The power of X

A power set is a set of all subsets of a set.

D is the set of all non-selfish natural numbers. It is not a power set.
For example if 1 and 2 were the only non-selfish natural numbers then D = {1,2}. If D was a power set of these then D = { {}, {1}, {2}, {1,2} }.

See the difference?

No D is a power set, but any D is a member of a power set and no power set exists without its members.

Cantor's proof is based on the existence of members of sets or powersets.

Try to do it by ignoring the members.

[latex]$$ D = \{ \, x \in X: x \not \in f(x) \, \} $$[/latex]​

D must be a subset of X and also a member of Power(X), and no D is disconnected from the other D's (or in other words, we define Power(X) as an inseparble part of a proof about Power(X), which is invalid).
 
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I do not think that a proof is a valid proof if we construct our examined objects, and also doing it in such a way that will lead us to some requested result.

Ok, fine. Why do you reject such proofs?
 
Cantor's proof is based on the existence of members of sets or powersets.

well duh! it's a proof about sets and power sets. How can one avoid the properties of sets and power sets when discussing the properties of sets and power sets?
 
In what way is it circular? Is it you don't understand proof by contradiction?
You have missed it.

We do not need any proof by contradiction here, because in this case Cantor directly provides us a way to define Power(X) in such a way that |Power(X)| > |X|, no matter if X is finite or not.

Furthermore, we do no have to conclude that a non-finite collection of distinct members is complete (we do not have to use the term "for all", but the term "for each" where |Power(X)| > |X| is a permanent state even if no one of the members of X or Power(X) is their final member).

By using this notion we also avoid Cantor's paradox of the power set of all power sets that is greater than itself, and we do not need any proper classes in order to avoid the paradox.

The largest cardinal does not exist because a set is an incomplete object.

At http://www.internationalskeptics.com/forums/showthread.php?t=108957 I provided the logical foundations of this incompleteness, which is based on simpler foundations that provide richer and more interesting results, at the moment that they understood.
 
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We do not need any proof by contradiction here, because in this case Cantor directly provides us a way to define Power(X) in such a way that |Power(X)| > |X|, no matter if X is finite or not.
Where did Cantor do this?

Furthermore, we do no have to conclude that a non-finite collection of distinct members is complete (we do not have to use the term "for all", but the term "for each" where |Power(X)| > |X| is a permanent state even if no one of the members of X or Power(X) is their final member).
What do you mean by 'complete', and 'permanent state' here. Also what has the order of set members have to do with anything (IIRC sets are unordered).

By using this notion we also avoid Cantor's paradox of the power set of all power sets that is greater than itself, and we do not need any proper classes in order to avoid the paradox.

Huh? Are you saying that for some sets |Power(X)| ≯ |X| ?

ETA:sigh, once again Doron persists in editing his posts after the fact ...

The largest cardinal does not exist because a set is an incomplete object.
In what way is {1} an incomplete object?

At http://www.internationalskeptics.com/forums/showthread.php?t=108957 I provided the logical foundations of this incompleteness.
no, you didn't. You spectacularly failed at convincing anyone and contradicted yourself at several points (where any meaning was apparent).
 
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Where did Cantor do this?
If you have somthing to say about what I wrote, than say it. Please avoid such "questions".

What do you mean by 'complete', and 'permanent state' here. Also what has the order of set members have to do with anything (IIRC sets are unordered).
How is talking here about ordinals? We are talking about Cardinals, and the largest cardinal does not exist ( http://en.wikipedia.org/wiki/Cantor's_paradox ).

About "permanent state", |Power(X)| > |X| no matter how many members there are.

Huh? Are you saying that for some sets |Power(X)| ≯ |X| ?
No at all. |Power(X)| > |X| no matter how many members there are.

In what way is {1} an incomplete object?
It does no have the completeness of Total symmetry at its self state.

no, you didn't. You spectacularly failed at convincing anyone and contradicted yourself at several points (where any meaning was apparent).
Yes I did.

Your abstract ability is too "noisy" and cannot get simplicity at its self state, which is naturally undefined, because any definition is a limitation of the totality (completeness) of simplicity at its self state.

As long as you unaware of this simplicity as the common basis of your thoughts, you will not get it.
 
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If you have somthing to say about what I wrote, than say it. Please avoid such "questions".

ok, so you refuse to say where Cantor 'define Power(X) in such a way that |Power(X)| > |X|, no matter if X is finite or not'.

How is talking here about ordinals? We are talking about Cardinals, and the largest cardinal does not exist ( http://en.wikipedia.org/wiki/Cantor's_paradox ).
Where did ordinals come into this?

About "permanent state", |Power(X)| > |X| no matter how many members there are.

so by 'permanent state' you mean '∀'? Is that it? What about 'complete'?

No at all. |Power(X)| > |X| no matter how many members there are.
ok, so how do we avoid 'Cantor's paradox of the power set of all power sets that is greater than itself'?

It does no have the completeness of Total symmetry at it self state.
oh good grief not that rubbish again. Please define 'completeness of Total symmetry at it self state'.

Yes I did.
yeah right <-- look two assertions combine to make a negation, like wow man!

Your abstract ability is too "noisy" and order to get simplicity at its self state, which is naturally undefined, because any definition is a limitation of the totality (completeness) of simplicity at its self state.

As long as you unaware of this simplicity as the common basis of your thoughts, you will not get it.
gibberish.
 
well duh! it's a proof about sets and power sets. How can one avoid the properties of sets and power sets when discussing the properties of sets and power sets?

Really.

The first trick of doing a mathematical proof is to either:

A) Go back to axioms you're working from.
B) Go back to the definition.

More often than not, you will go back to the definition. During some studying last night, I found myself repeating this.

"Go back to the definition of divisibility."

"Go back to the definition of divisibility."

"Use the properties."

"Use the properties."

*draws a little square.*

Actually, you will almost always be drawing from the definition or from properties that drop out of the definition with a few short steps, like say that a*0=0 when defining a field (among many, many other structures).
 
No D is a power set, but any D is a member of a power set and no power set exists without its members.

So what?

Cantor's proof starts with an arbitrary set, X. That's a given.

Given X, we know P(X) exists by the Axiom of Power Sets. So we have X and we now have P(X).

We can then image some mapping f(x) from X to P(X) that we will assume to be 1-1 and onto. We are now up to X, P(X), and f(x).

Given f(x), we can define the set, D. That puts us up to X, P(X), f(x), and D. There is no circular reasoning, no loops in definitions.

Cantor's proof is based on the existence of members of sets or powersets.

Well, yeah. Don't sets have members? (A power set is a set, by the way.)

D must be a subset of X and also a member of Power(X), and no D is disconnected from the other D's (or in other words, we define Power(X) as an inseparble part of a proof about Power(X), which is invalid).

No. There is only one set D. It is completely determined by the mapping, f(x). There are no other Ds. Moreover, the fact that D must be a member of P(X) does not define P(X). Quite the opposite. D's membership in P(X) is a direct consequence of the power set's definition.
 
We do not need any proof by contradiction here, because in this case Cantor directly provides us a way to define Power(X) in such a way that |Power(X)| > |X|, no matter if X is finite or not.


Ok, prove it. Cantor's proof is fairly compact. Show us your version of the proof with the shortcuts you suggest exist. No hand waving, no ill-defined rhetoric, just a rigorous proof, please.
 
No D is a power set, but any D is a member of a power set and no power set exists without its members.

Cantor's proof is based on the existence of members of sets or powersets.

Try to do it by ignoring the members.

latex.php
D must be a subset of X and also a member of Power(X), and no D is disconnected from the other D's (or in other words, we define Power(X) as an inseparble part of a proof about Power(X), which is invalid).

Is your D different from the D that is in the Wikipedia article?
where D is defined as "Let D be the set of all non-selfish natural numbers".

If it is can you tell us what proof of Cantor's theorem you are looking at?
 
Is your D different from the D that is in the Wikipedia article?
where D is defined as "Let D be the set of all non-selfish natural numbers".

If it is can you tell us what proof of Cantor's theorem you are looking at?

The D, here, is the same as the set B in the wikipedia article under "The Proof" heading. The non-selfish numbers stuff is in the "A detailed explanation" part.

(At least, that's what I meant by it, and doron copy/pasted the Latex from my post.)
 
ok, so you refuse to say where Cantor 'define Power(X) in such a way that |Power(X)| > |X|, no matter if X is finite or not'.


That's not a proof of anything, just a lot of hand waving.
It is not a proof but it is a direct way to define |Power(X)| > |X|.

No proof is needed here, and this is exactly my claim.
 
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It is not a proof but it is a direct way to define |Power(X)| > |X|.

No proof is needed here, and this is exactly my claim.


(a) You exhibit a collection of sets you call D (with only vague suggestions as to how a given D may have been generated), and you claim each D is a member of P(X).

Ok, so what? 2, 3 and 4 are each members of N, but that doesn't define the set N. Your D sets do not define P(X) either.

(b) You don't get to make up new definitions for already defined things. Power sets are defined (by the Axiom of Power Sets) not by your D sets.

(c) Nothing in your post deduces anything about cardinality.
 
It is not a proof but it is a direct way to define |Power(X)| > |X|.
How would you adjust your working in order to define |Power(X)| < |X|? After all, if it _is_ a definition, we can define it anyway we want.
 
(a) You exhibit a collection of sets you call D (with only vague suggestions as to how a given D may have been generated), and you claim each D is a member of P(X).

Ok, so what? 2, 3 and 4 are each members of N, but that doesn't define the set N. Your D sets do not define P(X) either.

(b) You don't get to make up new definitions for already defined things. Power sets are defined (by the Axiom of Power Sets) not by your D sets.

(c) Nothing in your post deduces anything about cardinality.

jsfisher do you remember this?

[latex]$$ D = \{ \, x \in X: x \not \in f(x) \, \} $$[/latex]​

D is not a single element but a direct way to constracut Power(X) in such a way that cannot be but |Power(X)| > |X|, no matter how many members are involved (finite or non-finite, it does not matter).

This time please read the all 3 posts one after the other, in order to get the right picture of it:

http://www.internationalskeptics.com/forums/showpost.php?p=3638296&postcount=1

http://www.internationalskeptics.com/forums/showpost.php?p=3641293&postcount=29

http://www.internationalskeptics.com/forums/showpost.php?p=3644296&postcount=39
 
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My D can be seen in http://www.internationalskeptics.com/forums/showpost.php?p=3644296&postcount=39 .

Please read this post this time instead of look at wiki's D.

It looks like your D is the B in the Wikipedia article:

B is not a power set. It is a subset of A such that for all x in A, xB if and only if xf(x) where f(x) is a function (mapping) of A to the power set of A.

Your definition of D is "the members of D are any X member that when mapped with some Power(X) member, this X member is not one of members this Power(X) member".
The mapping with some Power(X) member does not mean that the Power(X) member is a member of D - it is a condition on the inclusion of a member of X in D.
Thus D includes members of X but not subsets of X or Power(X) members. Therefore D is not a power set of X.
 
jsfisher do you remember this?

[latex]$$ D = \{ \, x \in X: x \not \in f(x) \, \} $$[/latex]​

D is not a single element but a direct way to constracut Power(X) in such a way that cannot be but |Power(X)| > |X|, no matter how many members are involved (finite or non-finite, it does not matter).

Umm, did you fail to notice that P(X) must already be defined in order to define the set D? Remember that f(x) thing? That is an onto mapping we assume exists from X to P(X). You can't turn around, then, and claim to construct P(X) from D, since P(X) had to exist in order to construct D.

In addition, Cantor had only one such set. Exactly one onto mapping f(x) (that is assumed to exist) yields exactly one set D.

But even ignoring all of that, how do you instantly conclude anything about cardinality? You "in such a way that cannot be but" is not a rigorous line of reasoning.
 
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This time please read the all 3 posts one after the other, in order to get the right picture of it:


By the way, your continued practice of telling people to reread your earlier posts is rather tiresome. Your prior posts are faulty; the faults have been pointed out; you haven't addressed the faults; there is no point rereading them.
 
How about we work from this as the reference model for Cantor's proof. The details for Steps 1 and 3 are omitted because they are easy enough to provide and don't add much to the "interesting" parts of the proof.

------------------------------------
  1. Let's take as given a standard definition of "A < B" based on one-to-one and onto mappings from A to B.

  2. Assume P(A) is defined by

    [latex]$$ \forall B \, [B \in P \iff \forall C \, (C \in B \Rightarrow C \in A)]$$[/latex]​

  3. Let's take as given that a one-to-one mapping exists from A to P(A). If no onto mapping from A to P(A) exists, then A < P(A).

  4. Assume an onto mapping exists from A to P(A). Call it f(x).

  5. Define

    [latex]$$ B = \{ \, x \in A: x \not \in f(x) \, \} $$[/latex]​

  6. Since the members of B are members if A, B must be a member of P(A).

  7. By definition of set B,

    [latex]$$ \forall x \in A: x \in B \iff x \not \in f(x)$$[/latex]​

    Therefore,

    [latex]$$ \forall x \in A: \, B \neq f(x) $$[/latex]​

  8. Therefore, a member of P(A) is not in the mapping f(x) from A to P(A).

  9. This contradicts the assumption f(x) was an onto mapping. Therefore, A < P(A).
------------------------------------

Now, doron, where in there other than Step #2 is power set defined (or definable)?

Where in there is it a matter of definition that A < P(A)?
 
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Umm, did you fail to notice that P(X) must already be defined in order to define the set D?

Not necessarily.

All we need is X and a map between some X member and some subset of X.

By doing this we get two things in a one ticket:

Thing 1: The collection of D versions is Power(X).

Thing 2: No D version (which is a Power(X) member) can be mapped can be mapped with any one of X members, so |Power(X)| > |X| is inevitable.
 
Your prior posts are faulty; the faults have been pointed out;

Not at all.

For example, in your last post you write:

"I must take an aside to lay my foundation so we are all on the same page."
( http://www.internationalskeptics.com/forums/showpost.php?p=3637338&postcount=777 )

I must say that I completely agree with you, because without a common basis no two thoughts are gathered into a one definition that can be used as a part of our research.

The difference between you and me is this:

You do not provide the common basis that enables two thoughts to be gathered into a one definition, as an explicit essence of your reasoning.

In that case your reasoning is based on the hidden assumption of the common basis.


On the contrary, my reasoning is aware of this common basis as the naturally undefined state of any given definition.

This common basis is naturally undefined because any thought cannot be but a particular aspect of the common basis, where the common basis at its self state is naturally free of any limitation (and a definition is -without a doubt- some categorical limitation).

Since Simplicity itself (the common basis of any given definition) is naturally undefined, we cannot use a definition in order to understand it.

In this case we have no choice but to use an analogy, but we must not mix between simplicity itself (which is naturally undefined) and some analogy of it (which is limited like any other definition).


Here is the analogy, but this time please try no to mix between the naturally undefined (simplicity at its self state) and this particular analogy, which is naturally limited like any other definition:


Simplicity at its self state is like a straight line that has no beginning, no end, and it is an atom (it is not made of any sub elements).

Any definition is like a fold along the straight line.

1. The fold is not a sub-element of the line, but it is a limited thing along the atomic line, where the atomic line is actually the basis of the fold and not vise versa.

2. The fold has no influence on the simplicity of the line as an atomic state, and on its naturally unlimited self state.

At the moment that 1 and 2 are directly understood (without any need of some fold (analogy, definition, thought, etc. ) as the simplest and naturally undefined basis of your consciousness, then and only then you are able to understand my framework.

Another analogy (which is, again, a naturally limited thing):

I try to show you a way to get a direct experience of a sea without waves, and you try to get it by making waves. By making waves, you will never get it because any wave is exactly not a sea without waves.

At the moments that you get the sea without waves, you are enable to get any wave right from the common state of any other wave, which is the sea without waves (the naturally non-limited atomic (and naturally undefined) state of any limited (and naturally defined) thing).
 
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In that case your reasoning is based on the hidden assumption of the common basis.

I think the underlying assumption everyone but you is making, is the desire to communicate effectively. You do not seem to want to communicate, yet you post here. To what purpose?
 
That isn't what Cantor did, so what's your point?
My point is this:

By providing D, Cantor does not need any proof that |Power(X)| > |X| and he also does not need to use the axiom of the power set, because he directly constructs Power(X) from X by providing D.

Simple and beautiful, isn't it?
 
Not necessarily.

All we need is X and a map between some X member and some subset of X.
So... we have X and some f(x) that produces subsets of X

It is trivial to show that at least one of these exists, e.g. f(x) = {x}, or even f(x) = {}. (i.e, constant). The problem is that without any additional conditions on the mapping, you're going to get a far larger variation of f(x) mappings than you have of actual D versions. In other words, the second part:

By doing this we get two things in a one ticket:

Thing 1: The collection of D versions is Power(X).
This is tautological - D versions are subsets of X, and Power(X) is the collection of all possible subsets of X. No need for the mappings to know this.

Thing 2: No D version (which is a Power(X) member) can be mapped can be mapped with any one of X members, so |Power(X)| > |X| is inevitable.
In other words, if we take F = the set of all possible f(x), then |F| > |X|. Quite, but it's also true that |F| > |Power(X)|, and this says nothing about the cardinalities of Power(X) and X. Just by shuffling the mapping a bit, I can easily map any arbitrary D to any member of X - and I've only treated *one* possible D here.
 
Since Simplicity itself (the common basis of any give definition) is naturally undefined, we cannot use a definition in order to understand it.
If you cannot provide a definition for a term you use, the term is meaningless. Why have you chosen to use the word 'Simplicity' for this undefinable thing? Why not use the word 'Potato'?

In this case we have no choice but to use an analogy, but we must not mix between simplicity itself (which is naturally undefined) and some analogy of it (which is limited like any other definition).
Incorrect. If 'simplicity' (is that different from 'Simplicity'? I can't tell, because you fail to define the two terms), is undefined, how can one know any analogy is accurate?

Simplicity at its self state is like a straight line that has no beginning, no end, and it is an atom (it is not made of any sub elements).

Why isn't Simplicity like a point? that has no beginning, no end and is not made of sub elements? What space is this straight line embedded in? Does it matter?

Any definition is like a fold along the straight line.
In what way can a straight line have a fold in it?

1. The fold is not a sub-element of the line, but it is a limited thing along the atomic line, where to atomic line is actually the basis of the fold and not vise versa.
Gibberish
2. The fold has no influence on the simplicity of the line as an atomic state, and on its naturally unlimited self state.
Gibberish
At the moment that 1 and 2 are directly understood (without any need of some fold (analogy, definition, thought, etc. ) as the simplest and naturally undefined basis of your consciousness, then and only then you are able to understand my framework.
Given that #1 and #2 are gibberish, I conclude your framework is gibberish.

I try to give you way to get a direct experience of a see[sea?] without waves, and you try to get it by making waves. By making waves, you will never get it because any wave is exactly not a sea without waves.
A calm sea is not exactly the same as a perturbed sea. ok.

At the moments that you get the sea without waves, you are enable to any wave from the common state of any other wave, which is the sea without waves (the naturally non-limited atomic (and naturally undefined) state of any limited (and naturally defined) thing).
This is not a sentence. Ergo, gibberish.
 
I simply take this concept from its simplest state.
No, you don't. You use sequences of words that have no meaning. That is not communication.

and to reiterate jsfisher's point. I'm getting tired of you continually telling people to reread your posts. Explain what you mean, or go away [to co-opt a well used phrase at the JREF]
 
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In other words, if we take F = the set of all possible f(x), then |F| > |X|. Quite, but it's also true that |F| > |Power(X)|, and this says nothing about the cardinalities of Power(X) and X. Just by shuffling the mapping a bit, I can easily map any arbitrary D to any member of X - and I've only treated *one* possible D here.

In that particular case F = Power(Power(X)) and |F| > |Power(X)|.

But I am talking about the genaral case of |Power(X)| > |X| where X is a placeholder of any set (and a powerset is a set (a collection of distecnt objects, where order is not important)).
 
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No, you don't. You use sequences of words that have no meaning. That is not communication.

Communication right from silence itself is the best communication, because any sound (definition) is totally clear when observed from silence itself.

No sound (definition, which is limited by nature) is a natural basis for real communication.

It can be understood only if you directly get silence itself as the simplest state of any thought.

Your struggle to find a common basis at the level of sound (definition) actually makes you tired.
 
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My point is this:

By providing D, Cantor does not need any proof that |Power(X)| > |X| and he also does not need to use the axiom of the power set, because he directly constructs Power(X) from X by providing D.

Cantor could have done many, many things, but he didn't construct the power set of X by providing D. He didn't provide the collection of sets D you claim. He needed exactly one set of that sort, so he exhibited only one

Go back to the version of Cantor's proof I posted. Please point out in it where you think the power set is "constructed" from the set D (named B in my posting).

Simple and beautiful, isn't it?

Cantor's proof? Yes, it is.

Your approach, on the other hand, is long and contorted. Moreover, you are supposed to be proving that A < P(A), so you must use the Axiom of the power set to give meaning to P(A).

The long and contorted part is where you'd need to prove that P(A) can be derived from all your D sets. You haven't done that.
 
In that particular case F = Power(Power(X)) and |F| > |Power(X)|.
Flatly incorrect. The functions f(x) are not subsets of Power(X) - even you should see that.

My point was that just because any "D version (which is a Power(X) member) [...] can be mapped with any one of X members", i.e. there are more possible f(x) definitions than members of X, this implies nothing about the cardinalities of Power(X) and X.


By the way, I had to paraphrase to make sense of the original post - your grammar is so bad that the original statement makes no sense at all.

This is what you wrote:
Thing 2: No D version (which is a Power(X) member) can be mapped can be mapped with any one of X members
If you meant that no D version can be mapped with any one X member, then D violates its own definition.
If you meant that any D version can be mapped with any one X member, well, that follows from the definition of f(x). And as noted above, implies nothing for the collection of possible D versions.
 

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