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Deeper than primes

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epix said:
Did you figure what that C in
http://en.wikipedia.org/wiki/Axiom_of_power_set
stands for, or are you going to twist around just the part that has been translated? That C is really Trojan Horse -- in your game of pretense.

Why don't you bless the forum with scholarly discourse on the difference between the version stated in Wiki and the one which is listed here:
http://mathworld.wolfram.com/AxiomofthePowerSet.html



What cardinality is it that, erm, set of yours? Pardon me? Let me see . . . Well, perhaps this could be right for you.

(wild horses, couldn't drag me away . . .)
:D
Click to expand...

http://www.fact-index.com/a/ax/axiom_of_power_set.htm :

∀ A, ∃ B, ∀ C, C ∈ B ↔ (∀ D, D ∈ C → D ∈ A);

or in words:

Given any set A, there is a set B such that, given any set C, C is a member of B if and only if, given any set D, if D is a member of C, then D is a member of A.
 
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Well, whatever else... this thread *is* a rigorous proof for Doron that there are an infinite number of posts points on a circle; we passed the same arguments and the same evasions a couple of times... I can see Epix doing what I did a while ago, and jsfisher... you *must* be into mental BDSM... ;)

But Doron... what is it that you hope to achieve? Convert anyone else to TM/OM? Do groundbreaking work on a new math that isn't a new math? What?
 
realpaladin said:
... jsfisher... you *must* be into mental BDSM... ;)
Click to expand...
He has been sucked into the vortex...

Is this the longest "Yes it is", "No it isn't" thread in history yet?

But Doron... what is it that you hope to achieve? Convert anyone else to TM/OM? Do groundbreaking work on a new math that isn't a new math? What?
Click to expand...
I remember asking that a long time ago :)
 
doronshadmi said:
Since subsets are not considered at all...
Click to expand...


I see. So, once again, you abandon well established definitions in favor of your own misunderstandings. What you are posting isn't Mathematics at all, just something else you invented to conceal your own ignorance.

If you ever want to get back to what's actually part of Mathematics, we can discuss things, but as long as you continue to explore the land of make-believe that is Doronetics, continue by yourself.
 
doronshadmi said:
http://www.fact-index.com/a/ax/axiom_of_power_set.htm :

∀ A, ∃ B, ∀ C, C ∈ B ↔ (∀ D, D ∈ C → D ∈ A);

or in words:

Given any set A, there is a set B such that, given any set C, C is a member of B if and only if, given any set D, if D is a member of C, then D is a member of A.
Click to expand...
My question concerned the difference between defining terms stated here
http://en.wikipedia.org/wiki/Axiom_of_power_set
and here
http://mathworld.wolfram.com/AxiomofthePowerSet.html

In other words, what is the difference between D being a member of A and D being a subset of A. Isn't it so that power set P(S) is made of subsets of S?

Clue: think of the empty set in P(S). Going against common sense has the same effect even on the hotshot number theorists. In that case . . .
"Subset is not used in the formal definition above because the axiom of power set is an axiom that may need to be stated without reference to the concept of subset."
or
"Note that the version given by Itô (1986, p. 147), is confusing, and possibly incorrect."

Your Trojan Horse was eased on the backstretch.
 
jsfisher said:
I see. So, once again, you abandon well established definitions in favor of your own misunderstandings. What you are posting isn't Mathematics at all, just something else you invented to conceal your own ignorance.

If you ever want to get back to what's actually part of Mathematics, we can discuss things, but as long as you continue to explore the land of make-believe that is Doronetics, continue by yourself.
Click to expand...
Mathematics is not a privet property of any dogmatic community.

your reply demonstrates your inability to get generalization about sets and power sets.
 
realpaladin said:
Well, whatever else... this thread *is* a rigorous proof for Doron that there are an infinite number of posts points on a circle; we passed the same arguments and the same evasions a couple of times... I can see Epix doing what I did a while ago, and jsfisher... you *must* be into mental BDSM... ;)

But Doron... what is it that you hope to achieve? Convert anyone else to TM/OM? Do groundbreaking work on a new math that isn't a new math? What?
Click to expand...

http://www.internationalskeptics.com/forums/showpost.php?p=6736887&postcount=13841
 
epix said:
Clue: think of the empty set in P(S).
Click to expand...
All we need is to use <0,1> as the common form of any non-empty set, whether it is finite or not, for example:

We start from the empty set {}.

The power set of {} is {0}.

The diagonal number that is not in the range of nothing, is 0.

The power set of set {0} is {0,1}.

The diagonal number (which has a single symbol, at this stage) that is not in the range of 0, is 1 (or 0, if 1 is considered as the single symbol).

The power set of set {0,1} is
{
00,
01,
10,
11
}
and the members of S are partial collections of any possible 2 distinct members that have a common <0,1> form with P(S) distinct members, for example:

S =
{
10,
11
}
or
{
00,
10
}


etc ... , where given any S version of 2 members , there is a diagonal object that is based on <0,1> form ( which is common for both S and P(S) ) that is not in the range of S, but it is in the range of P(S).

But also P(S) is a set that has a common <0,1> form with P(P(S)), for example:

0 1 0 1
0 0 1 1
-------
0 0 0 0
1 0 0 0
0 1 0 0
1 1 0 0
0 0 1 0
1 0 1 0
0 1 1 0
1 1 1 0
0 0 0 1
1 0 0 1
0 1 0 1
1 1 0 1
0 0 1 1
1 0 1 1
0 1 1 1
1 1 1 1

and in this case some 4 P(S) objects ( which are partial case of P(P(S)) ) are:

P(S)=
{
0 1 1 0,
1 1 1 0,
0 0 0 1,
1 0 0 1
}
or
{
0 1 0 0,
0 0 1 1,
1 1 1 1,
0 0 1 0
}

etc ... , where given any P(S) version of 4 members , there is a diagonal object that is based on <0,1> form ( which is common for both P(S) and P(P(S)) ) that is not in the range of P(S), but it is in the range of P(P(S)).



The same reasoning works also if S has an infinite size, as follows:

S=
{
.0 1 1 0 ...,
.1 1 1 0 ...,
.0 0 0 1 ...,
.1 0 0 1 ...,
...
}
or
{
.0 1 0 0 ...,
.0 0 1 1 ...,
.1 1 1 1 ...,
.0 0 1 0 ...,
...
}

etc ... are partial cases of P(S) where the diagonal member is not in the range of any S version ( although both S and P(S) have a common <0,1> form ).

By using <0,1> as a common form for both sets and powersets (finite or not), we discover that the ZFC axiom of powerset is actually a "Trojan horse" , which defines some object of ZFC that have the properties of a given set (it obeys the construction rules of the given framework) but it is not in the range of the given set (but can't be proved within this framework), exactly as Godel's first incompleteness theorem demonstrates.

epix said:
Isn't it so that power set P(S) is made of subsets of S?
Click to expand...
It is not a requirement epix, for example: No subsets of S are used as the members of P(S), because both of them have members of the same form.

Only the size of S and P(S) is different.

Here is again the example of this notion:

The power set of set {0,1} is
{
00,
01,
10,
11
}
and the members of S are partial collections of any possible 2 distinct members that have a common <0,1> form with P(S) distinct members, for example:

S =
{
10,
11
}
or
{
00,
10
}


etc ... , where given any S version of 2 members , there is a diagonal object that is based on <0,1> form ( which is common for both S and P(S) ) that is not in the range of S, but it is in the range of P(S).
 
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EDIT:

Let P=“This sentence has no proof” within a given framework.

If P is true, then there is P that has no proof within a given framework.

If P is false, then it is provable.

If P is provable then there is a proof that contradicts P, so also in this case it is true that there is P that has no proof within a given framework.

In the case of S and P(S), by using the diagonal method on S members, we define an object that has the same form of S members but it is not a member of S and we need P(S) in order to define its membership.

But also in the case of P(S) and P(P(S)), by using the diagonal method on P(S) members, we define an object that has the same form of P(S) members but it is not a member of P(S) and we need P(P(S)) in order to define its membership.

Etc. … ad infinitum.

It is equivalent to the case where some framework has true P but it is not provable within this framework and we need to extend the framework in order to prove it, but then there is a true P of the extended framework that is not provable within the extended framework, which has to be extended in order to prove it, but then ... etc. … ad infinitum.
 
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doronshadmi said:
Mathematics is not a privet property of any dogmatic community.

your reply demonstrates your inability to get generalization about sets and power sets.
Click to expand...


More of a complete disregard rather than a generalization. You seem to confuse those two concepts.

And as I have said before, you are welcome to invent whatever fantasies you like for Doronetics, just don't claim them to be real Mathematics.
 
doronshadmi said:
All we need is to use <0,1> as the common form of any non-empty set, whether it is finite or not, for example:

We start from the empty set {}.

The power set of {} is {0}.
Click to expand...

If and only if 0 represents the empty set. That is, 0 = {}. So, S = {} and P(S) = {{}}, or in Doron's extra special notation, S = 0 and P(S) = {0}. Hurray for useless notational substitutions.

The power set of set {0} is {0,1}.
Click to expand...

If and only if 1 represents {0}. Now, we have S = {0} and P(S) = {{},{0}}, or alternately, S = 1 and P(S) = {0,1}. Wow, this is moving fast, now.

The power set of set {0,1} is
{
00,
01,
10,
11
}
Click to expand...

Hmmm, well, S = {0,1} and therefore P(S) = {{}, {0}, {1}, {0,1}}, but it is anyones guess the correspondence between elements of P(S) and 00, 01, 10, and 11. Doron will deny there is any correspondence; after all, this is a generalization.

As already established, for Doron, generalization and complete misunderstanding are synonyms.

...and the members of S are partial collections of any possible 2 distinct members that have a common <0,1> form with P(S) distinct members, for example:

S =
{
10,
11
}
or
{
00,
10
}
Click to expand...

Well, isn't that interesting. We started with S = {0,1}, but through the special process of generalization we have allowed S to morph into either of two completely different possibilities.

Apparently, in Doronetics, constants aren't. How useful.
 
doronshadmi said:
The power set of set {0} is {0,1}.
Click to expand...
I see you've made some changes to the rule, which says that

if S = {0} then P(S) = {{},{0}}.

But that's a negligible adjustment in comparison to more profound alterations, which open the gate to many unexpected possibilities:

After several schools of the philosophy of mathematics ran into difficulties one after the other in the 20th century, the assumption that mathematics had any foundation that could be stated within mathematics itself began to be heavily challenged.

One attempt after another to provide unassailable foundations for mathematics was found to suffer from various paradoxes (such as Russell's paradox) and to be inconsistent: an undesirable situation in which every mathematical statement that can be formulated in a proposed system (such as 2 + 2 = 5) can also be proved in the system.
Click to expand...

the math theorists can ruin your life
coz two and two is sometimes five
sometimes three but seldom four
the ambulance is at your door.

coo-coo...........coo-coo...............coo-coo


smiley_lol.gif
 
jsfisher said:
If and only if 0 represents the empty set. That is, 0 = {}. So, S = {} and P(S) = {{}}, or in Doron's extra special notation, S = 0 and P(S) = {0}. Hurray for useless notational substitutions.
Click to expand...

jsfisher, you still do not get the generalization, where no subsets are used, and the diagonal number is a member of P(S) that is not in the range of S, where any non-empty S and its P(S) have members that have a common form.

Again:

We start from the empty set {}.

The power set of {} is {0}.

The diagonal number that is not in the range of nothing, is 0, and in this case there is no common form of members because nothing is not something, so our argument starts to work among non-empty sets and their power sets, as follows:

The power set of set {0} is {0,1} (no subset is used, and the S and P(S) have members that are based on a common form.

The diagonal number (which has a single symbol, at this stage) that is not in the range of 0, is 1 (or 0, if 1 is considered as the single symbol).

jsfisher said:
If and only if 1 represents {0}.
Click to expand...
Again, subsets are not used by this generalization, which according to it P(S) and S members are based on the same form.

jsfisher said:
Hmmm, well, S = {0,1} and therefore P(S) = {{}, {0}, {1}, {0,1}}, but it is anyones guess the correspondence between elements of P(S) and 00, 01, 10, and 11. Doron will deny there is any correspondence; after all, this is a generalization.
Click to expand...
Your mind can't rid of sub sets, isn't it jsfisher? You simply can't get <0,1> form such that P(S)={00,01,10,11} and S is some two members of P(S) (for example: {00,10} , {11,10} , etc ... where a given diagonal form is not in the range of a given S).

jsfisher said:
As already established, for Doron, generalization and complete misunderstanding are synonyms.
Click to expand...
As already established, for jsfisher, non-generalization and sub sets are synonyms.


jsfisher said:
Well, isn't that interesting. We started with S = {0,1}, but through the special process of generalization we have allowed S to morph into either of two completely different possibilities.
Click to expand...
By generalization, where both S and P(S) have the same form, all we care is the size of S and P(S).

jsfisher said:
Apparently, in Doronetics, constants aren't. How useful.
Click to expand...
It is very useful. By this generalization we are able to understand that incompleteness is an essential requirement of the consistency of any non-empty collection, whether it is finite or not.

This notion enables us to understand that completeness is an essential requirement of the consistency of Emptiness and Fullness.

--------------

Why do you ignore http://www.internationalskeptics.com/forums/showpost.php?p=6756356&postcount=13911 ?
 
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doronshadmi said:
Yes they do.

They are preliminary steps, which their goal is to develop a comprehensive framework where both objective and subjective aspects of the researched environment reinforce each other into a one organic realm.
Click to expand...

Good, so... stop writing nonsense and get on with the show.

Also... I see 'subjective' in there... why don't you accept the subjective truth of all co-posters in this thread that:

Doron Shadmi does not understand a single solid thing about math.

This is not attacking the arguer, it is just the bringing in of a subjective aspect of the researched environment (in casu, this thread).
 
jsfisher said:
No, Doron. You have not generalized anything. You have simply ignored the meaning of something and substituted something else.
Click to expand...

No, jsfisher. You have simply ignored generalization because you can't rid of sub sets.

By generalization, both S and P(S) have the same form (no sub sets are used), and all we care is the size of S and P(S).
 
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