| |
 |
|
|
Science Forum Index » Logic Forum » Infinite Arithmetic Sets - One-one Mapping
Page 1 of 5 Goto page 1, 2, 3, 4, 5 Next
|
| Author |
Message |
| Tony Thomas |
 Posted: Tue Jan 06, 2004 11:07 pm |
|
|
|
Guest
|
The principle of one-one correspondence is indispensable to the theory of
infinite sets.
That it is contentious goes without saying. If two infinite sets can be
shown to logically map one-one without appeal to the inductive principle of
one-one mapping then this is the preferable method.
Even if one-one mapping, as usually applied to infinite sets, cannot be
refuted as a procedure neither can it be asserted as necessarily valid. If,
however, it can be shown that the consequences of inductive one-one mapping
contradict a more defensible method, then it should be abandoned.
The following examples illustrate the weakness of Cantorian Set Theory and
its variants on this account.
D1: Z; The set of all integers
D2: Z+ The set of positive integers
D3: Z- The set of negative integers
D4: Ze The set of even integers
D5: Zo The set of odd integers
D6 {0} The unit set containing zero
D7: #S The cardinal of a general finite or infinite set S
D8: v Set union
D9: ^ Set intersection
D10: + Arithmetic addition
D11: - Arithmetic subtraction
Theorem: A + B = #(AvB) - #(A^B) where A and B are finite or infinite
sets*
Proposition (i): The cardinal of Z is odd
Z- v {0} v Z+ = Z
LHS accounts for all the integers.
#Z+ = #Z-
The cardinal of Z+ is equal to the cardinal of Z- without recourse to the
inductive principle of one-one mapping. Using the principle of symbolic
substitution "+" -> "-" demonstrated this assertion.
If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
If #Z+ has odd parity then #Z- has odd parity too.
In either case, #Z will have odd parity.
Proposition (i) has been demonstrated.
Proposition (ii): Either the set of even numbers or the set of odd integers
can be mapped to the set of all integers but not both.
Ze v Zo = Z
By theorem (i), The cardinal of Z is odd
The cardinals of both Ze and Zo must have different parities
If Ze can be mapped one-one to Z then Zo cannot and vice versa.
Proposition (ii) has been demonstrated.
* Objectors must show why this theorem cannot be applied to infinite sets. |
|
|
| Back to top |
|
| Barb Knox |
| Posted: Wed Jan 07, 2004 12:34 am |
|
|
|
Guest
|
In article <3ffb86be_1@news.iprimus.com.au>,
"Tony Thomas" <verdigris@iprimus.com.au> wrote:
Quote: The principle of one-one correspondence is indispensable to the theory of
infinite sets.
That it is contentious goes without saying. If two infinite sets can be
shown to logically map one-one without appeal to the inductive principle of
one-one mapping then this is the preferable method.
Even if one-one mapping, as usually applied to infinite sets, cannot be
refuted as a procedure neither can it be asserted as necessarily valid. If,
however, it can be shown that the consequences of inductive one-one mapping
contradict a more defensible method, then it should be abandoned.
The following examples illustrate the weakness of Cantorian Set Theory and
its variants on this account.
D1: Z; The set of all integers
D2: Z+ The set of positive integers
D3: Z- The set of negative integers
D4: Ze The set of even integers
D5: Zo The set of odd integers
D6 {0} The unit set containing zero
D7: #S The cardinal of a general finite or infinite set S
D8: v Set union
D9: ^ Set intersection
D10: + Arithmetic addition
D11: - Arithmetic subtraction
Theorem: A + B = #(AvB) - #(A^B) where A and B are finite or infinite sets*
I expect you mean something like #A + #B = #(AvB) + #(A^B),
since "^" and "v" only apply to sets and "+" only applies to numbers.
Quote: Proposition (i): The cardinal of Z is odd
Z- v {0} v Z+ = Z
LHS accounts for all the integers.
#Z+ = #Z-
The cardinal of Z+ is equal to the cardinal of Z- without recourse to the
inductive principle of one-one mapping. Using the principle of symbolic
substitution "+" -> "-" demonstrated this assertion.
What is your justification for this alleged "principle of symbolic
substitution"? And why bother, since #Z+ clearly = #Z- via the mapping
{1,2,3,...}<->{-1,-2,-3,...}.
Quote: If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
Agreed.
Quote: If #Z+ has odd parity then #Z- has odd parity too.
Agreed.
However, NEITHER Z+ nor Z- have even parity OR odd parity. You have not
shown that infinite sets have ANY parity, and in actual fact they don't.
Quote: In either case, #Z will have odd parity.
Not proven (and in fact false, since Z# has NO parity).
Quote: Proposition (i) has been demonstrated.
No.
Quote: Proposition (ii): Either the set of even numbers or the set of odd integers
can be mapped to the set of all integers but not both.
Ze v Zo = Z
By theorem (i), The cardinal of Z is odd
Except that (i) is bogus, since none of Z+, Z-, or Z have ANY parity.
Quote: The cardinals of both Ze and Zo must have different parities
If Ze can be mapped one-one to Z then Zo cannot and vice versa.
Proposition (ii) has been demonstrated.
No.
Quote: * Objectors must show why this theorem cannot be applied to infinite sets.
--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
----------------------------- |
|
|
| Back to top |
|
| ChrisX |
| Posted: Wed Jan 07, 2004 2:11 am |
|
|
|
Guest
|
"Barb Knox" <see@sig.below> wrote in message
news:btg5o9$uoh$1@lust.ihug.co.nz...
Quote: In article <3ffb86be_1@news.iprimus.com.au>,
"Tony Thomas" <verdigris@iprimus.com.au> wrote:
The principle of one-one correspondence is indispensable to the theory of
infinite sets.
That it is contentious goes without saying. If two infinite sets can be
shown to logically map one-one without appeal to the inductive principle
of
one-one mapping then this is the preferable method.
Even if one-one mapping, as usually applied to infinite sets, cannot be
refuted as a procedure neither can it be asserted as necessarily valid.
If,
however, it can be shown that the consequences of inductive one-one
mapping
contradict a more defensible method, then it should be abandoned.
The following examples illustrate the weakness of Cantorian Set Theory
and
its variants on this account.
D1: Z; The set of all integers
D2: Z+ The set of positive integers
D3: Z- The set of negative integers
D4: Ze The set of even integers
D5: Zo The set of odd integers
D6 {0} The unit set containing zero
D7: #S The cardinal of a general finite or infinite set S
D8: v Set union
D9: ^ Set intersection
D10: + Arithmetic addition
D11: - Arithmetic subtraction
Theorem: A + B = #(AvB) - #(A^B) where A and B are finite or infinite
sets*
I expect you mean something like #A + #B = #(AvB) + #(A^B),
since "^" and "v" only apply to sets and "+" only applies to numbers.
It seems false to say either
#A + #B = #(AvB) - #(A^B)
or
#A + #B = #(AvB) + #(A^B)
unless #(A^B) = 0
Quote:
Proposition (i): The cardinal of Z is odd
Z- v {0} v Z+ = Z
LHS accounts for all the integers.
#Z+ = #Z-
The cardinal of Z+ is equal to the cardinal of Z- without recourse to the
inductive principle of one-one mapping.
How?
Quote: Using the principle of symbolic
substitution "+" -> "-" demonstrated this assertion.
What is your justification for this alleged "principle of symbolic
substitution"? And why bother, since #Z+ clearly = #Z- via the mapping
{1,2,3,...}<->{-1,-2,-3,...}.
If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
Agreed.
I never saw the word parity used in that sense before. Perhaps my IT
background is confusing me.
Quote:
If #Z+ has odd parity then #Z- has odd parity too.
Agreed.
However, NEITHER Z+ nor Z- have even parity OR odd parity. You have not
shown that infinite sets have ANY parity, and in actual fact they don't.
In either case, #Z will have odd parity.
Not proven (and in fact false, since Z# has NO parity).
Proposition (i) has been demonstrated.
No.
Nor to me, I am afraid.
Quote:
Proposition (ii): Either the set of even numbers or the set of odd
integers
can be mapped to the set of all integers but not both.
Ze v Zo = Z
By theorem (i), The cardinal of Z is odd
Except that (i) is bogus, since none of Z+, Z-, or Z have ANY parity.
The cardinals of both Ze and Zo must have different parities
If Ze can be mapped one-one to Z then Zo cannot and vice versa.
Proposition (ii) has been demonstrated.
No.
* Objectors must show why this theorem cannot be applied to infinite
sets.
Love and respect
Chris |
|
|
| Back to top |
|
| John Jones |
| Posted: Wed Jan 07, 2004 2:57 am |
|
|
|
Guest
|
You would be better off by saying what mapping is, rather than assuming that
we can map and giving lists of what we say can be mapped.
Nothing is deduced from a mapping. All a mapping is, is a correlation of one
sign with another. We CANNOT make a mapping of even numbers to odd numbers,
such a mapping does not reflect or uphold the production of even or odd
numbers. There can be no mapping of even and odd numbers, BUT we can say
that we can map numerals or empty marks that result from the expression of
odd and even numbers.
We map numerals and signs, not numbers. Therefore, such mappings do not lead
to infinite sets or infinite anything, because signs are contingently made.
They are not out there waiting to be found. We also cannot speak of 'all'
integers.
JJ
"Tony Thomas" <verdigris@iprimus.com.au> wrote in message
news:3ffb86be_1@news.iprimus.com.au...
Quote: The principle of one-one correspondence is indispensable to the theory of
infinite sets.
That it is contentious goes without saying. If two infinite sets can be
shown to logically map one-one without appeal to the inductive principle
of
one-one mapping then this is the preferable method.
Even if one-one mapping, as usually applied to infinite sets, cannot be
refuted as a procedure neither can it be asserted as necessarily valid.
If,
however, it can be shown that the consequences of inductive one-one
mapping
contradict a more defensible method, then it should be abandoned.
The following examples illustrate the weakness of Cantorian Set Theory and
its variants on this account.
D1: Z; The set of all integers
D2: Z+ The set of positive integers
D3: Z- The set of negative integers
D4: Ze The set of even integers
D5: Zo The set of odd integers
D6 {0} The unit set containing zero
D7: #S The cardinal of a general finite or infinite set S
D8: v Set union
D9: ^ Set intersection
D10: + Arithmetic addition
D11: - Arithmetic subtraction
Theorem: A + B = #(AvB) - #(A^B) where A and B are finite or infinite
sets*
Proposition (i): The cardinal of Z is odd
Z- v {0} v Z+ = Z
LHS accounts for all the integers.
#Z+ = #Z-
The cardinal of Z+ is equal to the cardinal of Z- without recourse to the
inductive principle of one-one mapping. Using the principle of symbolic
substitution "+" -> "-" demonstrated this assertion.
If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
If #Z+ has odd parity then #Z- has odd parity too.
In either case, #Z will have odd parity.
Proposition (i) has been demonstrated.
Proposition (ii): Either the set of even numbers or the set of odd
integers
can be mapped to the set of all integers but not both.
Ze v Zo = Z
By theorem (i), The cardinal of Z is odd
The cardinals of both Ze and Zo must have different parities
If Ze can be mapped one-one to Z then Zo cannot and vice versa.
Proposition (ii) has been demonstrated.
* Objectors must show why this theorem cannot be applied to infinite sets.
|
|
|
| Back to top |
|
| William Elliot |
| Posted: Wed Jan 07, 2004 4:58 am |
|
|
|
Guest
|
On Wed, 7 Jan 2004, ChrisX wrote:
Quote: "Barb Knox" <see@sig.below> wrote in message
"Tony Thomas" <verdigris@iprimus.com.au> wrote:
D1: Z; The set of all integers
D2: Z+ The set of positive integers
D3: Z- The set of negative integers
D7: #S The cardinal of a general finite or infinite set S
D8: v Set union
D9: ^ Set intersection
D10: + Arithmetic addition
D11: - Arithmetic subtraction
It seems false to say either
#A + #B = #(AvB) - #(A^B)
This is false. Consider when A = B = Z
Then one has #Z + #Z = #Z - #Z
Quote: or
#A + #B = #(AvB) + #(A^B)
unless #(A^B) = 0
This is true. The commone elements of A and B get counted twice
once in #A and again in #B.
Quote: #Z+ = #Z-
The cardinal of Z+ is equal to the cardinal of Z- without recourse to the
inductive principle of one-one mapping.
Huh?
Quote: How?
for f:Z+ -> Z-, let f(n) = -n.
f is a bijection between Z+ and Z-.
Thus Z+ and Z- are equinumerous.
So induction isn't used.
Quote: If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
Agreed.
I never saw the word parity used in that sense before. Perhaps my IT
background is confusing me.
Parity of A means the (odd or even)-ness of #A.
There is no parity for infinite sets.
The rest of OP's posts misses this point. |
|
|
| Back to top |
|
| David C. Ullrich |
| Posted: Wed Jan 07, 2004 6:46 am |
|
|
|
Guest
|
On Wed, 7 Jan 2004 14:07:31 +1000, "Tony Thomas"
<verdigris@iprimus.com.au> wrote:
Quote: The principle of one-one correspondence is indispensable to the theory of
infinite sets.
That it is contentious goes without saying.
Only among crackpots on the internet.
Quote: If two infinite sets can be
shown to logically map one-one without appeal to the inductive principle of
one-one mapping then this is the preferable method.
Even if one-one mapping, as usually applied to infinite sets, cannot be
refuted as a procedure neither can it be asserted as necessarily valid. If,
however, it can be shown that the consequences of inductive one-one mapping
contradict a more defensible method, then it should be abandoned.
The following examples illustrate the weakness of Cantorian Set Theory and
its variants on this account.
D1: Z; The set of all integers
D2: Z+ The set of positive integers
D3: Z- The set of negative integers
D4: Ze The set of even integers
D5: Zo The set of odd integers
D6 {0} The unit set containing zero
D7: #S The cardinal of a general finite or infinite set S
D8: v Set union
D9: ^ Set intersection
D10: + Arithmetic addition
D11: - Arithmetic subtraction
Theorem: A + B = #(AvB) - #(A^B) where A and B are finite or infinite
sets*
This is false. In fact it's "not even false" - the _difference_ of two
infinite cardinals is undefined. Below you do not demonstrate
any problems with actual set theory - the problems you demonstrate
are just problems with your misunderstood version of set theory
(you give an illustration of _why_ the difference of two cardinals
is undefined.)
Quote: Proposition (i): The cardinal of Z is odd
There is also no notion of "odd" or "even" for infinite cardinals
in _actual_ set theory.
Quote: Z- v {0} v Z+ = Z
LHS accounts for all the integers.
#Z+ = #Z-
The cardinal of Z+ is equal to the cardinal of Z- without recourse to the
inductive principle of one-one mapping. Using the principle of symbolic
substitution "+" -> "-" demonstrated this assertion.
If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
If #Z+ has odd parity then #Z- has odd parity too.
In either case, #Z will have odd parity.
Proposition (i) has been demonstrated.
Proposition (ii): Either the set of even numbers or the set of odd integers
can be mapped to the set of all integers but not both.
Ze v Zo = Z
By theorem (i), The cardinal of Z is odd
The cardinals of both Ze and Zo must have different parities
If Ze can be mapped one-one to Z then Zo cannot and vice versa.
Proposition (ii) has been demonstrated.
* Objectors must show why this theorem cannot be applied to infinite sets.
No, first you need to say what these things mean. Give a _definition_
of the difference of two infinite cardinals, and then give a _proof_
of that "theorem".
************************
David C. Ullrich |
|
|
| Back to top |
|
| David C. Ullrich |
| Posted: Wed Jan 07, 2004 6:51 am |
|
|
|
Guest
|
On Wed, 7 Jan 2004 07:11:39 -0000, "ChrisX" <spambin@asarian-host.org>
wrote:
Quote:
"Barb Knox" <see@sig.below> wrote in message
news:btg5o9$uoh$1@lust.ihug.co.nz...
In article <3ffb86be_1@news.iprimus.com.au>,
"Tony Thomas" <verdigris@iprimus.com.au> wrote:
[...]
Theorem: A + B = #(AvB) - #(A^B) where A and B are finite or infinite
sets*
I expect you mean something like #A + #B = #(AvB) + #(A^B),
since "^" and "v" only apply to sets and "+" only applies to numbers.
It seems false to say either
#A + #B = #(AvB) - #(A^B)
That's "not even false", because the difference of two infinite
cardinals is undefined, so the assertion is simply meaningless.
(Elsewhere in the thread we see _why_ the difference is
undefined...)
Quote: or
#A + #B = #(AvB) + #(A^B)
Can you give an example where this fails?
(Hint: no you can't. If A and B are both finite it's true.
If A is infinite and B is finite then both sides are
equal to #A. If A and B are both infinite then both
sides equal max(#A, #B).)
************************
David C. Ullrich |
|
|
| Back to top |
|
| Aatu Koskensilta |
| Posted: Wed Jan 07, 2004 7:00 am |
|
|
|
Guest
|
David C. Ullrich wrote:
Quote: There is also no notion of "odd" or "even" for infinite cardinals
in _actual_ set theory.
But if we extend the usual definitions to give
1. A cardinal k is odd if and only if there is a cardinal number
l, s.t. k = 2*l+1
2. A cardinal k is even if and only if there is a cardinal number
l, s.t. k = 2*l
we can easily prove not only that every large cardinal is odd, but also
that all of them are even! In face of this horrid contradiction, the
whole edifice of contemporary large cardinal research crumbles.
--
Aatu Koskensilta (aatu.koskensilta@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus |
|
|
| Back to top |
|
| William Elliot |
| Posted: Wed Jan 07, 2004 8:36 am |
|
|
|
Guest
|
On Wed, 7 Jan 2004, Aatu Koskensilta wrote:
Quote: David C. Ullrich wrote:
There is also no notion of "odd" or "even" for infinite cardinals
in _actual_ set theory.
But if we extend the usual definitions to give
1. A cardinal k is odd if and only if there is a cardinal number
l, s.t. k = 2*l+1
2. A cardinal k is even if and only if there is a cardinal number
l, s.t. k = 2*l
we can easily prove not only that every large cardinal is odd, but also
that all of them are even! In face of this horrid contradiction, the
whole edifice of contemporary large cardinal research crumbles.
No no, you misunderstand.
An ordinal beta is even when some eta with beta = eta + eta
and odd when some eta with beta = eta + eta + 1.
How odd, all odd ordinals are successor ordinals.
Now cardinals are initial ordinals, thus they're even or odd accordingly
the initial ordinal is odd or even.
The first infinite cardinal omega_1, is neither odd or even.
Likewise all initial ordinals are neither odd or even.
Now omega+1 + omega+1 = omega + omega + 1
Also omega+1 + omega+1 is even
and omega + omega + 1 is odd.
Similarly
all infinite odd ordinals are even
Thus there's more even ordinals than odd ones.
If chi is a limit ordinal
chi + chi is even; chi+chi + 1 odd, also even
chi + chi + chi is neither odd nor even
chi + chi + chi + 1 is neither odd nor even
chi+1, chi+2, chi+3, ... neither odd nor even
chi + chi + n, for finite n>1, odd and even
if xi < chi
chi+xi + chi+xi = chi+chi + xi is even |
|
|
| Back to top |
|
| ChrisX |
| Posted: Wed Jan 07, 2004 9:41 am |
|
|
|
Guest
|
"William Elliot" <marsh@xyzt.org> wrote in message
news:20040107014400.K98684@agora.rdrop.com...
Quote: On Wed, 7 Jan 2004, ChrisX wrote:
"Barb Knox" <see@sig.below> wrote in message
"Tony Thomas" <verdigris@iprimus.com.au> wrote:
D1: Z; The set of all integers
D2: Z+ The set of positive integers
D3: Z- The set of negative integers
D7: #S The cardinal of a general finite or infinite set S
D8: v Set union
D9: ^ Set intersection
D10: + Arithmetic addition
D11: - Arithmetic subtraction
It seems false to say either
#A + #B = #(AvB) - #(A^B)
This is false. Consider when A = B = Z
Then one has #Z + #Z = #Z - #Z
So it is, thank you for correcting me William. Tony's theorem is even more
awfully false than I thought, isn't it.
Quote:
or
#A + #B = #(AvB) + #(A^B)
unless #(A^B) = 0
This is true. The commone elements of A and B get counted twice
once in #A and again in #B.
#Z+ = #Z-
The cardinal of Z+ is equal to the cardinal of Z- without recourse to
the
inductive principle of one-one mapping.
Huh?
How?
for f:Z+ -> Z-, let f(n) = -n.
f is a bijection between Z+ and Z-.
Thus Z+ and Z- are equinumerous.
So induction isn't used.
Thank you for introducing me gently to a new word William. Now I've looked
up
bijection and think I understand it.
What I still don't understand is, how to make sense of the claim that a
function
is a bijection between infinite sets, without recourse to the inductive
principle. I
can understand clearly that f(n) = -n is a bijection for some finite sets,
and I
readily believe that it is a bijection for some infinite sets, but I do not
understand
how that second belief makes sense without invoking recursion.
Quote:
If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
Agreed.
I never saw the word parity used in that sense before. Perhaps my IT
background is confusing me.
Parity of A means the (odd or even)-ness of #A.
There is no parity for infinite sets.
The rest of OP's posts misses this point.
Thank you William. Like you I cannot think how it would be useful to say
whether the cardinality of an infinite set is odd or even.
In Information Technology a number has even parity exactly when its binary
representation has an even number of 1's. This is clearly not the same
concept
at all, since in the sense I am used to both 5 and 6 have even parity
whereas 4
and 7 both have odd parity. I must admit I still can't quite understand why
one
would want a definition of 'parity' that simply substitutes 'has even
parity' for the
less verbose phrase 'is even' , but I won't mention that in case anyone
thinks
(rightly) that I am being picky.
But could you explain to me how one evaluates bijection over infinite sets
without using recursion? Or perhaps you can point me to the right area of
study to work it out for myself?
My sincere thanks
Love and respect
for Tony, Barb and William, if ok
from Chris |
|
|
| Back to top |
|
| ChrisX |
| Posted: Wed Jan 07, 2004 9:47 am |
|
|
|
Guest
|
"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
news:eesnvvsd079qdmhn7li3h1kp5d8hmtaqp7@4ax.com...
Quote: On Wed, 7 Jan 2004 07:11:39 -0000, "ChrisX" <spambin@asarian-host.org
wrote:
"Barb Knox" <see@sig.below> wrote in message
news:btg5o9$uoh$1@lust.ihug.co.nz...
In article <3ffb86be_1@news.iprimus.com.au>,
"Tony Thomas" <verdigris@iprimus.com.au> wrote:
[...]
Theorem: A + B = #(AvB) - #(A^B) where A and B are finite or infinite
sets*
I expect you mean something like #A + #B = #(AvB) + #(A^B),
since "^" and "v" only apply to sets and "+" only applies to numbers.
It seems false to say either
#A + #B = #(AvB) - #(A^B)
That's "not even false", because the difference of two infinite
cardinals is undefined, so the assertion is simply meaningless.
(Elsewhere in the thread we see _why_ the difference is
undefined...)
Thank you David, I should not have said it seems false, I should have
said it does not seem true.
Quote:
or
#A + #B = #(AvB) + #(A^B)
Can you give an example where this fails?
(Hint: no you can't. If A and B are both finite it's true.
If A is infinite and B is finite then both sides are
equal to #A. If A and B are both infinite then both
sides equal max(#A, #B).)
I take your point regarding infinite sets David. With finite sets I already
gave an example where it fails:
because for finite sets, if A^B is not empty, #(A^B) < #A + #B
Wishing you love and respect
Chris |
|
|
| Back to top |
|
| ChrisX |
| Posted: Wed Jan 07, 2004 10:03 am |
|
|
|
Guest
|
"ChrisX" <spambin@asarian-host.org> wrote in message
news:bth6i6$hmo$2@news8.svr.pol.co.uk...
Quote: because for finite sets, if A^B is not empty, #(A^B) < #A + #B
when he should have written
Quote: because for finite sets, if A^B is not empty, #(AvB) < #A + #B
and he's very sorry for being so careless
Love and respect
Chris |
|
|
| Back to top |
|
| George Greene |
| Posted: Wed Jan 07, 2004 12:54 pm |
|
|
|
Guest
|
: In article <3ffb86be_1@news.iprimus.com.au>,
: "Tony Thomas" <verdigris@iprimus.com.au> wrote:
:
: >The principle of one-one correspondence is indispensable to the theory of
: >infinite sets.
Right.
: >Even if one-one mapping, as usually applied to infinite sets, cannot be
: >refuted as a procedure neither can it be asserted as necessarily valid. If,
: >however, it can be shown that the consequences of inductive one-one mapping
: >contradict a more defensible method, then it should be abandoned.
This obligates Tony to CAREFULLY describe an alternative method.
: >The following examples illustrate the weakness of Cantorian Set Theory and
: >its variants on this account.
: >
: >D1: Z; The set of all integers
: >D2: Z+ The set of positive integers
: >D3: Z- The set of negative integers
: >D4: Ze The set of even integers
: >D5: Zo The set of odd integers
: >D6 {0} The unit set containing zero
: >D7: #S The cardinal of a general finite or infinite set S
: >D8: v Set union
: >D9: ^ Set intersection
: >D10: + Arithmetic addition
: >D11: - Arithmetic subtraction
: >
: >Theorem: A + B = #(AvB) - #(A^B) where A and B are finite or infinite sets*
Barb Knox <see@sig.below> writes:
: I expect you mean something like #A + #B = #(AvB) + #(A^B),
: since "^" and "v" only apply to sets and "+" only applies to numbers.
Right.
But why is this a "Theorem:"? Theorems are supposed to have PROOFS!
Where are the axioms or premises from which this is being derived?
At the moment, this looks to me more like an Axiom or a Definition than
a Theorem.
: >Proposition (i): The cardinal of Z is odd
: >
: >Z- v {0} v Z+ = Z
: >
: >LHS accounts for all the integers.
: >
: >#Z+ = #Z-
: >
: >The cardinal of Z+ is equal to the cardinal of Z- without recourse to the
: >inductive principle of one-one mapping. Using the principle of symbolic
: >substitution "+" -> "-" demonstrated this assertion.
"The principle of symbolic substitution" is not a reasonable alternative
to "the existence or use of the inductive principle of 1-1 mapping".
I think you need to pause to define "the inductive principle".
But I want to defend Tony vs. Barb in what follows:
Barb Knox <see@sig.below> writes:
: What is your justification for this alleged "principle of symbolic
: substitution"?
It does NOT need justification. It's part of the AXIOMATIC framework.
AXIOMS are by definition JUST ASSUMED. His justification is that
YOU can't derive a contradiction from it (YES, the burden of proof
is on YOU, since this is an AXIOM; it's on HIM for the THEOREMS) and
that it doesn't share the shortfalls and flaws of competing approaches.
: And why bother, since #Z+ clearly = #Z- via the mapping
: {1,2,3,...}<->{-1,-2,-3,...}.
Because "mappings" DON'T NECESSARILY *exist*, THAT'S why.
Because the machinery you have to invoke to define "mapping"
is both a) unnecessarily huge and complex; I mean, you need ZFC,
and before you know it, you're all the way up to inaccessible cardinals,
while you can't even decide the continuum hypothesis; and b) known
to give counter-intuitive results, such as that a proper superset of
some infinite set, the set, and a proper subset of it are all the same size.
This cannot happen for finite sets. The goal here is to decrease the
number of surprising differences between finite and infinite sets.
Another counter-intuitive result is that mappings that you KNOW must
exist can get LEFT OUT OF countable models.
: >If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
:
: Agreed.
:
: >If #Z+ has odd parity then #Z- has odd parity too.
:
: Agreed.
Then that is just the end of it! PLEASE!
: However, NEITHER Z+ nor Z- have even parity OR odd parity.
IN CLASSICAL LOGIC, YES THEY DO.
: You have not
: shown that infinite sets have ANY parity,
He has, however, shown that IT DOESN'T MATTER, and that IS
sufficient! With enough other axioms, of course.
: and in actual fact they don't.
You are claiming that Ex[ N(x) & par(Z+)=x ] leads to
a contradiction in his system. In that case, the burden of proof
is on YOU to derive that contradiction. N(x) above means tht
x is a cardinal number.
: >In either case, #Z will have odd parity.
:
: Not proven (and in fact false, since Z# has NO parity).
You can't prove THAT, either, Barb.
: >Proposition (i) has been demonstrated.
:
: No.
In classical first-order languages, all functions and predicates are total.
If Ax[par(x)=0 v par(x)=1] is an axiom of this system, then Z+
DOES have a parity, DESPITE the fact that he can't prove which one it is.
And from this it DOES follow that Z's parity is odd. Of course, combined
with all his other guidelines and axioms and assumptions, it may also
follow that P&~P. But it is up to his OPPONENTS to show THAT.
: >Proposition (ii): Either the set of even numbers or the set of odd integers
: >can be mapped to the set of all integers but not both.
: >
: >Ze v Zo = Z
: >
: >By theorem (i), The cardinal of Z is odd
:
: Except that (i) is bogus, since none of Z+, Z-, or Z have ANY parity.
:
: >The cardinals of both Ze and Zo must have different parities
: >
: >If Ze can be mapped one-one to Z then Zo cannot and vice versa.
: >
: >Proposition (ii) has been demonstrated.
:
: No.
:
: >* Objectors must show why this theorem cannot be applied to infinite sets.
Objectors need show only that Tony wouldn't know what a Theorem was if
it walked up to him and slapped him. Before you can have theorems,
you have to have axioms. For this treatment, Tony has to DEFINE
"integer", "cardinal number" (not just the operator, but its result-type),
and "the principle of symbolic substitution". That principle cannot just
be alluded to in natural language; it has to be EXPRESSED in formal language.
That is going to take you very far afield from traditional logic, which
does NOT care about the names you use for things. If you decide to define
numbers as strings then you are going to have to import a whole string
theory into this in order to express the principle. YOU WILL Be every
bit as ugly as ZFC before this is over. It is NOT worth it.
--
---
"It's difficult ... you need to be united to have any
strength, but internal issues have to be addressed."
--- E. Ray Lewis, on liberalism in America |
|
|
| Back to top |
|
| George Greene |
| Posted: Wed Jan 07, 2004 12:58 pm |
|
|
|
Guest
|
: On Wed, 7 Jan 2004 14:07:31 +1000, "Tony Thomas"
: <verdigris@iprimus.com.au> wrote:
: ...
: >If #Z+ has even parity then #Z- has even parity, since #Z+ = #Z-.
: >If #Z+ has odd parity then #Z- has odd parity too.
: >In either case, #Z will have odd parity.
: >
: >Proposition (i) has been demonstrated.
: >
: >
: >
: >Proposition (ii): Either the set of even numbers or the set of odd integers
: >can be mapped to the set of all integers but not both.
: >
: >Ze v Zo = Z
: >
: >By theorem (i), The cardinal of Z is odd
: >
: >The cardinals of both Ze and Zo must have different parities
: >
: >If Ze can be mapped one-one to Z then Zo cannot and vice versa.
: >
: >Proposition (ii) has been demonstrated.
: >
: >* Objectors must show why this theorem cannot be applied to infinite sets.
David C. Ullrich <ullrich@math.okstate.edu> writes:
: No, first you need to say what these things mean. Give a _definition_
: of the difference of two infinite cardinals, and then give a _proof_
: of that "theorem".
I just wanted to reply to that with "Thank you very much".
It is important to encourage new people to realize that
there are standard ways of conducting inquiries of this type.
I have just endured a HORRIBLE flamewar arising out of somebody's
reaction to my insistence that he give clear axioms and definitions
by saying that I had succumbed to the "definist fallacy".
Let us pray that Tony will not invoke anything like THAT.
--
---
"It's difficult ... you need to be united to have any
strength, but internal issues have to be addressed."
--- E. Ray Lewis, on liberalism in America |
|
|
| Back to top |
|
| Tony Thomas |
| Posted: Wed Jan 07, 2004 7:24 pm |
|
|
|
Guest
|
Thanks Autu, you get the point.
Aleph1, Aleph2 and successive cardinals must be of even parity because they
are powersets.
The high priests dare not admit that infinite cardinals have parity because
it introduces antinomies.
Tony Thomas
"Aatu Koskensilta" <aatu.koskensilta@xortec.fi> wrote in message
news:5vSKb.2670$fI5.783@reader1.news.jippii.net...
Quote: David C. Ullrich wrote:
There is also no notion of "odd" or "even" for infinite cardinals
in _actual_ set theory.
But if we extend the usual definitions to give
1. A cardinal k is odd if and only if there is a cardinal number
l, s.t. k = 2*l+1
2. A cardinal k is even if and only if there is a cardinal number
l, s.t. k = 2*l
we can easily prove not only that every large cardinal is odd, but also
that all of them are even! In face of this horrid contradiction, the
whole edifice of contemporary large cardinal research crumbles.
--
Aatu Koskensilta (aatu.koskensilta@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
|
|
|
| Back to top |
|
| |
Page 1 of 5 Goto page 1, 2, 3, 4, 5 Next
All times are GMT - 5 Hours
The time now is Sun Sep 07, 2008 4:56 am
|
|