommy1729 or
myself, or as he himself pointed out, to those who ask
questions with obvious answers. (His own example was, why aren't
bridges made out of cheese?)
But here Mark-T is not a so-called "crank," asking a reasonable
question (which is most likely classified as an economics or
manufacturing question, rather than a science question) whose
answer isn't apparent, yet Ullrich still felt the need to give
Mark-T a joke answer as well.
My point is that I wonder why so many sci.math posters defend
Ullrich as a great teacher, when he is apparently almost as
likely to give a joke response or sarcastic reply as opposed
to a legitimate answer, even when the person asking isn't an
established "crank."