card(P(N)) = card(N), an elementary proof
ju...@diegidio.name
here and there).
--- Preliminaries:
'P(A)' stands for the "powerset" of set 'A'.
'~' stands for the existence of a bijection, which will be, in this
context, equivalent to stating that two sets are equinumerous.
Infinity of the set 'N' (the set of natural/ordinal numbers) is taken
to be defined as usual, by the existence of bijections between N and
its (infinite) subsets. No other kind of "infinities" are going to be
assumed: the only tool I have used is transfinite induction over the
naturals/ordinals, with "omega" as the limit ordinal.
'oo' stands for "omega".
--- Finite case:
card(A) = n <=>def
A ~ { 0, 1, 2, .., n-1 }
card(P(A)) = 2^n <=>
P(A) ~ { 1, 2, 4, ..., 2^n-1 }
--- Infinite case (transf. induction, notation a bit fuzzy here):
(1) n=oo => card(A) = card(N) = oo <=>def
A ~ N := { 0, 1, 2, ..., n }, n->oo
(2) n=oo => card(P(A)) = card(P(N)) = 2^oo <=>
P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo
--- Theorem:
card(P(N)) = card(N)
--- Proof:
(i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card'
(ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~'
(iii) A bijection 2^N ~ N trivially exists:
2^N ~ N
--------
1 ~ 0
2 ~ 1
4 ~ 2
...
Follows the thesis.
QDE.
It's THEN easy to get interesting lemmas. We get that omega is an
absolute upper bound and all infinities are equivalent. The diagonal
argument then assumes another shape, which will be the subject of
another post of mine, unless someone finds a flaw in the above quite
elementary argument.
Thanks for any feedback.
Julio
status: OPEN
sender: Julio Di Egidio (aka LudovicoVan)
sender-email: ju...@diegidio.aleph.name (del 'aleph.')
copyright: 2008 (on behalf of) sci.math, sci.logic
All rights reserved.
(I've heard they are gonna patent Fibonacci... shouldn't the flowers
and actually the whole nature have a... "natural" right to it? Fuck
them all, this is more true than it sounds: they *are* patenting
nature, and through it exploiting and literally enslaving whole
countries, although out of any media report. It is the very self-
legality of the global empire of multinationals, through the WTO
institution: fuck them all, *they* must be retarded, there is no other
explanation for this insult to life and intelligence, and basically to
this mass-level induced self-destruction. Sick they must be, deeply
sick. But there is a cure, they are really just invoking it. It's the
madness of the whole power class of the world: they want to be
protagonist, and they always get what they deserve, that very hell
they have so contributed to build. --- Sorry for the digression.)
ju...@diegidio.name
> card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> here and there).
>
> --- Preliminaries:
>
> 'P(A)' stands for the "powerset" of set 'A'.
>
> '~' stands for the existence of a bijection, which will be, in this
> context, equivalent to stating that two sets are equinumerous.
> Infinity of the set 'N' (the set of natural/ordinal numbers) is taken
> to be defined as usual, by the existence of bijections between N and
> its (infinite) subsets. No other kind of "infinities" are going to be
> assumed: the only tool I have used is transfinite induction over the
> naturals/ordinals, with "omega" as the limit ordinal.
>
> 'oo' stands for "omega".
>
> --- Finite case:
>
> card(A) = n <=>def
> A ~ { 0, 1, 2, .., n-1 }
>
> card(P(A)) = 2^n <=>
> P(A) ~ { 1, 2, 4, ..., 2^n-1 }
Sorry, should of course read:
P(A) ~ { 0, 1, 2, ..., 2^n-1 }
and below:
P(A) ~ P(N) ~ 2^N := { 0, 1, 2, ..., 2^n }, n->oo
-LV
José Carlos Santos
> card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> here and there).
>
> --- Preliminaries:
>
> 'P(A)' stands for the "powerset" of set 'A'.
>
> '~' stands for the existence of a bijection, which will be, in this
> context, equivalent to stating that two sets are equinumerous.
Why? Is there some context in which equinumerous is *not* equivalent to
the existence of a bijection?
> Infinity of the set 'N' (the set of natural/ordinal numbers) is taken
> to be defined as usual, by the existence of bijections between N and
> its (infinite) subsets.
Actually, the usual definition is about the existence of bijections
between the set and *some* of its infinite subsets.
> No other kind of "infinities" are going to be
> assumed: the only tool I have used is transfinite induction over the
> naturals/ordinals, with "omega" as the limit ordinal.
>
> 'oo' stands for "omega".
>
> --- Finite case:
"Finite case" of *what*?
> card(A) = n <=>def
> A ~ { 0, 1, 2, .., n-1 }
>
> card(P(A)) = 2^n <=>
> P(A) ~ { 1, 2, 4, ..., 2^n-1 }
My guess is that you meant 2^{n - 1} instead of 2^n - 1.
> --- Infinite case (transf. induction, notation a bit fuzzy here):
"Infinite case" of *what*?
> (1) n=oo => card(A) = card(N) = oo <=>def
> A ~ N := { 0, 1, 2, ..., n }, n->oo
>
> (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <=>
> P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo
>
> --- Theorem:
>
> card(P(N)) = card(N)
>
> --- Proof:
>
> (i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card'
>
> (ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~'
>
> (iii) A bijection 2^N ~ N trivially exists:
>
> 2^N ~ N
> --------
> 1 ~ 0
> 2 ~ 1
> 4 ~ 2
Cute. Under this bijection, which element of N corresponds to, say,
the set of odd numbers?
> Follows the thesis.
Try to do the same thing without using a "notation a bit fuzzy".
> QDE.
In your dreams.
Best regards,
Jose Carlos Santos
ju...@diegidio.name
Given the previous correction, this bijection is not well written, and
not at all trivial, at least to me. The very symbol '2^N' must be re-
thought. Hmm...
-LV
ju...@diegidio.name
> > '~' stands for the existence of a bijection, which will be, in this
> > context, equivalent to stating that two sets are equinumerous.
>
> Why? Is there some context in which equinumerous is *not* equivalent to
> the existence of a bijection?
The fact is I go conservative: reality is, do not ask *me* about
*that*.
> > card(P(A)) = 2^n <=>
> > P(A) ~ { 1, 2, 4, ..., 2^n-1 }
>
> My guess is that you meant 2^{n - 1} instead of 2^n - 1.
No, just the same blatant mistake in my construction.
Now, I am gonna think better before posting further "corrections".
Thanks in any case for the feedback, very well appreciated.
-LV
ju...@diegidio.name
that, together, they will complement the eventual ambiguities.
----- 1st formulation
(i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
Let's simply call P(A) the set on the RHS of (i), keeping in mind that
the argument is going to hold due to this bijection and the
transitivity of the '~' relation.
(ii) P(A) c= N, AneN
(iii) P(N) is infinite (trivial)
We pass from P(A) to P(N) by transfinite induction, then:
(iv) P(N) ~ N, from (ii) and (iii)
QDE. (1)
----- 2nd formulation
Same comments as above.
(i) card(P(A)) = 2^n, AneN
(ii) card(P(A)) <= card(N), AneN
(iii) card(P(N)) >= oo (meaning: 'oo' as lower bound)
(iv) card(P(N)) = card(N)
QDE. (2)
-----
Hmm, correct? The basic idea behind this argument is that P(A) is
always a subset of N in the finite case. Then, by transfinite
induction, while P(N) becomes infinite, its cardinality's upper bound
remains N's cardinality. (These informal expositions get me into more
troubles than they help, but I have to run the risk: no pain, no
gain!)
Good luck.
Julio
status: OPEN
revision: 1
sender: Julio Di Egidio (aka LudovicoVan)
sender-email: ju...@diegidio.aleph.name (del 'aleph.')
copyright: 2008 (on behalf of) sci.math, sci.logic
All rights reserved.
On 26 Jul, 22:52, ju...@diegidio.name wrote:
Brian Chandler
I compliment you on starting a new thread, though given the enormity
of the task you have set yourself, it may take at least another 8000
posts.
I'm responding to your third post, even though you seem to have "minor
doubts" already...
> On 26 Jul, 23:03, ju...@diegidio.name wrote:
> > On 26 Jul, 22:52, ju...@diegidio.name wrote:
> > >
> > > card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> > > here and there).
> >
> > > --- Preliminaries:
> >
> > > 'P(A)' stands for the "powerset" of set 'A'.
Are you sure you _really_ understand what "powerset" means? It rather
seems not...
> > > '~' stands for the existence of a bijection, which will be, in this
> > > context, equivalent to stating that two sets are equinumerous.
> > > Infinity of the set 'N' (the set of natural/ordinal numbers) is taken
> > > to be defined as usual, by the existence of bijections between N and
> > > its (infinite) subsets. No other kind of "infinities" are going to be
> > > assumed: the only tool I have used is transfinite induction over the
> > > naturals/ordinals, with "omega" as the limit ordinal.
> >
> > > 'oo' stands for "omega".
> >
> > > --- Finite case:
> >
> > > card(A) = n <=>def
> > > � � A ~ { 0, 1, 2, .., n-1 }
> >
> > > card(P(A)) = 2^n <=>
> > > � � P(A) ~ { 1, 2, 4, ..., 2^n-1 }
> >
> > Sorry, should of course read:
> >
> > �P(A) ~ { 0, 1, 2, ..., 2^n-1 }
Absolutely not. It _should_ read something like:
P(A) = { {}, {0}, {1}, {2}, ... {n}, {0,1}, {0,2}, {0,3}, ...{0,n},
{1,2}, {1,3}, ...{1,n}, {2,3}, {2,4}, ...{2,n}, ...{n-2,n-1}, {n-2,n},
{n-1,n}, ... {0,1,2,...n-2}, ...{0,1,2,4,...n}, {0,1,3,4,...n},
{0,2,3,...n}, {1,2,3,...n}, {0,1,2,3...n}}
(Actually it's totally impractical to "spell out" like this, but since
you seem to be missing the point, I've tried.)
> >
> > and below:
> >
> > �P(A) ~ P(N) ~ 2^N := { 0, 1, 2, ..., 2^n }, n->oo
Not at all. You thanked me for trying to show you what the powerset is
last time, and said this was the first useful feedback you had
received. But Alan Smaill had already pointed out the same thing, and
I was only repeating my previous post, so this is the fourth time
really.
> > > --- Infinite case (transf. induction, notation a bit fuzzy here):
> >
> > > (1) n=oo => card(A) = card(N) = oo <=>def
> > > � � A ~ N := { 0, 1, 2, ..., n }, n->oo
Right
> > > (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <=>
> > > � � P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo
Wrong
> > > --- Theorem:
> >
> > > card(P(N)) = card(N)
> >
> > > --- Proof:
> >
> > > (i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card'
> >
> > > (ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~'
> >
> > > (iii) A bijection 2^N ~ N trivially exists:
> >
> > > 2^N ~ N
> > > --------
> > > � 1 ~ 0
> > > � 2 ~ 1
> > > � 4 ~ 2
> > > ...
>
>
> Given the previous correction, this bijection is not well written, and
> not at all trivial, at least to me. The very symbol '2^N' must be re-
> thought. Hmm...
No, the problem is not the _symbol_, except in that it is helping to
confuse you.
You might like to do some groundwork first: here's a (very simple!)
proof that there is no bijection. Can you see any flaw in it? sci.math
is overflowing with "proofs" to the contrary; look for them, and see
if you can see flaws. Then ask why _your_ attempt is going to be
Really Different.
-----------------------------------------------------------------
Theorem
If T is a set, and P(T) is the powerset of T, there is no function
mapping T onto P(T).
Remark
We suppose that f:T->P(T) is _any_ function mapping T to P(T), then
show that there is an element of P(T) to which f does not map
anything, thus proving that f is not an onto mapping (surjection).
Whether T is a finite or infinite set makes no difference to the
proof. (Also note that while we could make this look like a proof by
contradiction, by supposing that f _is_ onto, it's really better the
other way, because we construct a specific element to which f does not
map anything, so (I think?) the proof is OK for constructivists too.)
Proof
Suppose f:T->P(T) is a function from T to its powerset. Then if a e T,
f(a) e P(T), so f(a) is a _subset_ of T.
Note that there are two possibilities regarding f(a): it might be any
subset, and thus may or may not include a itself as an element. So the
elements of T naturally fall into two classes, we'll call the idio-ts
and the alio-ts.
Definitions
a e T is an idio-t if a e f(a)
a e T is an alio-t if a is not an element of f(a)
These are mutually exclusive cases, so every a is either an idio-t or
an alio-t.
Now consider the subset C of T which is the set of alio-ts.
Note: we could write everything so far in one (less readable) line
C = { x | x e T, x /e f(x) } //... /e means "not an element of"
We now show that there is no element b of T such that f: b -> C
(1) If b were an idio-t, it would have to be an element of C, but C is
the set of alio-ts so it isn't; contradiction.
(2) If b were an alio-t, if would not be an element of C, but C is the
set of alio-ts so it is; contradiction.
Therefore f is not an onto mapping. But this is true of _every_
mapping f, so we have proved the nonexistence of an onto mapping, or
surjection if you want to be latinamerican about it.
QED
-----------------------------------------------------------------
Just to get you started, here are some demonstrations of other people
who can't understand the above proof:
http://groups.google.com/group/sci.math/browse_frm/thread/5ecf405c1e064579/86a1a20092038c0b?lnk=gst&q=%22infinite+subsets%22+surjection+#
(Tony Orlow, Albrecht and others)
http://groups.google.com/group/sci.math/browse_frm/thread/dfe0193302960aef/c76f95b8612fdbdb?lnk=gst&q=%22infinite+subsets%22+surjection+#
(I _think_ the OP in this thread has the same confusion as you about
"2^n")
HEre's the Google search I used: there must be hundred's of threads
full of the same confusion:
http://groups.google.com/group/sci.math/search?group=sci.math&q=%22infinite+subsets%22+surjection+&qt_g=Search+this+group
HTH
Brian Chandler
TOT:
> > > (I've heard they are gonna patent Fibonacci... shouldn't the flo
No need to get worked up about this. The US Patent Office is staffed
with people with *gasp* worse understanding of mathematics than most
cranks. They will issue a patent on just about anything: famous
examples include "entertaining a cat", "making a list", indeed
"writing a program" basically. This is a threat mostly to US business,
but nothing to do with the real world.
ju...@diegidio.name
> ju...@diegidio.name wrote:
>
> I compliment you on starting a new thread, though given the enormity
> of the task you have set yourself, it may take at least another 8000
> posts.
It seems it was not enough to open a new thread, and that you guys do
mean to make it an 8000+ thing at all costs.
Now, this is ridiculous.
Your feedback has been taken into account, and a NEW PROOF has been
submitted, should anyone have missed that. (This is like the Kafka's
Castle, but *I* am Kafka.) So it is COMPLETELY UNUSEFUL THAT YOU KEEP
CORRECTING WHAT I HAVE ALREADY AKNOWLEDGED WAS INCORRECT.
Let's move on. Now I have bounded P(N)!!!!! Go read the argument. At
least, AFAICT, it diserves a mention to the infamous medal for "some
*new* inane rubbish", and not just this reiterated and reiterated
bubbling. LET'S MOVE ON!
AND TO ALL THE MORONS AROUND:
READ THE THREADS BEFORE ANSWERING, BECAUSE I'M GONNA KICK YOU BADLY IN
THE ASS OTHERWISE!
And now, please, just goodnight.
-LV
P.S. And, for your information, I do even know my combinatorics. I'll
send you my detailed list of skill should any of you guys really need.
I have also a master digree from the actors studio, should you wander.
And another one coming from Bruce Lee in person.
Brian Chandler
> Second attempt. I'll try two equivalent formulations, in the hope
> that, together, they will complement the eventual ambiguities.
Hmm. So this is the new version? Did you take the trouble to read the
proof I gave you that there is no bijection from N to P(N) (where N is
the set of all naturals, or actually any other set)? Do you understand
it? Do you see any flaw in it? (There is a tiny flaw: I forgot to add
the empty set as a special case.)
You do realise that if there is no significant flaw in my proof, and
your proof is also correct, that pretty much all of modern mathematics
crumbles to dust? So you have set yourself a pretty big task, no?
> ----- 1st formulation
>
> (i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
>
> Let's simply call P(A) the set on the RHS of (i), keeping in mind that
> the argument is going to hold due to this bijection and the
> transitivity of the '~' relation.
>
> (ii) P(A) c= N, AneN
>
> (iii) P(N) is infinite (trivial)
>
> We pass from P(A) to P(N) by transfinite induction, then:
>
> (iv) P(N) ~ N, from (ii) and (iii)
>
> QDE. (1)
> ----- 2nd formulation
>
> Same comments as above.
>
> (i) card(P(A)) = 2^n, AneN
>
> (ii) card(P(A)) <= card(N), AneN
> (iii) card(P(N)) >= oo (meaning: 'oo' as lower bound)
>
> (iv) card(P(N)) = card(N)
>
> QDE. (2)
>
> -----
>
> Hmm, correct?
Hmm, no.
> The basic idea behind this argument is that P(A) is
> always a subset of N in the finite case. Then, by transfinite
> induction, ....
Er no. What you are using appears to be "induction-in-the-infinite-
case", a mechanism much loved by cranks, but not by anyone else,
because it is invalid. Check "transfinite induction" in the Wikipedia
article, and see if you can see the difference. Yours is just: "look
p(x) is true for finite x, so - wave hands - must be true for infinite
x as well".
> status: OPEN
> revision: 1
Right, 7998 to go. Do tell me: do you actually think you _have_ this
proof (effectively that mathematics is inconsistent)? Or are you
merely sure you'll find one in a day or two if you look around?
Brian Chandler
David C. Ullrich
>Second attempt. I'll try two equivalent formulations, in the hope
>that, together, they will complement the eventual ambiguities.
>
>----- 1st formulation
>
>(i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
>
>Let's simply call P(A) the set on the RHS of (i),
The slight problem with this is that the right side
of (i) simply is _not_ P(A). (Nor is it equinumerous
with P(A), in case that's what you had in mind.)
>keeping in mind that
>the argument is going to hold due to this bijection and the
>transitivity of the '~' relation.
>
>(ii) P(A) c= N, AneN
>
>(iii) P(N) is infinite (trivial)
>
>We pass from P(A) to P(N) by transfinite induction, then:
>
>(iv) P(N) ~ N, from (ii) and (iii)
>
>QDE. (1)
>
>----- 2nd formulation
>
>Same comments as above.
>
>(i) card(P(A)) = 2^n, AneN
>
>(ii) card(P(A)) <= card(N), AneN
>(iii) card(P(N)) >= oo (meaning: 'oo' as lower bound)
>
>(iv) card(P(N)) = card(N)
>
>QDE. (2)
>
>-----
>
>Hmm, correct? The basic idea behind this argument is that P(A) is
>always a subset of N in the finite case. Then, by transfinite
>induction, while P(N) becomes infinite, its cardinality's upper bound
>remains N's cardinality. (These informal expositions get me into more
>troubles than they help, but I have to run the risk: no pain, no
>gain!)
This informal exposition is simply _wrong_. Yes, P(A) is
countable if A is finite - if you want to apply some bijection,
pretend it's the identity, and say that P(A) is a subset of
N in the finite case that's fine. Or at least that's not the
big problem.
The problem is that the bound on the size of the power
set simply does _not_ remain valid when you pass
from the finite case to the infinite case.
In more precise notation: Say
A_n = {1, 2, ... n}.
Then P(A_n) has cardinality 2^n. And N is
in some sense the limit of A_n as n tends to infinity.
But it simply does not follow that card(N) is the
limit of card(A_n). That's just not true. Not every
function is continuous - P is _discontinuous_.
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.)
hagman
> card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> here and there).
>
> --- Preliminaries:
>
> 'P(A)' stands for the "powerset" of set 'A'.
>
> '~' stands for the existence of a bijection, which will be, in this
> context, equivalent to stating that two sets are equinumerous.
> Infinity of the set 'N' (the set of natural/ordinal numbers) is taken
> to be defined as usual, by the existence of bijections between N and
> its (infinite) subsets. No other kind of "infinities" are going to be
> assumed: the only tool I have used is transfinite induction over the
> naturals/ordinals, with "omega" as the limit ordinal.
>
> 'oo' stands for "omega".
>
> --- Finite case:
>
> card(A) = n <=>def
> A ~ { 0, 1, 2, .., n-1 }
>
> card(P(A)) = 2^n <=>
> P(A) ~ { 1, 2, 4, ..., 2^n-1 }
>
> --- Infinite case (transf. induction, notation a bit fuzzy here):
>
> (1) n=oo => card(A) = card(N) = oo <=>def
> A ~ N := { 0, 1, 2, ..., n }, n->oo
What is this supposed to mean?
What kind of limit process do you have in mind?
Possibly
lim_{n->oo} A_n = U{A_n | n in N} provided A_n subset A_{n+1} for all
n ?
What properties are preserved ("continuous") with respect to your
notion of limit?
How about the property "is not equinumerous to a proper subset of
itself"?
How about "is not equinumerous to its poewerset"?
How about "is equinumerous with a subset of N"?
>
> (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <=>
> P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo
Is the Powerset of the limit really the limit of the powersets?
José Carlos Santos
>> Second attempt. I'll try two equivalent formulations, in the hope
>> that, together, they will complement the eventual ambiguities.
>>
>> ----- 1st formulation
>>
>> (i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
>>
>> Let's simply call P(A) the set on the RHS of (i),
>
> The slight problem with this is that the right side
> of (i) simply is _not_ P(A). (Nor is it equinumerous
> with P(A), in case that's what you had in mind.)
Yes, that is what he had in mind. Read his first post.
Brian Chandler
Um, what _would_ P(A) be, anyway?? A seems to be a free variable,
unless there's something I've missed?
Brian Chandler
ju...@diegidio.name
> ju...@diegidio.name wrote:
> > Second attempt. I'll try two equivalent formulations, in the hope
> > that, together, they will complement the eventual ambiguities.
>
> Hmm. So this is the new version? Did you take the trouble to read the
> proof I gave you that there is no bijection from N to P(N) (where N is
> the set of all naturals, or actually any other set)? Do you understand
> it? Do you see any flaw in it? (There is a tiny flaw: I forgot to add
> the empty set as a special case.)
No, I didn't get the flaw, but I am quite aware (am I wrong?) you have
given a formulation of Cantor's Theorem, as well as I am aware that
the fact that there is not a bijection P(N)~N is the salt of most
arguments around this topic. But *that* is the very culprit: as I'd
put it, you keep proving Cantor with Cantor. So, I have taken the
beast the other way 'round...
> You do realise that if there is no significant flaw in my proof, and
> your proof is also correct, that pretty much all of modern mathematics
> crumbles to dust? So you have set yourself a pretty big task, no?
I do realize (so to say) that there is a flaw in *your* argument, that
is, in the standard way any objections (the cranks!) get treated. My
intuition is that you cannot demonstrate that such an injection does
not exist, unless you give for granted what you had to demostrate in
the very first place. Then, I won't claim this and my above argument
entail any flow in Cantor's Theorem and related themselves, that is
simply out of the question at the moment, although might be an obvious
next step.
> > The basic idea behind this argument is that P(A) is
> > always a subset of N in the finite case. Then, by transfinite
> > induction, ....
>
> Er no.
Hmm, you are arguing against the informal statement, and have ignored
the formal one... that brings secure loss for *me* in any case.
(Thinking out loud.)
> What you are using appears to be "induction-in-the-infinite-
> case", a mechanism much loved by cranks, but not by anyone else,
> because it is invalid. Check "transfinite induction" in the Wikipedia
> article, and see if you can see the difference. Yours is just: "look
> p(x) is true for finite x, so - wave hands - must be true for infinite
> x as well".
No, what I am using is what I am using, or at least what I do believe
i am using, and although I might be using it badly and/or in the wrong
context, that's really what I meant to be using: trasfinite induction.
That is, AFAICT, needed to show that the property:
card(P(A)) <= card(N)
indeed holds for every n in N, "up to infinity included". And so we go
from an upper bound on the cardinality of P(A) to the *same* upper
bound for P(N). By transfinite induction rather than simple induction
just so that I can say "omega" with no fear, and nothing is left.
Now, *that* is far from rigorous and of course you might be right and
there might be the very flaw in my construction, but it's not through
a critic on the informal or the crank stuff and counter-stuff that I
am going to buy it (so to say).
> > status: OPEN
> > revision: 1
>
> Right, 7998 to go. Do tell me: do you actually think you _have_ this
> proof (effectively that mathematics is inconsistent)? Or are you
> merely sure you'll find one in a day or two if you look around?
I am not going to show that mathematics is inconsistent, I do not
believe so. I am going to show that all mathematicians are liers. ;)
Seriously: thanks again for the feedback.
-LV
>
> Brian Chandler
ju...@diegidio.name
> >----- 1st formulation
>
> >(i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
>
> >Let's simply call P(A) the set on the RHS of (i),
>
> The slight problem with this is that the right side
> of (i) simply is _not_ P(A). (Nor is it equinumerous
> with P(A), in case that's what you had in mind.)
I will be reading all the messages, I am not trying to start any
debate at once. Just I don't understand that remark of yours.
The two sets are not supposed to be the same set: that is a bijection
and it trivially holds, it just comes from the definitions:
card(A) = n <=> EXISTS( A ~ { 0, 1, 2, ..., n-1 } ), AneN
Then, it's a shortcut to be writing, say:
P(A) c= N
rather than:
( 0, 1, 2, ..., (2^n)-1 } c= N
-LV
José Carlos Santos
>>> ----- 1st formulation
>>> (i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
>>> Let's simply call P(A) the set on the RHS of (i),
>> The slight problem with this is that the right side
>> of (i) simply is _not_ P(A). (Nor is it equinumerous
>> with P(A), in case that's what you had in mind.)
>
> I will be reading all the messages, I am not trying to start any
> debate at once. Just I don't understand that remark of yours.
>
> The two sets are not supposed to be the same set: that is a bijection
> and it trivially holds, it just comes from the definitions:
>
> card(A) = n <=> EXISTS( A ~ { 0, 1, 2, ..., n-1 } ), AneN
And what does *this* mean, exactly?
> Then, it's a shortcut to be writing, say:
>
> P(A) c= N
Is "c=" the same thing as "~"?
> rather than:
>
> ( 0, 1, 2, ..., (2^n)-1 } c= N
If it is, then this is trivially false.
José Carlos Santos
>>>> Second attempt. I'll try two equivalent formulations, in the hope
>>>> that, together, they will complement the eventual ambiguities.
>>>>
>>>> ----- 1st formulation
>>>>
>>>> (i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
>>>>
>>>> Let's simply call P(A) the set on the RHS of (i),
>>> The slight problem with this is that the right side
>>> of (i) simply is _not_ P(A). (Nor is it equinumerous
>>> with P(A), in case that's what you had in mind.)
>> Yes, that is what he had in mind. Read his first post.
>
> Um, what _would_ P(A) be, anyway?? A seems to be a free variable,
> unless there's something I've missed?
I was just saying that the OP had already said in his first post that
"A~B" is equivalent to "A and B are equinumerous".
ju...@diegidio.name
> On 27-07-2008 16:42, ju...@diegidio.name wrote:
>
> >>> ----- 1st formulation
> >>> (i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
> >>> Let's simply call P(A) the set on the RHS of (i),
> >> The slight problem with this is that the right side
> >> of (i) simply is _not_ P(A). (Nor is it equinumerous
> >> with P(A), in case that's what you had in mind.)
>
> > I will be reading all the messages, I am not trying to start any
> > debate at once. Just I don't understand that remark of yours.
>
> > The two sets are not supposed to be the same set: that is a bijection
> > and it trivially holds, it just comes from the definitions:
>
> > card(A) = n <=> EXISTS( A ~ { 0, 1, 2, ..., n-1 } ), AneN
>
> And what does *this* mean, exactly?
"card(A) = n <=>def there exists a bijection between the set A and the
set of the first n ordinals". I could give a link to Wikipedia for
something already better...
> > Then, it's a shortcut to be writing, say:
>
> > P(A) c= N
>
> Is "c=" the same thing as "~"?
'c=' stands for set inclusion: "for every n in N: P(A) is an
(improper) subset of N."
'~' stands for the existence of a bijection, or otherwise for
equinumerosity.
>
> > rather than:
>
> > ( 0, 1, 2, ..., (2^n)-1 } c= N
>
> If it is, then this is trivially false.
But it isn't, trivially.
-LV
hagman
> On 27 Jul, 13:04, Brian Chandler <imaginator...@despammed.com> wrote:
>
> > ju...@diegidio.name wrote:
> > > Second attempt. I'll try two equivalent formulations, in the hope
> > > that, together, they will complement the eventual ambiguities.
>
> > Hmm. So this is the new version? Did you take the trouble to read the
> > proof I gave you that there is no bijection from N to P(N) (where N is
> > the set of all naturals, or actually any other set)? Do you understand
> > it? Do you see any flaw in it? (There is a tiny flaw: I forgot to add
> > the empty set as a special case.)
>
> No, I didn't get the flaw, but I am quite aware (am I wrong?) you have
> given a formulation of Cantor's Theorem, as well as I am aware that
> the fact that there is not a bijection P(N)~N is the salt of most
> arguments around this topic. But *that* is the very culprit: as I'd
> put it, you keep proving Cantor with Cantor. So, I have taken the
> beast the other way 'round...
No.
Actually, one might say that the proof of Cantor starts with assuming
~Cantor.
That is: The assumption of the existence of a surjective map A -> P(A)
leads to a contradiction.
If you think, making this assumption is not valid, please not that you
think
that you have actually proved that there is such a surjective map
A -> P(A) for some A (namely for A=N and as the surjective map we may
take the bijection which is claimed to exist according
to your "theorem" N ~ P(N)).
Thus in fact *using* this *proven*(?) to exist *bijective* mapping f:N-
>P(N)
we find a subset M := { x e N | ~(x e f(x)) } c N such that there is
no
n e N with f(n) = M (for n e M would imply ~(n e f(n)=M) and vice
versa).
Isn't that funny? A bijection that fails to be surjective?
>
> > You do realise that if there is no significant flaw in my proof, and
> > your proof is also correct, that pretty much all of modern mathematics
> > crumbles to dust? So you have set yourself a pretty big task, no?
>
> I do realize (so to say) that there is a flaw in *your* argument, that
> is, in the standard way any objections (the cranks!) get treated. My
> intuition is that you cannot demonstrate that such an injection does
> not exist, unless you give for granted what you had to demostrate in
> the very first place.
See above.
Or point out a single wrong step in the proof.
> Then, I won't claim this and my above argument
> entail any flow in Cantor's Theorem and related themselves, that is
> simply out of the question at the moment, although might be an obvious
> next step.
>
> > > The basic idea behind this argument is that P(A) is
> > > always a subset of N in the finite case. Then, by transfinite
> > > induction, ....
>
> > Er no.
>
> Hmm, you are arguing against the informal statement, and have ignored
> the formal one... that brings secure loss for *me* in any case.
> (Thinking out loud.)
It is just that you did not make proper use of transfinite induction.
Remember that (in the case of ordinals) transfinite induction
goes like this:
You prove for some property Q
(*) "If x is an ordinal and Q(y) is true for all ordinals with y<x,
then Q(x) is true"
and are then allowed to conclude that
(**) "For all ordinals x, Q(x) is true"
What is the property Q you wor with?
Where is your proof of (*) for that property?
What is (**) in your case anyway?
>
> > What you are using appears to be "induction-in-the-infinite-
> > case", a mechanism much loved by cranks, but not by anyone else,
> > because it is invalid. Check "transfinite induction" in the Wikipedia
> > article, and see if you can see the difference. Yours is just: "look
> > p(x) is true for finite x, so - wave hands - must be true for infinite
> > x as well".
>
> No, what I am using is what I am using, or at least what I do believe
> i am using, and although I might be using it badly and/or in the wrong
> context, that's really what I meant to be using: trasfinite induction.
> That is, AFAICT, needed to show that the property:
>
> card(P(A)) <= card(N)
>
> indeed holds for every n in N, "up to infinity included".
Go ahead, see above.
> And so we go
> from an upper bound on the cardinality of P(A) to the *same* upper
> bound for P(N). By transfinite induction rather than simple induction
> just so that I can say "omega" with no fear, and nothing is left.
>
> Now, *that* is far from rigorous and of course you might be right and
> there might be the very flaw in my construction, but it's not through
> a critic on the informal or the crank stuff and counter-stuff that I
> am going to buy it (so to say).
Of course one may start to develop some ideas in an informal way
and hope that it is possible to somehow find a proof of these ideas.
But still hoping so in the face of a well-known and simple proof of
the exact opposite is at least bold.
José Carlos Santos
>>>>> ----- 1st formulation
>>>>> (i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
>>>>> Let's simply call P(A) the set on the RHS of (i),
>>>> The slight problem with this is that the right side
>>>> of (i) simply is _not_ P(A). (Nor is it equinumerous
>>>> with P(A), in case that's what you had in mind.)
>>> I will be reading all the messages, I am not trying to start any
>>> debate at once. Just I don't understand that remark of yours.
>>> The two sets are not supposed to be the same set: that is a bijection
>>> and it trivially holds, it just comes from the definitions:
>>> card(A) = n <=> EXISTS( A ~ { 0, 1, 2, ..., n-1 } ), AneN
>> And what does *this* mean, exactly?
>
>
> "card(A) = n <=>def there exists a bijection between the set A and the
> set of the first n ordinals". I could give a link to Wikipedia for
> something already better...
I was talking about the expression "AneN".
>>> Then, it's a shortcut to be writing, say:
>>> P(A) c= N
>> Is "c=" the same thing as "~"?
>
>
> 'c=' stands for set inclusion: "for every n in N: P(A) is an
> (improper) subset of N."
>
> '~' stands for the existence of a bijection, or otherwise for
> equinumerosity.
>
>>> rather than:
>>> ( 0, 1, 2, ..., (2^n)-1 } c= N
>> If it is, then this is trivially false.
>
> But it isn't, trivially.
Indeed. Then the statement is trivially true.
ju...@diegidio.name
> On 27 Jul., 17:34, ju...@diegidio.name wrote:
> > On 27 Jul, 13:04, Brian Chandler <imaginator...@despammed.com> wrote:
> > No, I didn't get the flaw, but I am quite aware (am I wrong?) you have
> > given a formulation of Cantor's Theorem, as well as I am aware that
> > the fact that there is not a bijection P(N)~N is the salt of most
> > arguments around this topic. But *that* is the very culprit: as I'd
> > put it, you keep proving Cantor with Cantor. So, I have taken the
> > beast the other way 'round...
>
> No.
> Actually, one might say that the proof of Cantor starts with assuming
> ~Cantor.
You are technically right, I was complaining about some "old" stories
there. I'll try to be more careful, sorry.
Yes, I'll dare say the substance of Cantor's Theorem is quite clear to
me, at least as I have seen exposed around, including Wikipedia.
Should you tell me that, say, the treatement on Wikipedia is seriously
flawed, that would be big news to me, as that's what at the moment I
would be inclined to think, should I tell you.
> That is: The assumption of the existence of a surjective map A -> P(A)
> leads to a contradiction.
> If you think, making this assumption is not valid, please not that you
> think
> that you have actually proved that there is such a surjective map
> A -> P(A) for some A (namely for A=N and as the surjective map we may
> take the bijection which is claimed to exist according
> to your "theorem" N ~ P(N)).
But I prove that bijection, and so the equinumerosity, through a
*completely different way*. So to say, I have bounded the cardinality
of the powerset! I have not even attempted a confrontation with cantor-
like arguments and counter-arguments.
> See above.
> Or point out a single wrong step in the proof.
I don't need to. Mine is not a "counter-proof". I just prove a result
on its own. Then it's maybe up to "cantorians" (whoever they are) to
show that my simple inductive argument is flawed, and that their magic
sets exist.
> Of course one may start to develop some ideas in an informal way
> and hope that it is possible to somehow find a proof of these ideas.
> But still hoping so in the face of a well-known and simple proof of
> the exact opposite is at least bold.
That well known and simple proof has given rise to endless debates to
say the least. I am not that bold... or maybe I am. ;)
All that said, the rest of your post and your remarks on induction
I'll be meditating. Thank you.
-LV
ju...@diegidio.name
> On 27-07-2008 16:59, ju...@diegidio.name wrote:
> >>> card(A) = n <=> EXISTS( A ~ { 0, 1, 2, ..., n-1 } ), AneN
> >> And what does *this* mean, exactly?
>
> > "card(A) = n <=>def there exists a bijection between the set A and the
> > set of the first n ordinals". I could give a link to Wikipedia for
> > something already better...
>
> I was talking about the expression "AneN".
"For each n in N", though that one I supposed was the only one clear,
poor me... and you! :)
> >>> ( 0, 1, 2, ..., (2^n)-1 } c= N
> >> If it is, then this is trivially false.
>
> > But it isn't, trivially.
>
> Indeed. Then the statement is trivially true.
Indeed it is!
-LV
ju...@diegidio.name
> On 27 Jul, 18:01, José Carlos Santos <jcsan...@fc.up.pt> wrote:
>
> > On 27-07-2008 16:59, ju...@diegidio.name wrote:
> > >>> card(A) = n <=> EXISTS( A ~ { 0, 1, 2, ..., n-1 } ), AneN
> > >> And what does *this* mean, exactly?
>
> > > "card(A) = n <=>def there exists a bijection between the set A and the
> > > set of the first n ordinals". I could give a link to Wikipedia for
> > > something already better...
>
> > I was talking about the expression "AneN".
To be more specific: there I meant that that statement holds for any
finite A, where n is the cardinality of A.
-LV
Brian Chandler
> On 27 Jul, 13:04, Brian Chandler <imaginator...@despammed.com> wrote:
> > ju...@diegidio.name wrote:
> > > Second attempt. I'll try two equivalent formulations, in the hope
> > > that, together, they will complement the eventual ambiguities.
> >
> > Hmm. So this is the new version? Did you take the trouble to read the
> > proof I gave you that there is no bijection from N to P(N) (where N is
> > the set of all naturals, or actually any other set)? Do you understand
> > it? Do you see any flaw in it? (There is a tiny flaw: I forgot to add
> > the empty set as a special case.)
>
>
> No, I didn't get the flaw, but I am quite aware (am I wrong?) you have
> given a formulation of Cantor's Theorem, as well as I am aware that
> the fact that there is not a bijection P(N)~N is the salt of most
> arguments around this topic.
Oh dear, lots of problems here. Grammar first: I get to "...P(N)~N",
and you seen to have said "I understand there is no bijection P(N) to
N, and find no flaw in the proof". But then there's another main verb
and clause ("is the salt..."), and I don't understand the last bit.
It sounds a bit as though you prefer poetry. I gave a proof (not a
"formulation", in usual terminology) of what is apparently called
"Cantor's theorem" (a most unfortunate name, IMO), and though the
proof is "neat" (IMO, and of course I didn't invent it), in many ways
it's just a humdrum bit of mathematical argument. It really isn't
"salt" at all, but it is true, which means that most mathematically
inclined people don't spend much time looking for counterexamples.
> But *that* is the very culprit: as I'd
> put it, you keep proving Cantor with Cantor. So, I have taken the
> beast the other way 'round...
Skipping the poetic bits, I don't understand this at all. I gave a
proof. Obviously the proof does not use its own truth - for that would
not be a proof at all. How is this "proving Cantor with Cantor"? Is a
proof of Pythagoras's theorem "proving Pythagoras with Pythagoras"?
>
> > You do realise that if there is no significant flaw in my proof, and
> > your proof is also correct, that pretty much all of modern mathematics
> > crumbles to dust? So you have set yourself a pretty big task, no?
>
>
> I do realize (so to say) that there is a flaw in *your* argument, that
> is, in the standard way any objections (the cranks!) get treated.
Make a valid objection, and I (and many others will be *very*
interested). Spew out the usual crank nonsense, and what can I do but
yawn.
> My intuition is that you cannot demonstrate that such an injection does
> not exist, unless you give for granted what you had to demostrate in
> the very first place.
Look, I just gave you a proof. Either show some reason to suspect a
flaw in it, or shut up. No, that won't work, knock yourself out with
the usual stuff. I shan't bother to post it again.
> Then, I won't claim this and my above argument
> entail any flow in Cantor's Theorem and related themselves, that is
> simply out of the question at the moment, although might be an obvious
> next step.
>
>
> > > The basic idea behind this argument is that P(A) is
> > > always a subset of N in the finite case. Then, by transfinite
> > > induction, ....
> >
> > Er no.
>
>
> Hmm, you are arguing against the informal statement, and have ignored
> the formal one... that brings secure loss for *me* in any case.
> (Thinking out loud.)
>
>
> > What you are using appears to be "induction-in-the-infinite-
> > case", a mechanism much loved by cranks, but not by anyone else,
> > because it is invalid. �Check "transfinite induction" in the Wikipedia
> > article, and see if you can see the difference. Yours is just: "look
> > p(x) is true for finite x, so - wave hands - must be true for infinite
> > x as well".
>
>
> No, what I am using is what I am using, or at least what I do believe
> i am using, and although I might be using it badly and/or in the wrong
> context, that's really what I meant to be using: trasfinite induction.
Well, you are wrong. Read the Wikipedia description of transfinite
induction. (I did, the other day) It is induction extended to any set
of ordinals, in which ***having proved*** that if p(i) is true for
*any* ordinal i less than an ordinal j *it follows that* p(j) is true,
you can say that p(i) is true for any ordinal i. (Roughly)
You are trying to use it as follows:
Show that p(i) is true for any finite i. Assert that p(omega) must be
true. This is not what transfinite induction is.
Demonstrating that crankinduction is invalid happens about once a day
on sci.math. Let p(i) be the property "i is finite". Observe that the
fact that p(i) is true for any finite i does not imply that p(i) is
true for any infinite i, and of course it isn't. There is a standard
slew of crank nonsense that follows from this point, but I'll let you
do that yourself.
> That is, AFAICT, needed to show that the property:
>
> card(P(A)) <= card(N)
>
> indeed holds for every n in N, "up to infinity included".
What are A and N, here? Your expression is incoherent, because it has
too many free variables.
<snip rest of confusion>
> I am not going to show that mathematics is inconsistent, I do not
> believe so. I am going to show that all mathematicians are liers. ;)
Another curious statement. Mathematics has no hidden information, so
claiming that someone making a mathematical statement is a "liar" is
very strange indeed. Of course I could lie about what I proved (or
understood, or wrote in an examination) but when I write something
down - like my proof that card(A) /= card(P(A)) for any A - there may
be errors, there may be confusion, but I really don't see how there
can be lies.
However, carry on.
Brian Chandler
Brian Chandler
> On 27 Jul, 17:48, hagman <goo...@von-eitzen.de> wrote:
> > On 27 Jul., 17:34, ju...@diegidio.name wrote:
> > > On 27 Jul, 13:04, Brian Chandler <imaginator...@despammed.com> wrote:
>
> > > No, I didn't get the flaw, but I am quite aware (am I wrong?) you have
> > > given a formulation of Cantor's Theorem, as well as I am aware that
> > > the fact that there is not a bijection P(N)~N is the salt of most
> > > arguments around this topic. But *that* is the very culprit: as I'd
> > > put it, you keep proving Cantor with Cantor. So, I have taken the
> > > beast the other way 'round...
> >
> > No.
> > Actually, one might say that the proof of Cantor starts with assuming
> > ~Cantor.
>
>
> You are technically right,
No he isn't. Here is the proof again:
Kindly indicate where I "assume ~Cantor". In the Remark I explicitly
pointed out that it is *not* a proof by contradiction.
Brian Chandler
Albrecht
ju...@diegidio.name schrieb:
Without reading your attempt of proof in detail, the problem is and
stays always the fact that set theoretics claim to have a big bunch of
elements in a powerset of an infinite set which are not describable.
The gag is now this, that they wonder about the fact, that they can
not describe this elements (e.g. in a list) and claim therefore that
there must be more elements as is describable. Now, since only a
describable amount of elements is describable, and this amount is
denumerable, they claim that the powerset of an infinite set must be
nondenumerable.
Something like this.
Good wishes in trying to understand all this stuff and in, maybe,
finding the "gaps" in this considerations.
Albrecht S. Storz
Virgil
<017dce3e-1625-40ca...@2g2000hsn.googlegroups.com>,
ju...@diegidio.name wrote:
> On 27 Jul, 18:39, ju...@diegidio.name wrote:
> > On 27 Jul, 18:01, José Carlos Santos <jcsan...@fc.up.pt> wrote:
> >
> > > On 27-07-2008 16:59, ju...@diegidio.name wrote:
> > > >>> card(A) = n <=> EXISTS( A ~ { 0, 1, 2, ..., n-1 } ), AneN
> > > >> And what does *this* mean, exactly?
> >
> > > > "card(A) = n <=>def there exists a bijection between the set A and the
> > > > set of the first n ordinals". I could give a link to Wikipedia for
> > > > something already better...
> >
> > > I was talking about the expression "AneN".
>
>
> To be more specific: there I meant that that statement holds for any
> finite A, where n is the cardinality of A.
When you use A far a set and also as a quantifier in the same sentence,
you unnecessarily increase ambiguity.
And since for any finite set, S, Card(S) < Card(P(S)), why do you think
that this suddenly inequality fails for infinite sets?
ju...@diegidio.name
> > card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> > here and there).
<snip>
> Without reading your attempt of proof in detail
Please, let me anyway point out that you are quoting the wrong
argument. The correct one is here:
http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1
> the problem is and
> stays always the fact that set theoretics claim to have a big bunch of
> elements in a powerset of an infinite set which are not describable.
You are refering to uncountable numbers and similar concepts, right?
> The gag is now this, that they wonder about the fact, that they can
> not describe this elements (e.g. in a list) and claim therefore that
> there must be more elements as is describable. Now, since only a
> describable amount of elements is describable, and this amount is
> denumerable, they claim that the powerset of an infinite set must be
> nondenumerable.
>
> Something like this.
I won't claim I know what "set theoretics claims", but yes, I can see
the form of this paralogism: "proving Cantor with Cantor" is just a
slogan.
> Good wishes in trying to understand all this stuff and in, maybe,
> finding the "gaps" in this considerations.
It really seems to boil down to the same few primitive notions we have
been accustomed to for at least the last few thousend years.
I see no problem of understanding here, the problem nowadays seems to
be lowering the level of noise. And there is nothing trivial in this.
If anything, my effort takes this issue as a priority. Although nobody
is perfect...
Cheers,
Julio
>
> Albrecht S. Storz
lwa...@lausd.net
> card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> here and there).
Just by glancing at the rest of this post, I can already tell
that this will be another one of those arguments that gives
a bijection between N and the set of all _finite_ subsets of N.
There are several ways to come up with a rigorous set
theory in which card(P(N)) = card(N). The simplest is
to deny the Axiom of Infinity and allow there to be
proper classes. Then N would be a proper class --
indeed, it would be exactly the Burali-Forti class of all
ordinals -- and P(N) would also be a proper class. The
elements of P(N) would be (finite) sets, since classes
can only have _sets_ as elements -- and in the intended
model of this theory, all sets are finite -- and the OP's
proof would be valid in this theory:
ZF-Infinity+"all sets are finite"+"there exists infinite classes"
There are other ways to make card(P(N)) = card(N). In
general, one can either make card(N) larger by adding
nonstandard natural numbers, or else make card(P(N))
smaller by only allowing for constructible (or computable,
or something similar) subsets.
I am also considering a set theory influenced by some of
the posting of a few other so-called "cranks" to come up
with another theory in which card(P(N)) = card(N).
ju...@diegidio.name
> On Jul 26, 2:52 pm, ju...@diegidio.name wrote:
>
> > card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> > here and there).
>
> Just by glancing at the rest of this post, I can already tell
> that this will be another one of those arguments that gives
> a bijection between N and the set of all _finite_ subsets of N.
I don't think so but I am not sure, I guess I have just tried another
way.
Sorry, that fact is that's the wrong one, we are at rev. 1 already:
http://groups.google.co.uk/group/sci.math/msg/7013e48b205a9ee1
I guess I should have changed the subject, sorry again. Next time!
In any case, your post is very interesting and quite on the spot. You
must be magic.
I'll say nothing more as I am still "meditating" the various
contributions.
-LV
David C. Ullrich
Whatever. You ignored the second half of my post,
by the way, which explains the actual problem in
your thinking.
Another way to explain the problem: Say
A_n = {1,2,...n}. Then P(A_n) does not
converge to P(N) as n -> infinity. The reason
is that although N is the union of the A_n,
P(N) is not the union of P(A_n). For example,
let S be the set of even integers. Then S is a
subset of N but S is not a subset of A_n for
any n.
In very bad more or less incomprehensible
notation you've given a proof that the set of
_finite_ subsets of N is countable. This is a
well known fact.
>-LV
hagman
But you did *not* give such a proof.
(Unless "through a completely different way" means "by summoning
transfinite induction as a buzz-word").
What set theory is yuor proof based upon?
Is there in your theory the concept of power set as the set
of all subsets?
Is there an axiom (scheme) of separation?
Can mappings from one set to another be viewd as sets?
>
> > See above.
> > Or point out a single wrong step in the proof.
>
> I don't need to. Mine is not a "counter-proof". I just prove a result
> on its own.
... which happens to be the negation of a well-established result.
> Then it's maybe up to "cantorians" (whoever they are) to
> show that my simple inductive argument is flawed, and that their magic
> sets exist.
You did not give any inductive argument.
You merely claimed that something be true by transfinite induction.
However, I see no proof or even mentioning of a statement of the form
If x is an ordinal and P(y) is true for all ordinals y < x,
then P(x) is true.
ju...@diegidio.name
> >I will be reading all the messages, I am not trying to start any
> >debate at once. Just I don't understand that remark of yours.
> Whatever. You ignored the second half of my post,
> by the way, which explains the actual problem in
> your thinking.
I didn't, please read more carefully.
I have been "meditating" the variuos feedbacks. It seems I shall make
the inductive steps explicit... I mean, I'll try.
-LV
ju...@diegidio.name
> On 26 Jul., 23:52, ju...@diegidio.name wrote:
> > card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> > here and there).
> > (1) n=oo => card(A) = card(N) = oo <=>def
> > A ~ N := { 0, 1, 2, ..., n }, n->oo
>
> What is this supposed to mean?
You are still arguing with the wrong argument.
That's nothing, just fuzzy notation.
-LV
ju...@diegidio.name
Sorry, some confusion.
-LV
MoeBlee
> card(P(N)) = card(N), an elementary proof
> the only tool I have used is transfinite induction over the
> naturals/ordinals, with "omega" as the limit ordinal.
Unless you show me a proven schema otherwise, the only transfinite
induction I know of is over ALL the ordinals, and omega is not the
only limit ordinal.
> 'oo' stands for "omega".
>
> --- Finite case:
Already, this not transfinite induction.
> card(A) = n <=>def
> A ~ { 0, 1, 2, .., n-1 }
>
> card(P(A)) = 2^n <=>
> P(A) ~ { 1, 2, 4, ..., 2^n-1 }
So your proposition doesn't hold in the finite case.
> --- Infinite case (transf. induction, notation a bit fuzzy here):
This is not transfinite induction; or please tell me exactly what
proven transfinite induction schema you are using.
> (1) n=oo => card(A) = card(N) = oo <=>def
> A ~ N := { 0, 1, 2, ..., n }, n->oo
Maybe what you mean is this:
For all natural numbers n, if A~n, then
card(A) = card(N) iff A~N~{1 2 ... n}.
> (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <=>
> P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo
Maybe what you mean is this:
For all natural numbers n, if A~n, then
card(PA) = card(PA) iff PA~PN~{1 2 ... 2^n}.
> --- Theorem:
>
> card(P(N)) = card(N)
Which contradicts, even if in the case where N is finite:
It is not the case that N~PN.
> --- Proof:
>
> (i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card'
>
> (ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~'
>
> (iii) A bijection 2^N ~ N trivially exists:
>
> 2^N ~ N
> --------
> 1 ~ 0
> 2 ~ 1
> 4 ~ 2
WRONG. You skipped, for example, 3 being a member of 2^n.
Look, try it, for example, where n=2. Then 2^n=4. There is no
bijection between 4 and 2.
MoeBlee
ju...@diegidio.name
That argument is indeed badly flawed.
Not that this answers all your objections, but anyway we are at
revision 1 already, and with something quite different than the usual
approach: I have tried to give an upper bound to the cardinality of
the powerset!
Here it is:
http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1?hl=en
Given all the feedback so far, later during the day I will try to
provide still another formulation where the inductive steps are
explicit. But should you have any further comments in the meantime,
please feel free to proceed.
Again, sorry.
-LV
MoeBlee
> http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1?hl=en
I looked at that. It's not a proof that a set is 1-1 with its power
set.
If you are sincere in wanting to communicate mathematics and give
correct mathematical proofs, then you'd study at least enough of the
subject so that you actually UNDERSTOOD what things such as
'transfinite induction' are. Right now, you are utterly groping in the
dark. Why not read a book on predicate logic and one on set theory?
MoeBlee
Mike Kelly
> the argument is going to hold due to this bijection and the
> transitivity of the '~' relation.
>
> (ii) P(A) is a subset of N, AneN
>
> (iii) P(N) is infinite (trivial)
>
> We pass from P(A) to P(N) by transfinite induction, then:
How does that step work, exactly?
> (iv) P(N) ~ N, from (ii) and (iii)
ju...@diegidio.name
> On Jul 28, 10:34 am, ju...@diegidio.name wrote:
>
> >http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1?hl=en
>
> I looked at that. It's not a proof that a set is 1-1 with its power
> set.
Indeed, but I'd be missing your point.
> If you are sincere in wanting to communicate mathematics and give
> correct mathematical proofs, then you'd study at least enough of the
> subject so that you actually UNDERSTOOD what things such as
> 'transfinite induction' are. Right now, you are utterly groping in the
> dark. Why not read a book on predicate logic and one on set theory?
I am going to write the inductive steps explicitly later during the
day, and so I will find my mistake, if any.
For the rest, I wander how you can be so sure that I have no idea
about what I mention - say, about transf. induction.
There is only one thing I am really groping in the dark for, and that
is learning how to communicate with *you*.
-LV
> MoeBlee
ju...@diegidio.name
> On 28 Jul, 18:47, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 28, 10:34 am, ju...@diegidio.name wrote:
>
> > >http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1?hl=en
>
> > I looked at that. It's not a proof that a set is 1-1 with its power
> > set.
>
> Indeed, but I'd be missing your point.
Sorry, that read: I might be missing your point.
-LV
Dave L. Renfro
> If you are sincere in wanting to communicate mathematics
> and give correct mathematical proofs, then you'd study at
> least enough of the subject so that you actually UNDERSTOOD
> what things such as 'transfinite induction' are. Right now,
> you are utterly groping in the dark. Why not read a book
> on predicate logic and one on set theory?
This is something I've wondered about for years (and have
posted about several times before), namely why do some
people, who appear to be very interested in a subject,
not just get a few books on the subject and read up on
the subject? I know this is something that takes more
than a few weeks, even a few months depending on the
book(s), but for many long-time posters this is far less
effort than they've devoted to posting and arguing on
sci.math over a several year period. I can only conclude
that they really *aren't* interested in the subject,
but instead are just interested in arguing with people
on usenet. That's one reason I've mostly stayed out of
these kinds of threads. Give me some indication that you're
truly interested in a topic (calvin seems as if he might
be such a person) and, if I know anything about it, I'll
bend over backwards to do what I can. But I'm just not
interested in trying to figure out what someone is saying
in order to help when they're not willing to do anything.
Exceptions occur, of course, such as when I have a lot
of free time and feel like "grading/correcting someone's
post" or when I think something I could say would have
wider interest than just to the person I'm responding to.
Dave L. Renfro
MoeBlee
> On 28 Jul, 18:47, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 28, 10:34 am, ju...@diegidio.name wrote:
>
> > >http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1?hl=en
>
> > I looked at that. It's not a proof that a set is 1-1 with its power
> > set.
>
> Indeed, but I'd be missing your point.
>
> > If you are sincere in wanting to communicate mathematics and give
> > correct mathematical proofs, then you'd study at least enough of the
> > subject so that you actually UNDERSTOOD what things such as
> > 'transfinite induction' are. Right now, you are utterly groping in the
> > dark. Why not read a book on predicate logic and one on set theory?
>
> I am going to write the inductive steps explicitly later during the
> day, and so I will find my mistake, if any.
>
> For the rest, I wander how you can be so sure that I have no idea
> about what I mention - say, about transf. induction.
I'm not absolutely sure, but from the way you've been trying to
invoke, it seems pretty clear that you don't understand that there are
particular schemata for transfinite induction that we prove are
correct and that, more importantly, you don't know these schemata.
Or, I'd be happy to see that you do understand it by your stating
exactly which transfinite induction schema you are using and putting
your argument correctly in that form.
>
> There is only one thing I am really groping in the dark for, and that
> is learning how to communicate with *you*.
No, other posters have mentioned that they find your notation and
arguments to be incomprehensible. If you wish to communicate with me
or just about ANYONE on these threads, then the best approach would be
to first learn the concepts and notation of the subject matter we're
discussing.
I've suggested how you can do that. And elsewhere you appealed for
help, so I offered that I'd help if you'd just begin at a certain
starting point - the very first point where you're confused,
uninformed, or ill-informed. So, I don't know why you won't at least
just say why you don't read at least an introductory textbook.
MoeBlee
MoeBlee
> On 28 Jul, 18:56, ju...@diegidio.name wrote:
>
> > On 28 Jul, 18:47, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > On Jul 28, 10:34 am, ju...@diegidio.name wrote:
>
> > > >http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1?hl=en
>
> > > I looked at that. It's not a proof that a set is 1-1 with its power
> > > set.
>
> > Indeed, but I'd be missing your point.
>
> Sorry, that read: I might be missing your point.
How can my point be more clear? Just as I said, what I find at your
link, you new formulation, is not a proof that a set is 1-1 with its
power set. It's not a proof that card(x) = card(Px).
MoeBlee
MoeBlee
what I find at your
> link, you new formulation, is not a proof that a set is 1-1 with its
> power set. It's not a proof that card(x) = card(Px).
Nor even in the specific case where x=N=the set of natural numbers.
MoeBlee
Chris Menzel
<ju...@diegidio.name> said:
> I apologise for not having changed the subject.
>
> That argument is indeed badly flawed.
>
> Not that this answers all your objections, but anyway we are at
> revision 1 already, and with something quite different than the usual
> approach: I have tried to give an upper bound to the cardinality of
> the powerset!
I suppose you mean a least upper bound. Let me save you some work.
That would be 2^n, where card(A) = n.
Albrecht
ju...@diegidio.name schrieb:
> On 27 Jul, 20:25, Albrecht <albst...@gmx.de> wrote:
> > ju...@diegidio.name schrieb:
>
> > > card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> > > here and there).
>
> <snip>
>
> > Without reading your attempt of proof in detail
>
> Please, let me anyway point out that you are quoting the wrong
> argument. The correct one is here:
>
> http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1
>
> > the problem is and
> > stays always the fact that set theoretics claim to have a big bunch of
> > elements in a powerset of an infinite set which are not describable.
>
> You are refering to uncountable numbers and similar concepts, right?
And uncountable sets of natural numbers.
The "uncountable part" of the whole of the elements of the powerset of
the naturals is the part which consists of nondescribable sets of
natural numbers.
Stay clear
Albrecht S. Storz
ju...@diegidio.name
> MoeBlee wrote:
> > If you are sincere in wanting to communicate mathematics
> > and give correct mathematical proofs, then you'd study at
> > least enough of the subject so that you actually UNDERSTOOD
> > what things such as 'transfinite induction' are. Right now,
> > you are utterly groping in the dark. Why not read a book
> > on predicate logic and one on set theory?
>
> This is something I've wondered about for years (and have
> posted about several times before), namely why do some
> people, who appear to be very interested in a subject,
> not just get a few books on the subject and read up on
> the subject?
Your interpretation, at least as applied to "my case", is simply off
mark.
I do have read *many* introductory and less introductory books on
these matters, and I keep doing. That is simply *not* enough to write
down arguments correctly in the language of a specific discipline such
as mathematics.
Now you can complain as much as you like, but the fact is: I do work,
I am neither a student nor a prof (no pun intended). That kind of
practice I have to do *here*, I have no other place to practice that:
no profs, no tutors, no collegues, no nothing, I have got *this* group
to practice and check my mathematics.
Books alone are not enough. I have spent 30+ years of my recent life
reading and studing books, not just mathematics. It is not enough to
be a pro: you need to complement with practice!
So, I am not idiot, nor I am clueless, or anything like that. Rather
it's your wandering about all this that is a bit clueless. With all
respect.
BTW, I even understand that there *are* lots of morons around, and
that I manage to be myself quite moronish at times: you guys need a
holiday.
-LV
Dave L. Renfro
> Books alone are not enough. I have spent 30+ years of my
> recent life reading and studing books, not just mathematics.
> It is not enough to be a pro: you need to complement
> with practice!
>
> So, I am not idiot, nor I am clueless, or anything like
> that. Rather it's your wandering about all this that is
> a bit clueless. With all respect.
I think few people who haven't done graduate work in
mathematics have really *read* any math books, and
this is something that takes quite a bit of practice,
which I've learned from my own teaching and from my
observation of others. Not from first hand experience,
since I never got the memo everyone else apparently gets
in middle school or high school that says you aren't
allowed to actually *read* your math texts, and certainly
not well ahead of the class. For instance, I never took
a single-variable or multi-variable calculus course, in high
school or in college, because I decided to read ahead while
in high school and later wound up getting credit-by-exam
for the 3-4 semester elementary calculus sequence.
I learned early in my high school years, after quite
a few false starts with texts too advanced for me at
the time, that you need to find a text at the appropriate
level. If it's too elementary and/or easy (these two things
are not always the same, by the way), then you're mostly
doing busy work. If it's too advanced and/or hard, then
progress will be very slow (if at all), painful, and
demoralizing. Of the two, however, too easy/elementary
is best, because you'll always find something you thought
you understood but didn't -- kind of like learning a subject
you thought you knew very well by teaching it or by tutoring
someone in it.
Recently, in another thread, I posted some undergraduate
level set theory texts:
Enderton, "Elements of Set Theory"
Hrbacek/Jech, "Introduction to Set Theory"
Halmos, "Naive Set Theory"
Moschovakis, "Notes on Set Theory"
Devlin, "The Joy of Sets"
Suppes, "Axiomatic Set Theory"
Vaught, "Set Theory: An Introduction"
Roitman, "Introduction to Modern Set Theory"
Shen/Verschagin, "Basic Set Theory"
Just/Weese, "Discovering Modern Set Theory" (Volume I)
For independent study, I'd most strongly recommend
Enderton's, Devlin's, and Just/Weese's texts (and perhaps
Suppes' text as well, although it's a little on the dry side).
However, these might be just a tad bit too advanced,
depending on your background. If this seems to be
the case, then I'd recommend beginning with one of
those sophomore/junior "transition to advance mathematics"
texts, such as:
http://www.amazon.com/dp/0534382142
http://www.amazon.com/dp/0201710900
http://www.amazon.com/dp/003098338X
Also, if you're especially interested in the fundational
aspects of set theory, you should work through some logic
texts.
At the most basic level, something that should be a quick
read but also be very helpful in getting you to the next
level without tutors and teachers, I recommend:
Lemmon, "Beginning Logic"
http://www.amazon.com/dp/0915144506
and
Mates, "Elementary Logic"
http://www.amazon.com/dp/019501491X
After these, I'd next recommend the following.
Some people might be able to begin with these
next two, but I included the above two books
so that you'll have a good beginning foundation
on which to build.
Boolos/Jeffrey, "Computability and Logic"
http://www.amazon.com/dp/0521389232
Enderton, "A Mathematical Introduction to Logic"
http://www.amazon.com/dp/0122384520
One of the problems I see in your posts (the few that
I've read, so there may be others as well, or these
may be over-represented in the posts I've read) is that
there's a lot of jargon being tossed around inappropriately.
For example, you brought up transfinite induction, but
this is something that is well beyond the issues you're
dealing with (at least, when you then invoke something
like n --> oo). I also see a lot of posters responding
to you at various levels, some with basic comments and
some with fairly advanced comments, but I would think
it'd be difficult to sort out what's basic and what isn't
if you don't already know a fair amount about the subject.
For example, things about bijections with the natural
numbers and such are basic (naturals are bijectively
equivalent with the rationals, but not with the reals),
while issues involving the axiom of choice or the ZF axioms
aren't basic.
The best use of your time (you did say you work; I happen
to work full-time, and not in academics either (as of the
last three years)), in my opinion, would be to start with
one of the above texts and, when you come to something you
don't understand, post exactly what the author says and
exactly what you think the problem is (or simply that it
seems incomprehensible to you). I think you'll get a lot
more people willing to go out of their way to help,
especially if you begin showing signs of clear progress
over the next few months.
Dave L. Renfro
Virgil
<2f718f76-674a-4fb0...@a1g2000hsb.googlegroups.com>,
Albrecht <albs...@gmx.de> wrote:
> ju...@diegidio.name schrieb:
> > On 27 Jul, 20:25, Albrecht <albst...@gmx.de> wrote:
> > > ju...@diegidio.name schrieb:
> >
> > > > card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> > > > here and there).
> >
> > <snip>
> >
> > > Without reading your attempt of proof in detail
> >
> > Please, let me anyway point out that you are quoting the wrong
> > argument. The correct one is here:
> >
> > http://groups.google.co.uk/group/sci.logic/msg/7013e48b205a9ee1
> >
> > > the problem is and
> > > stays always the fact that set theoretics claim to have a big bunch of
> > > elements in a powerset of an infinite set which are not describable.
> >
> > You are refering to uncountable numbers and similar concepts, right?
>
> And uncountable sets of natural numbers.
Do you mean an uncountable set of sets of natural numbers?
There cannot be an uncountable set of natural numbers.
MoeBlee
> On 28 Jul, 19:27, "Dave L. Renfro" <renfr...@cmich.edu> wrote:
>
> > MoeBlee wrote:
> > > If you are sincere in wanting to communicate mathematics
> > > and give correct mathematical proofs, then you'd study at
> > > least enough of the subject so that you actually UNDERSTOOD
> > > what things such as 'transfinite induction' are. Right now,
> > > you are utterly groping in the dark. Why not read a book
> > > on predicate logic and one on set theory?
>
> > This is something I've wondered about for years (and have
> > posted about several times before), namely why do some
> > people, who appear to be very interested in a subject,
> > not just get a few books on the subject and read up on
> > the subject?
>
> Your interpretation, at least as applied to "my case", is simply off
> mark.
>
> I do have read *many* introductory and less introductory books on
> these matters, and I keep doing.
Then please tell me what textbook on the predicate calculus you use,
so I can help you with the parts you've not understood.
> That is simply *not* enough to write
> down arguments correctly in the language of a specific discipline such
> as mathematics.
Sure, some practical experience helps. But you need the systematic
study too. Whatever reading you've done, either it was not systematic,
or it was but you've forgotten too much important material.
> Now you can complain as much as you like, but the fact is: I do work,
> I am neither a student nor a prof (no pun intended). That kind of
> practice I have to do *here*, I have no other place to practice that:
> no profs, no tutors, no collegues, no nothing, I have got *this* group
> to practice and check my mathematics.
Fine. And to get anything USEFUL out of people checking your work
here, you need to at least bring some coherent work to these people.
You're in a terrible confusion about some fundamentals. You need to
get those fundamentals taken care of BEFORE you start misspeaking all
over the place about more advanced topics such as transfinite
induction.
> Books alone are not enough. I have spent 30+ years of my recent life
> reading and studing books, not just mathematics. It is not enough to
> be a pro: you need to complement with practice!
Yes, but you AT LEAST need to PROPERLY do the reading.
> So, I am not idiot,
I didn't say you are.
> nor I am clueless, or anything like that.
On the fundamentals of set theory you are virtually clueless.
> Rather
> it's your wandering about all this that is a bit clueless. With all
> respect.
No, I don't want wandering about. I'd rather get STRAIGHT to the very
first point at which your misunderstandings begin.
But you're eluding that each time you post.
Since you asked for help, I offered to give help. The best help I can
give is for you to start by getting a good book on the predicate
calculus, and I would answer your questions or look at your exercise
answers, as time permits.
Otherwise, until you get solid on the basics of this subject, you're
wasting time and energy just posting a bunch of confusions for which
the proper response is to deal with them at their most basic level - a
response you continue to reject.
MoeBlee
Chris Menzel
<renf...@cmich.edu> said:
> MoeBlee wrote:
>
>> If you are sincere in wanting to communicate mathematics and give
>> correct mathematical proofs, then you'd study at least enough of the
>> subject so that you actually UNDERSTOOD what things such as
>> 'transfinite induction' are. Right now, you are utterly groping in
>> the dark. Why not read a book on predicate logic and one on set
>> theory?
>
> This is something I've wondered about for years (and have posted about
> several times before), namely why do some people, who appear to be
> very interested in a subject, not just get a few books on the subject
> and read up on the subject?
Equal parts hubris and ignorance, each in boundless supply. Due to
their ignorance, they think they are onto something Deep. Due to their
hubris, they think they are smart enough to figure it all out themselves
from scratch -- despite helpful warnings (obviously, from lesser minds)
that they are trying to do something impossible or, at the least,
extremely difficult.
Whatever the reason, it is completely bizarre.
Dave L. Renfro
> One of the problems I see in your posts (the few that
> I've read, so there may be others as well, or these
> may be over-represented in the posts I've read) is that
> there's a lot of jargon being tossed around inappropriately.
Less ambiguous (I hope) rewrite:
One of the problems I see in your posts (the few posts that
I've read, so there may be other problems as well, or these
problems may be over-represented in the posts I've read) is
that there's a lot of jargon being tossed around inappropriately.
Dave L. Renfro
ju...@diegidio.name
-------------------------------------------------
I shall only try to prove a preliminary property, from which it should
then be trivial to prove the thesis, so I will leave the final steps
out for now.
The proof below is based on transfinite induction (see Wikipedia). All
should be quite trivial apart from one of the last passages, which I
have marked.
To simplify the notation for the limit case, I shall take the set of
the extended naturals as reference. To be explicit:
N* := N U { 0, oo }
'A' is for-each
'U' is set-union
'e' is set-membership
'c=' is set-inclusion (improper)
Let P(n) be the set:
P(n) := { k e N* | k < 2^n }, n e N*
By transfinite induction, we shall prove the following property:
Prop.) A n e N* : P(n) c= N*
-- Zero case: n = 0
P(0) = { 0 } c=
c= N*
-- Successor case: n+1, assuming P(n) c= N*, n e N
P(n+1) = P(n) U { k e N* | 2^n <= k < 2^(n+1) } c=
c= N* U { k e N* | 2^n <= k < 2^(n+1) } =
= N*
-- Limit case: n = oo, assuming A m e N : P(m) c= N*
P(oo) = P(n) U { k e N* | 2^n <= k < 2^oo } c=
c= N* U { k e N* | 2^n <= k < 2^oo }
(The following passage should be the culprit:)
Since k e N*, where N* := N U { 0, oo }, it must be:
{ k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k < oo }
Then:
P(oo) c= N* U { k e N* | 2^n <= k < 2^oo } =
= N* U { k e N* | 2^n <= k < oo } =
= N*
Follows the thesis. QDE.
A possible interpretation for the whole matter, in the form of a fast
slogan: "however fast, it's just going to infinity".
Anyway, right or wrong, I hope we have narrowed it down... and that
there just isn't some blatant mistake!
Thanks again for all the feedback and contributions.
Julio
status: OPEN
revision: 2
sender: Julio Di Egidio (aka LudovicoVan)
sender-email: ju...@diegidio.aleph.name (del 'aleph.')
copyright: 2008 (on behalf of) sci.math, sci.logic
All rights reserved.
On 27 Jul, 04:32, ju...@diegidio.name wrote:
> Second attempt. I'll try two equivalent formulations, in the hope
> that, together, they will complement the eventual ambiguities.
>
> ----- 1st formulation
>
> (i) P(A) ~ { 0, 1, 2, ..., 2^n - 1}, AneN
>
> Let's simply call P(A) the set on the RHS of (i), keeping in mind that
> the argument is going to hold due to this bijection and the
> transitivity of the '~' relation.
>
> (ii) P(A) c= N, AneN
>
> (iii) P(N) is infinite (trivial)
>
> We pass from P(A) to P(N) by transfinite induction, then:
>
> (iv) P(N) ~ N, from (ii) and (iii)
>
> QDE. (1)
>
> ----- 2nd formulation
>
> Same comments as above.
>
> (i) card(P(A)) = 2^n, AneN
>
> (ii) card(P(A)) <= card(N), AneN
> (iii) card(P(N)) >= oo (meaning: 'oo' as lower bound)
>
> (iv) card(P(N)) = card(N)
>
> QDE. (2)
>
> -----
>
> Hmm, correct? The basic idea behind this argument is that P(A) is
> always a subset of N in the finite case. Then, by transfinite
> induction, while P(N) becomes infinite, its cardinality's upper bound
> remains N's cardinality. (These informal expositions get me into more
> troubles than they help, but I have to run the risk: no pain, no
> gain!)
>
> Good luck.
>
> Julio
>
> status: OPEN
> revision: 1
> sender: Julio Di Egidio (aka LudovicoVan)
> sender-email: ju...@diegidio.aleph.name (del 'aleph.')
> copyright: 2008 (on behalf of) sci.math, sci.logic
> All rights reserved.
>
> On 26 Jul, 22:52, ju...@diegidio.name wrote:
>
>
> > card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> > here and there).
>
> > --- Preliminaries:
>
> > 'P(A)' stands for the "powerset" of set 'A'.
>
> > '~' stands for the existence of a bijection, which will be, in this
> > context, equivalent to stating that two sets are equinumerous.
> > Infinity of the set 'N' (the set of natural/ordinal numbers) is taken
> > to be defined as usual, by the existence of bijections between N and
> > its (infinite) subsets. No other kind of "infinities" are going to be
> > assumed: the only tool I have used is transfinite induction over the
> > naturals/ordinals, with "omega" as the limit ordinal.
>
> > 'oo' stands for "omega".
>
> > --- Finite case:
>
> > card(A) = n <=>def
> > A ~ { 0, 1, 2, .., n-1 }
>
> > card(P(A)) = 2^n <=>
> > P(A) ~ { 1, 2, 4, ..., 2^n-1 }
>
> > --- Infinite case (transf. induction, notation a bit fuzzy here):
>
> > (1) n=oo => card(A) = card(N) = oo <=>def
> > A ~ N := { 0, 1, 2, ..., n }, n->oo
>
> > (2) n=oo => card(P(A)) = card(P(N)) = 2^oo <=>
> > P(A) ~ P(N) ~ 2^N := { 1, 2, 4, ..., 2^n }, n->oo
>
> > --- Theorem:
>
> > card(P(N)) = card(N)
>
> > --- Proof:
>
> > (i) card(P(N)) = card(N) <=> P(N) ~ N, by def. of 'card'
>
> > (ii) card(P(N)) = card(N) <=> 2^N ~ N, by (2) and trans. of '~'
>
> > (iii) A bijection 2^N ~ N trivially exists:
>
> > 2^N ~ N
> > --------
> > 1 ~ 0
> > 2 ~ 1
> > 4 ~ 2
> > ...
>
> > Follows the thesis.
>
> > QDE.
>
> > It's THEN easy to get interesting lemmas. We get that omega is an
> > absolute upper bound and all infinities are equivalent. The diagonal
> > argument then assumes another shape, which will be the subject of
> > another post of mine, unless someone finds a flaw in the above quite
> > elementary argument.
>
> > Thanks for any feedback.
>
> > Julio
>
> > status: OPEN
> > sender: Julio Di Egidio (aka LudovicoVan)
> > sender-email: ju...@diegidio.aleph.name (del 'aleph.')
> > copyright: 2008 (on behalf of) sci.math, sci.logic
> > All rights reserved.
Virgil
<9c41a2d0-e899-44b4...@p25g2000hsf.googlegroups.com>,
ju...@diegidio.name wrote:
> card(P(N)) = card(N), rev 2: a preliminary result
> -------------------------------------------------
>
> I shall only try to prove a preliminary property, from which it should
> then be trivial to prove the thesis, so I will leave the final steps
> out for now.
>
> The proof below is based on transfinite induction (see Wikipedia). All
> should be quite trivial apart from one of the last passages, which I
> have marked.
>
> To simplify the notation for the limit case, I shall take the set of
> the extended naturals as reference. To be explicit:
>
> N* := N U { 0, oo }
One presumes that this N* is intended to be an ordered set with the
usual order of N u {0} and with x < oo for all x =/= oo.
In such things, it is better t be explicit about what one is assuming.
>
> 'A' is for-each
> 'U' is set-union
> 'e' is set-membership
> 'c=' is set-inclusion (improper)
>
> Let P(n) be the set:
>
> P(n) := { k e N* | k < 2^n }, n e N*
What has this to do with power sets?
The point for power sets would be to have something like:
for each n in N*, P(n) = {k : (k c= N*) and AxeN*(x e k => x <= n)}
That would mean that each P(n), for any n on N*, is the power set for
the set of x in N* no greater than that n.
Since you do not have that, you are not showing anything about power
sets at all.
ju...@diegidio.name
Sorry, there is a typo there, better read:
P(oo) = P(m) U { k e N* | 2^m <= k < 2^oo } c=
and so on...
-LV
ju...@diegidio.name
> In article
> <9c41a2d0-e899-44b4-9134-d4e9eff59...@p25g2000hsf.googlegroups.com>,
> ju...@diegidio.name wrote:
> > To simplify the notation for the limit case, I shall take the set of
> > the extended naturals as reference. To be explicit:
>
> > N* := N U { 0, oo }
>
> One presumes that this N* is intended to be an ordered set with the
> usual order of N u {0} and with x < oo for all x =/= oo.
>
> In such things, it is better t be explicit about what one is assuming.
That's why I have given it explicitly:
N* := N U { 0, oo }
which is different than what you are presuming above.
> > Let P(n) be the set:
>
> > P(n) := { k e N* | k < 2^n }, n e N*
>
> What has this to do with power sets?
So, does the property hold?
If it holds, the rest should be trivial.
-LV
Virgil
<6c52d0dd-d29c-4646...@z72g2000hsb.googlegroups.com>,
ju...@diegidio.name wrote:
Since as stated your symbols say nothing at all about power sets, what
is it about power sets that is supposed to be trivial?
ju...@diegidio.name
> Since as stated your symbols say nothing at all about power sets, what
> is it about power sets that is supposed to be trivial?
Are you kidding me?
Does the property hold?
The rest is trivial, but I won't get into the next step until this one
has been cleared.
-LV
ju...@diegidio.name
Another one:
Looking at the above equivalence, I'd maybe better have written:
... = { k e N* | 2^n <= k <= oo }
I am not 100%, since here, as said, 'N*' helps in simplifying the
notation for the limit case. The substance of the argument should
remain unchanged.
This said, I'll be waiting for feedback.
-LV
Norman Megill
[...]
> I do have read *many* introductory and less introductory books on
> these matters, and I keep doing. That is simply *not* enough to write
> down arguments correctly in the language of a specific discipline such
> as mathematics.
>
> Now you can complain as much as you like, but the fact is: I do work,
> I am neither a student nor a prof (no pun intended). That kind of
> practice I have to do *here*, I have no other place to practice that:
> no profs, no tutors, no collegues, no nothing, I have got *this* group
> to practice and check my mathematics.
>
> Books alone are not enough. I have spent 30+ years of my recent life
> reading and studing books, not just mathematics. It is not enough to
> be a pro: you need to complement with practice!
There is a free on-line book that I think is a good introduction to
propositional and predicate calculus, Hirst & Hirst's "A Primer for
Logic and Proof":
http://www.mathsci.appstate.edu/~jlh/primer/hirst.pdf
This will show you the correct notation to use when you discuss proofs
here.
If you are impatient, the actual axioms of propositional calculus,
predicate calculus, and equality respectively are found on pp. 15-16,
51, & 64 (PDF pp. 21-22, 57, & 70). (Of course that doesn't mean one
shouldn't read the rest of the book to grasp the axioms intuitively and
learn how to use them in proofs.) These are the "foundation" of
classical first-order logic, from which everything else in logic is
derived and mathematical proofs are based on.
To understand the pred. calc. axioms, the tricky concepts for a beginner
are free variables (pp. 36-37, PDF pp. 42-43) and "free
for"/substitution (pp. 48-51, PDF pp. 54-57). These are not difficult
or deep, just a little tedious, but one must learn them to understand
the axioms and do proofs correctly.
Some useful theorems derived from the axioms are L1 through L24 for
prop. calc. and K1 through K36 for pred. calc. There are plenty of
exercises also.
Set theory consists of additional axioms on top of these and is not
given formally in this book. But predicate calculus is a prerequisite
for it.
There are other first-order logic systems based mainly on inference
rules, called natural deduction and Gentzen systems, that some people
prefer because they can be easier to work with directly in proofs.
These are equivalent to and can be derived from the simpler system
presented in this book.
--
Norm http://us.metamath.org/email.html (Reply to author
at this URL. The "from" address in this post is not valid.)
MoeBlee
Unless I made a huge mistake (and therefore would need new glasses
right away), it's trivial that it holds that P(n) subset of N*. You
DEFINED P(n) explicitly to be a subset of N* when you put keN* in
{keN* | ANYTHING_HERE!}.
Sheesh.
MoeBlee
MoeBlee
> card(P(N)) = card(N), rev 2: a preliminary result
> -------------------------------------------------
>
> I shall only try to prove a preliminary property, from which it should
> then be trivial to prove the thesis, so I will leave the final steps
> out for now.
>
> The proof below is based on transfinite induction (see Wikipedia). All
> should be quite trivial apart from one of the last passages, which I
> have marked.
>
> To simplify the notation for the limit case, I shall take the set of
> the extended naturals as reference. To be explicit:
>
> N* := N U { 0, oo }
Usually we take N (w, for omega) to be {0 1 2 ...}.
But I take it that you have N={1 2 3 .. }. Okay.
Meanwhile, I take that your 'oo' stands for w = {0 1 2 ... }.
So N* = wu{w}.
> 'A' is for-each
> 'U' is set-union
> 'e' is set-membership
> 'c=' is set-inclusion (improper)
> Let P(n) be the set:
>
> P(n) := { k e N* | k < 2^n }, n e N*
> By transfinite induction, we shall prove the following property:
>
> Prop.) A n e N* : P(n) c= N*
Your proof is a waste of energy, since, BY DEFINITION,
P(n) subset of N*.
Sheesh. P(n) = { k e N* | k < 2^n }, so every member of P(n) is a
member of N* since every member of P(n) is some k such that k e N*
End of proof.
By the way, just in case it comes up, keep in mind that for n in w or
n=w, we have:
2^n = card({f | f is a function from n into {0 1}}).
Unless you state otherwise, that is the definition of cardinal
exponenitation (which, in the finite case, is equivalent with ordinary
exponentiation on natural numbers).
Anyway, it's not clear why you're thinking of transfinite induction.
Transfinite induction proves something holds for all ordinals. The
only things you want to prove for are the finite ordinals and w
itself.
> -- Zero case: n = 0
>
> P(0) = { 0 } c=
> c= N*
Fine, but unneeded.
> -- Successor case: n+1, assuming P(n) c= N*, n e N
>
> P(n+1) = P(n) U { k e N* | 2^n <= k < 2^(n+1) } c=
> c= N* U { k e N* | 2^n <= k < 2^(n+1) } =
> = N*
Fine, but unneeded.
> -- Limit case: n = oo,
NO, that's wrong. The limit case is an arbitrary limit ordinal, not
just w.
But let's see what you do to try for w itself.
However, it's unneeded.
> assuming A m e N : P(m) c= N*
We don't need to assume it. We HAVE it.
But, anyway, that's not how transfinite induction works.
Transfinite induction, in the form YOU referenced, allows assuming for
an ARBITRARY limit ordinal that all its members have the property and
then showing that that arbitrary limit ordinal has the property. It
does NOT allow choosing a PARTICULAR limit ordinal.
Here are some transfinite induction schemata:
1.
F0 &
Ak(k is a successor ordinal -> Fk) &
Ak(k is a limit ordinal -> Fk)
->
Ak(k is an ordinal -> Fk)
2.
Ak((k is an ordinal & Am(m<k -> Fm)) -> Fk)
->
Ak(k is an ordinal -> Fk)
3
F0 &
Ak((k is an ordinal & Fk) -> Fk+1) &
Ak((k is a limit ordinal & Am(m<k -> Fm)) -> Fk)
->
Ak(k is an ordinal -> Fk)
Please either put your transfinite induction in one of those forms, or
PROVE another form.
But anyway, your argument is not transfinite induction, and doesn't
need to be.
What you wanted to prove is IMMEDIATE just from you definitions.
> P(oo) = P(n) U { k e N* | 2^n <= k < 2^oo } c=
> c= N* U { k e N* | 2^n <= k < 2^oo }
Fine, but unneeded.
> (The following passage should be the culprit:)
> Since k e N*, where N* := N U { 0, oo }, it must be:
>
> { k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k < oo }
No.
It is actually:
{ k e N* | 2^n <= k < 2^oo } = { k e N* | 2^n <= k <= oo }
The very last inequality is '<=' not '<', since 'oo' is a member of
N*.
> Then:
>
> P(oo) c= N* U { k e N* | 2^n <= k < 2^oo } =
> = N* U { k e N* | 2^n <= k < oo } =
> = N*
What a ridiculously convoluted proof just to show what is obvious.
To prove P(w) subset of N*, it suffices just to observe that, BY
DEFINITION, for all n in N* we have P(n) subset of N*.
And so where is your proof that N is 1-1 with PN?
MoeBlee
MoeBlee
> There are other first-order logic systems based mainly on inference
> rules, called natural deduction and Gentzen systems, that some people
> prefer because they can be easier to work with directly in proofs.
> These are equivalent to and can be derived from the simpler system
> presented in this book.
Norm, don't you agree that he'd do a lot better to start right from
the beginning with an easy to understand natural deduction system such
as in the Kalish, Montague, and Mar book? The web primer is okay for
what it is, but for a really great treatment for beginners, I don't
know of anything that beats Kalish, Montague, and Mar.
MoeBlee
ju...@diegidio.name
> On Jul 28, 3:34 pm, ju...@diegidio.name wrote:
>
> What a ridiculously convoluted proof just to show what is obvious.
Thank you very much for the in depth review, that's simply great.
Again, now I need some pondering.
The next step should be as trivial, but we'll see. The basic idea is
there is a straightforward bijection between the powerset and our set
P(n) above. Anyway, we'll see.
-LV
Norman Megill
Actually I am not familiar with that book - from what you say it sounds
good, and I'll make a point to look it up when I go to the library.
I agree that natural deduction is a more practical working tool. My
idea was partly that with a Hilbert-style system, there is less to learn
if one just wants to see what the "foundation" or "starting point" is,
in as simple a system as possible. Whether that is a good idea I guess
is a matter of opinion.
Also, an on-line book that anyone can access makes it easy for anyone to
answer questions about the book and provides a consistent reference if
questions of logical syntax or what constitutes a correct proof arise,
both of which have been problems with his posts. Is there a good
on-line introduction to natural deduction similar to the Hirst/Hirst
book (or the Kalish, Montague, and Mar one)?
ju...@diegidio.name
Yes! :) I guess that's what was known by the back of my mind.
The argument is indeed subtle: it seems to me it depends on how we
define our domain. One only limit ordinal w is not in itself a problem
(is it?), actually it seems we are sort of giving it substance with
this argument.
Anyway, never mind, thinking out loud. I'll try the next step.
-LV
>
> Sheesh.
>
> MoeBlee
Ralf Bader
What is that 2^oo appearing in ANYTHING_HERE? His statement is not trivially
true, because it doesn't even make any sense. It is going as usual with
such cranky stuff: an endless repetition of attempts, each a greater mess
than the previous one. The next step will be to smuggle that 2^oo which is
floating around loosely into N*, and then to declare it countable because
there is no place for anything uncountable in N*; and somehow this 2^oo is
going to represent the power set of N. This then is the next version of
this "proof". While the initial problem in the previous version was the
inappropriate usage of transfinite induction, now an additional and bigger
problem arises: a basic misunderstanding about sound definitions. And this
opens the door for another endless discussion about nonsense.
R.B.
Brian Chandler
I would recommend doing the first step first. You started all this
with the claim in the title that suggests to mathematicians that you
are saying something about powersets, yet powersets appear nowhere in
your arguments.
So, please start by saying -- in words only -- what you mean by the
"powerset" of an arbitrary set X. You might also like to give a simple
example by showing the powerset of the following simple set:
A = {r, s, t}
If we could see that _you_ mean the same by "powerset" as everyone
else, we could try to identify the discrepancy between your argument
and, um, the normal mathematical one. Otherwise, we could perhaps try
to thrash out mutually acceptable terminology for referring to what
_you_ mean.
Brian Chandler
ju...@diegidio.name
> ju...@diegidio.name wrote:
> > On 29 Jul, 02:29, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > Unless I made a huge mistake (and therefore would need new glasses
> > > right away), it's trivial that it holds that P(n) subset of N*. You
> > > DEFINED P(n) explicitly to be a subset of N* when you put keN* in
> > > {keN* | ANYTHING_HERE!}.
>
> > Yes! :) I guess that's what was known by the back of my mind.
>
> > The argument is indeed subtle: it seems to me it depends on how we
> > define our domain. One only limit ordinal w is not in itself a problem
> > (is it?), actually it seems we are sort of giving it substance with
> > this argument.
>
> > Anyway, never mind, thinking out loud. I'll try the next step.
>
> I would recommend doing the first step first. You started all this
> with the claim in the title that suggests to mathematicians that you
> are saying something about powersets, yet powersets appear nowhere in
> your arguments.
I am afraid you must have missed some bits.
> So, please start by saying -- in words only -- what you mean by the
> "powerset" of an arbitrary set X. You might also like to give a simple
> example by showing the powerset of the following simple set:
>
> A = {r, s, t}
You call:
A = ( 1, 2, 3 }
P(A) = {
{},
{ 1 }, { 2 }, { 3 },
{ 1, 2 }, { 1, 3 }, { 2, 3 },
{ 1, 2, 3 }
}
IMHO, to lower the level of noise, a good strategy might be to take
people for serious and just tell what there is to tell: although one
might straight discover there was nothing worth, the sooner the better
in any case, isn't it?
-LV
Brian Chandler
<snip>
> > I would recommend doing the first step first. You started all this
> > with the claim in the title that suggests to mathematicians that you
> > are saying something about powersets, yet powersets appear nowhere in
> > your arguments.
>
>
> I am afraid you must have missed some bits.
>
>
> > So, please start by saying -- in words only -- what you mean by the
> > "powerset" of an arbitrary set X.
You haven't done this. Why not? It's quite simple, actually...
>> .., You might also like to give a simple
> > example by showing the powerset of the following simple set:
> >
> > A = {r, s, t}
>
> You call:
>
> A = ( 1, 2, 3 }
>
> P(A) = {
> {},
> { 1 }, { 2 }, { 3 },
> { 1, 2 }, { 1, 3 }, { 2, 3 },
> { 1, 2, 3 }
> }
Why do you say "You call"?? Are you not quite sure if you agree?
>
> IMHO, to lower the level of noise, a good strategy might be to take
> people for serious and just tell what there is to tell: although one
> might straight discover there was nothing worth, the sooner the better
> in any case, isn't it?
That was my first reaction. I posted a proof of the opposite of what
you claim to be trying to prove. Lots of people have given you
straight comments, ranging from "Wrong" to "Nonsense". What more do
you want?
Brian Chandler
Albrecht
Virgil schrieb:
No, I don't mean _an_ uncountable set of natural numbers since
everybody knows that this would be nonsense by definition. My sentence
is very sloppy fomulated, I see. My apology for that.
Informell, as I had written, better would be: "And uncountable many
sets of natural numbers."
David C. Ullrich
>On 28 Jul, 11:31, David C. Ullrich <dullr...@sprynet.com> wrote:
>
>> >I will be reading all the messages, I am not trying to start any
>> >debate at once. Just I don't understand that remark of yours.
>
>> Whatever. You ignored the second half of my post,
>> by the way, which explains the actual problem in
>> your thinking.
>
>I didn't, please read more carefully.
>
>I have been "meditating" the variuos feedbacks. It seems I shall make
>the inductive steps explicit... I mean, I'll try.
Why? The result you're trying to prove is _false_ - you're not
going to be able to make a correct proof, because it's impossible
to prove something false.
There are many big problems with the way you state things.
Those problems can be fixed. But the biggest problem
_cannot_ be fixed: Again, say A_n = {1,2,...n}.
It is true that card(P(A_n)) <= alpeh_0 for every natural
number n. But it simply does not follow that
card(P(N)) <= aleph_0. That's the most important
step in your "proof" and it's simply wrong.
>-LV
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.)
ju...@diegidio.name
> Why do you say "You call"?? Are you not quite sure if you agree?
We'll never agree, you are stupid.
-LV
ju...@diegidio.name
> Why? The result you're trying to prove is _false_ - you're not
> going to be able to make a correct proof, because it's impossible
> to prove something false.
FUCK OFF!
You and your stupid remarks.
-LV
ju...@diegidio.name
Bye bye.
-LV
> > card(P(N)) = card(N), an elementary proof (and some fuzzy notation
> > here and there).
>
Ralf Bader
ju...@diegidio.name
> It is remarkable how the habits of set theory cranks are always the same.
Yes, it's remarkable how you do keep the status quo.
-LV
MoeBlee
> The argument is indeed subtle: it seems to me it depends on how we
> define our domain. One only limit ordinal w is not in itself a problem
> (is it?), actually it seems we are sort of giving it substance with
> this argument.
It's not even subtle. It's as obvious as the Eiffel Tower. You can
take the following as a theorem schema:
For any term T and any formula F, all closures of
{keT | F} c= T
are theorems.
MoeBlee
ju...@diegidio.name
> It's not even subtle. It's as obvious as the Eiffel Tower.
Yes, thanks for the confirmation.
Still waiting to learn where you guys see the overall flaw.
Oh yes, you wouldn't have tried it at all. That one I know.
-LV
MoeBlee
> Still waiting to learn where you guys see the overall flaw.
We're waiting to see your argument that N is 1-1 with PN.
MoeBlee
Mariano Suárez-Alvarez
The overall flaw is that your argument is trying to
prove something that is false (in `usual' set
theory)
-- m
ju...@diegidio.name
Now you are driving me crazy with this non-synchronized cross-posting.
No, I didn't even attempt that!!
(Although it might be a consequence.)
How many times shall I still repeat this?
-LV
>
> MoeBlee
ju...@diegidio.name
<mariano.suarezalva...@gmail.com> wrote:
> The overall flaw is that your argument is trying to
> prove something that is false (in `usual' set
> theory)
Thanks for clarifing the whole matter up.
It is just "indirectly" false, BTW.
You have not even means to really disprove me.
It's a funny thing.
-LV
>
> -- m
MoeBlee
The TITLE of this thread begins:
card(PN) = card(N)
and that is what you keep saying you're going to prove.
That is equivalent to
N is 1-1 with PN.
Either way it is said:
"card(PN) = card(N)"
or
"N is 1-1 with PN"
we're waiting for you new argument. (It was made clear to you how your
previous argument for "card(PN) = card(N)" was incorrect.)
MoeBlee
Mariano Suárez-Alvarez
> On 29 Jul, 18:00, Mariano Suárez-Alvarez
>
> <mariano.suarezalva...@gmail.com> wrote:
> > The overall flaw is that your argument is trying to
> > prove something that is false (in `usual' set
> > theory)
>
> Thanks for clarifing the whole matter up.
>
> It is just "indirectly" false, BTW.
It is false in so far as one can actually prove the
negation of your statement.
I have no idea what `indirectly false' might mean.
> You have not even means to really disprove me.
I do not know what this means.
> It's a funny thing.
Not really.
-- m
MoeBlee
> I agree that natural deduction is a more practical working tool. My
> idea was partly that with a Hilbert-style system, there is less to learn
> if one just wants to see what the "foundation" or "starting point" is,
> in as simple a system as possible. Whether that is a good idea I guess
> is a matter of opinion.
Right, I had more in mind a book that would give him directly
applicable techniques for forming mathematical arguments and for
knowing that an argument is a valid first order argument. After that
is acheived, I think it then becomes pertinent to learn about Hilbert
style systems and mathematical logic in more generality.
> Is there a good
> on-line introduction to natural deduction similar to the Hirst/Hirst
> book (or the Kalish, Montague, and Mar one)?
I don't know of one.
MoeBlee
ju...@diegidio.name
> On 29 Jul, 17:48, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 28, 8:54 pm, Ralf Bader <ba...@nefkom.net> wrote:
> > > What is that 2^oo appearing in ANYTHING_HERE? His statement is not trivially
> > > true, because it doesn't even make any sense.
>
> > I take it that 'oo' stands for omega (w). So '2^oo' stands for
> > cardinal exponentiation:
> > card({f | f is a function from w into {0 1}}).
>
> > If he means otherwise, he can say so. But as I've described it, I
> > don't see a problem.
I didn't even know what cardinal exponentiation is.
What I believe is that the argument is over N, the naturals, and, that
there is anything beyond and/or beside w, is rather a result we are
supposed to get, not a starting point.
So, to me, 2^oo just tells about a lower bound. The rest is what there
is to prove. And my argument by induction seems to show there is an
upper bound too. That is enough for it, and we have that there is only
one 'w' or 'oo', and that it is -say- just:
2^oo = oo
as it is:
2*oo = oo
as it is:
1+oo = oo
and so on.
And then have a back-look at that "1+oo=oo" because there is the
diagonal argument, as well as the halting problem, as well as much of
the rest. (BTW, those who won't get this one, please don't get mad.)
At least, as I see it. No pretence.
-LV
On 29 Jul, 18:10, ju...@diegidio.name wrote:
> On 29 Jul, 17:48, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 28, 8:54 pm, Ralf Bader <ba...@nefkom.net> wrote:
> > > MoeBlee wrote:
> > > > On Jul 28, 4:31 pm, ju...@diegidio.name wrote:
> > > >> On 29 Jul, 00:20, Virgil <Vir...@gmale.com> wrote:
>
> > > >> > Since as stated your symbols say nothing at all about power sets, what
> > > >> > is it about power sets that is supposed to be trivial?
>
> > > >> Are you kidding me?
>
> > > >> Does the property hold?
>
> > > >> The rest is trivial, but I won't get into the next step until this one
> > > >> has been cleared.
>
> > > > Unless I made a huge mistake (and therefore would need new glasses
> > > > right away), it's trivial that it holds that P(n) subset of N*. You
> > > > DEFINED P(n) explicitly to be a subset of N* when you put keN* in
> > > > {keN* | ANYTHING_HERE!}.
>
> > > What is that 2^oo appearing in ANYTHING_HERE? His statement is not trivially
> > > true, because it doesn't even make any sense.
>
> > I take it that 'oo' stands for omega (w). So '2^oo' stands for
> > cardinal exponentiation:
> > card({f | f is a function from w into {0 1}}).
>
> > If he means otherwise, he can say so. But as I've described it, I
> > don't see a problem.
>
> I didn't even know what cardinal exponentiation is.
>
> What I believe is that the argument is over N, the naturals, and, that
> there is anything beyond and/or beside w, is rather a result we are
> supposed to get, not a starting point.
>
> So, to me, 2^oo just tells about a lower bound. The rest is what there
> is to prove. And my argument by induction seems to show there is an
> upper bound too. That is enough for it, and we have that there is only
> one 'w' or 'oo', and that it is -say- just:
>
> 2^oo = oo
>
> as it is:
>
> 2*oo = oo
>
> as it is:
>
> 1+oo = oo
>
> and so on.
>
> And then have a back-look at that "1+oo=oo" because there is the
> diagonal argument, as well as the halting problem, as well as much of
> the rest. (BTW, those who won't get this one, please don't get mad.)
>
> At least, as I see it. No pretence.
>
> -LV
>
> > MoeBlee
Mike Kelly
> On 29 Jul, 17:45, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > It's not even subtle. It's as obvious as the Eiffel Tower.
>
> Yes, thanks for the confirmation.
>
> Still waiting to learn where you guys see the overall flaw.
I think you are trying to use transfinite induction incorrectly. If
you explain what you think transfinite induction is then we can see.
Try to be as clear as possible! For example if I was explaining
induction on the naturals then I would state something like this:
If we have a property, P and we have
* P(0) is true and
* For any natural number, k, if P is true of k then P is true of k+1
Then P is true of all natural numbers
Can you explain your understanding of transfinite induction in a
similar way? That is, the conditions and the conclusion..
ju...@diegidio.name
> On Jul 29, 5:51 pm, ju...@diegidio.name wrote:
>
> > On 29 Jul, 17:45, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > It's not even subtle. It's as obvious as the Eiffel Tower.
>
> > Yes, thanks for the confirmation.
>
> > Still waiting to learn where you guys see the overall flaw.
>
> I think you are trying to use transfinite induction incorrectly.
And I don't think so, but I might be mistaken.
Everything has already been said, but here is again the substance of
the argument, hopefully even more explicit (I just won't restate again
the notation).
I assume MoeBlee is correct in telling that the following property is
trivial:
A n e N* : P(n) c= N*
Then there is an as trivial bijection between the set P(n) above and
P(A_n), the powerset of set A_n, where n is the cardinality of some
set A:
A n e N* : P(A_n) ~ P(n)
Is that correct? (I think that bijection *must* hold, otherwise -say-
we could not have our "parings" to state Cantor's Theorem.)
I'd say the thesis just follows.
(Was transfinite or even plain induction unneeded? You'll tell me.)
> If
> you explain what you think transfinite induction is then we can see.
C'mon...
-LV
Virgil
<b7781248-dc24-4b54...@d45g2000hsc.googlegroups.com>,
ju...@diegidio.name wrote:
Does Julio then claim that it IS possible to prove something false?
That is certainly what his remarks would seem to indicate.
MoeBlee
> > > I take it that 'oo' stands for omega (w). So '2^oo' stands for
> > > cardinal exponentiation:
> > > card({f | f is a function from w into {0 1}}).
>
> > > If he means otherwise, he can say so. But as I've described it, I
> > > don't see a problem.
>
> I didn't even know what cardinal exponentiation is.
Then you need to say that you mean by '2^oo'.
Also, please tell me whether you're using 'oo' to stand for w, which
is the set of natural numbers {0 1 2 ... }, or for something else.
> What I believe is that the argument is over N, the naturals, and, that
> there is anything beyond and/or beside w, is rather a result we are
> supposed to get, not a starting point.
>
> So, to me, 2^oo just tells about a lower bound.
PLEASE STOP RIGHT THERE. There can be no coherent, meaningful
discussion with you about your argument unless your define your
terminology.
And "tells about" is not a form of definition.
MoeBlee
ju...@diegidio.name
> In article
> <b7781248-dc24-4b54-be67-5044cffe8...@d45g2000hsc.googlegroups.com>,
>
> ju...@diegidio.name wrote:
> > On 29 Jul, 12:01, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> > > Why? The result you're trying to prove is _false_ - you're not
> > > going to be able to make a correct proof, because it's impossible
> > > to prove something false.
>
> > FUCK OFF!
>
> > You and your stupid remarks.
>
> Does Julio then claim that it IS possible to prove something false?
No, he doesn't. How do *you* infer *that*?
> That is certainly what his remarks would seem to indicate.
That is the opposite of certain.
BTW, have you finished with Mr WM? Julio is still waiting for you to
address even one of the points he made. There, here and elsewhere.
-LV
MoeBlee
> I assume MoeBlee is correct in telling that the following property is
> trivial:
>
> A n e N* : P(n) c= N*
Why do you have to assume? Why don't you see for yourself that
{keT | F} is a subset of T
for any term T and any formula F.
> Then there is an as trivial bijection between the set P(n) above and
> P(A_n), the powerset of set A_n,
STOP RIGHT THERE. You're using 'P' now for two DIFFERENT THINGS. That
only bodes confusion. In your latest posts you had P serve as
something quite different from power set.
Please use 'P' for just one thing here. Since 'P' is common for power
set, I suggest, you not use 'P' for your definition:
P(n) := { k e N* | k < 2^n }, n e N*
But rather use 'Q' (or whatever other than 'P'):
For neN*, we put Q(n) = {keN* | k<2^n}.
MoeBlee
Mike Kelly
You seem to be using P() to denote two different things. One of them
is the powerset, and one of them is something else to do with powers
of natural numbers. Is this right? If so, could you rewrite the above
so the two different things have different symbols denoting them? It's
very confusing for me at the moment. Sorry.
Virgil
<c35b47e2-cee1-4aa1...@m3g2000hsc.googlegroups.com>,
ju...@diegidio.name wrote:
> On 29 Jul, 17:56, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 29, 9:51 am, ju...@diegidio.name wrote:
> >
> > > Still waiting to learn where you guys see the overall flaw.
> >
> > We're waiting to see your argument that N is 1-1 with PN.
>
> Now you are driving me crazy with this non-synchronized cross-posting.
>
> No, I didn't even attempt that!!
Then it is not at all clear what you are attempting.
Except that it appears to be incoherent.
Virgil
<a05b917a-a4b0-4f61...@34g2000hsf.googlegroups.com>,
ju...@diegidio.name wrote:
> BTW, have you finished with Mr WM? Julio is still waiting for you to
> address even one of the points he made. There, here and elsewhere.
I do not find that any points have been made.
If, for an arbitrary set S, P(S) is to denote the set of all subsets of
S and for any two sets S and T, card(S) = card(T) requires the existence
of a bijection between S and T, as is usual in standard set theories,
then I do not find that Julio has made any points re his claim that
card(P(N)) = card(N) that are sufficiently coherent as to require
addressing.
Virgil
<1d885fea-ca61-49a2...@a70g2000hsh.googlegroups.com>,
ju...@diegidio.name wrote:
> On 29 Jul, 18:40, Mike Kelly <mikekell...@googlemail.com> wrote:
> > On Jul 29, 5:51 pm, ju...@diegidio.name wrote:
> >
> > > On 29 Jul, 17:45, MoeBlee <jazzm...@hotmail.com> wrote:
> >
> > > > It's not even subtle. It's as obvious as the Eiffel Tower.
> >
> > > Yes, thanks for the confirmation.
> >
> > > Still waiting to learn where you guys see the overall flaw.
> >
> > I think you are trying to use transfinite induction incorrectly.
>
> And I don't think so, but I might be mistaken.
>
> Everything has already been said, but here is again the substance of
> the argument, hopefully even more explicit (I just won't restate again
> the notation).
>
> I assume MoeBlee is correct in telling that the following property is
> trivial:
>
> A n e N* : P(n) c= N*
It is trivially false if P(S) is to mean the set of all subsets of S.
Do you perhaps mean something like Card(P(n)) e N* ?
Note that 'P(x)' as a predicate should have the same meaning for all
arguments 'x', but you seem to be giving it two conflicting meanings
depending on what sort of argument it is taking.
That is certainly an abuse of notation sufficient to make it impossible
for anyone else to be sure of what you are trying to say.
Peter Webb
<ju...@diegidio.name> wrote in message
news:c1820d43-1072-44c6...@z72g2000hsb.googlegroups.com...
Bye bye.
-LV
If you want to use transfinite induction, you have to show that the result
is true for n=1, n+1, and the limit ordinal of n. The first two apply only
to "normal" induction; the third is required for transfinite induction. You
have not shown this third requirement is met in your proof, and nor can you,
as its not.
Consider the following "proof" which is similar in form to yours:
All ordinals are finite.
proof:
1 is finite.
If n is finite, so is n+1.
Therefore all numbers are finite.
Of course, this is wrong, because w is not finite; the proof fails because w
is not a successor to any other ordinal (its a limit ordinal), and the
requirements of transfinite induction are not met.
David C. Ullrich
>On 29 Jul, 12:01, David C. Ullrich <dullr...@sprynet.com> wrote:
>
>> Why? The result you're trying to prove is _false_ - you're not
>> going to be able to make a correct proof, because it's impossible
>> to prove something false.
>
>FUCK OFF!
Tee-hee. You know, that's not a very convincing proof.
>You and your stupid remarks.
Stupid remarks, right.
It's been known for more than a century that the
result you're trying to prove is false, by the way.
You're claiming that you're right and _every_
mathematician for more than a hundred years
has been wrong. And not wrong about something
deep and complicated, wrong about a very simple
proof. You _really_ think you understand all this
better than _anyone_?
Wow.
>-LV
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.)