The power of X

doronshadmi

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doronshadmi
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By using a proof by contradiction Cantor proved that Power(X) > X.

Let us carefully examine this proof.

A question:

If –as a part of our proof– we define E and F in such a way that E > F, can we still claim that our proof holds?

Some facts:

1) A set is a collection of distinct objects (each object is included once and only once in a given set).

2) X is a set, and so is Power(X)

3) Let set D be the result of the mapping between Power(X) and X according to this rule:

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, for example:

Code: 
X      Power(X)

a <--> {c,d}
b <--> {a,b}
c <--> {a,b,d,e,f,g,…}
d <--> {d,@,*}
…
In this particular example (out of infinitely many examples) D = {a, c, …}

There are two extreme versions of D which are:

D={} or D={a, b, c, d, …} , where each one of them is the result of infinitely many different mappings between Power(X) and X, but the general thing is this:

D={} only if each X member that is mapped with some arbitrary Power(X) member, is also a member of this arbitrary Power(X) member.

D={a, b, c, d, …} only if each X member that is mapped with some arbitrary Power(X) member, is not a member of this arbitrary Power(X) member.

Between D={} and D={a, b, c, d, …} we can define the infinitely many D results which are not D={} or D={a, b, c, d, …}.

So Cantor's method actually defines Power(X)={{}, … , {a, b, c, d, …}} as a part of a proof that says something about Power(X) (and in this case: Power(X) > X).

Furthermore, the result of how Power(X) is constructed by Cantor, cannot be but
Power(X) > X, because given any D (which is a Power(X) member) there is no X member that is mapped with D.

Some conclusions and questions:

1) We do not need a proof by contradiction in order to conclude that Power(X) > X,
because Cantor actually defines Power(X) (by D's) in such a way that the result cannot be but
Power(X) > X.

2) Is a proof is still a proof if we actually determine the result, as a pert of the proof?

3) Do we need the ZF axiom of the power set (after all Cantor's method actually defines it, without any need of this axiom)?
 
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doronshadmi said:
By using a proof by contradiction Cantor proved that Power(X) > X.

Let us carefully examine this proof.

A question:

If –as a part of our proof– we define E and F in such a way that E > F, can we still claim that our proof holds?
Click to expand...

Since your proof nowhere uses the symbols E or F, their properties are irrelevant, and the proof still holds.

Consider the following:

All men are mortal.
Socrates is a man.
The best flavor of pie is cherry.
Therefore, Socrates is mortal.

The third axiom is irrelevant to the proof, and therefore cannot possibly affect the validity of the underlying argument.


Unfortunately, not only is your question irrelevant to your proof, but your "proof" is irrelevant to Cantor's mathematics. You need at a minimum to define "Power(X)," and if you intend to treat it as a set, you will need to show that it is a set. (That's why you need the Power Set axiom, which defines Power(X) as a set if X is a set.)
 
drkitten said:
Since your proof nowhere uses the symbols E or F, their properties are irrelevant, and the proof still holds.

Consider the following:

All men are mortal.
Socrates is a man.
The best flavor of pie is cherry.
Therefore, Socrates is mortal.

The third axiom is irrelevant to the proof, and therefore cannot possibly affect the validity of the underlying argument.


Unfortunately, not only is your question irrelevant to your proof, but your "proof" is irrelevant to Cantor's mathematics. You need at a minimum to define "Power(X)," and if you intend to treat it as a set, you will need to show that it is a set. (That's why you need the Power Set axiom, which defines Power(X) as a set if X is a set.)
Click to expand...

drkitten,

E and F are some placeholders of an idea, and there is no proof here, but only an examination of Cantor's theorem.

If you get that, then please read again my first post.

Thank you.
 
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doronshadmi said:
Some facts:

1) A set is a collection of distinct objects (each object is included once and only once in a given set).

2) X is a set, and so is Power(X)

3) Let set D be the result of the mapping between Power(X) and X according to this rule:

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, for example:

Code: 
X      Power(X)

a <--> {c,d}
b <--> {a,b}
c <--> {a,b,d,e,f,g,…}
d <--> {d,@,*}
…
In this particular example (out of infinitely many examples) D = {a, c, …}

There are two extreme versions of D which are:

D={} or D={a, b, c, d, …} , where each one of them is the result of infinitely many different mappings between Power(X) and X, but the general thing is this:

D={} only if each X member that is mapped with some arbitrary Power(X) member, is also a member of this arbitrary Power(X) member.

D={a, b, c, d, …} only if each X member that is mapped with some arbitrary Power(X) member, is not a member of this arbitrary Power(X) member.

Between D={} and D={a, b, c, d, …} we can define the infinitely many D results which are not D={} or D={a, b, c, d, …}.

So Cantor's method actually defines Power(X)={{}, … , {a, b, c, d, …}} as a part of a proof that says something about Power(X) (and in this case: Power(X) > X).

Furthermore, the result of how Power(X) is constructed by Cantor, cannot be but
Power(X) > X, because given any D (which is a Power(X) member) there is no X member that is mapped with D.
Click to expand...


You know, if someone bet me $1000 that I couldn't write mathematics this badly and this incorrectly, I'm not sure I would be able even deliberately to pack this much misunderstanding and misinformation into a forum post. It's one of those posts where the fundamental errors are almost entirely masked by the superficial and notational errors.

Let's start with the basics.

X is a set, by assumption. The power set of X, Power(X) exists and is also a set -- but this is specifically by the Axiom of Power Sets, so you can't get away with this. D, however, is not a set, but a mapping. And < has no definition until you give it one. The problem isn't that E and F are placeholders, but that they're placeholders for an idea that you fail to express. And, in fact, Cantor's theorem is primarily about the definition of <.

To show you how "real" mathematics works, I will offer a few definitions. A function f from A to B is a subset of AxB such that each element of A occurs at most once in f. A function f is 1:1 if and only if for every A, there is a corresponding B (=f(A)) and if distinct domain elements correspond to distinct range elements (i.e. f(A) = f(A') -> A = A'). A function is onto if for every B, there is an A such that f(A) = B. If I were doing this as a class lecture, I would give lots of examples at this point. Ask if you need them.

Two sets have the same cardinality if there is a function F between the sets that is both 1:1 and onto. If there is no possible 1:1 and onto map between A and B, but there is a 1:1 (but not onto) map from A to B, then we say (with some abuse of notation) that A < B or B>A.

Now, given a set X and the set Power(X), we let D be an arbitrary mapping between them. First, I will show (by constructing a specific D*) that D can be 1:1. Let D* map each element x \in X to the set {x}. {x} is clearly a member of Power(X), and two distinct elements x and y will generate two distincts mapped sets {x} and {y}. Hence X can be mapped 1:1 (but not onto) into the set Power(X)>

Now, I will show that D is never an onto map.

Consider an arbitary element x and its mapping y (i.e. (x,y) \in D). There are two possibilities here -- either x \in y or x \not\in y. Consider the set Z of all x (\in X) for which x \not\in y. (This is clearly a set by the Axiom of separation, and clearly a subset of X,
and hence a member of the power set of X. I claim that for no value of z (\in X) is (z,Z) a member of D. (if there were such an element, then it would introduce a contradiction). Hence D does not include the subset Z in its range and is not onto.

Since D cannot be onto, but can be 1:1, we conclude that X < Power(X).
 
drkitten said:
You know, if someone bet me $1000 that I couldn't write mathematics this badly and this incorrectly, I'm not sure I would be able even deliberately to pack this much misunderstanding and misinformation into a forum post. It's one of those posts where the fundamental errors are almost entirely masked by the superficial and notational errors.

Let's start with the basics.

X is a set, by assumption. The power set of X, Power(X) exists and is also a set -- but this is specifically by the Axiom of Power Sets, so you can't get away with this. D, however, is not a set, but a mapping. And < has no definition until you give it one. The problem isn't that E and F are placeholders, but that they're placeholders for an idea that you fail to express. And, in fact, Cantor's theorem is primarily about the definition of <.

To show you how "real" mathematics works, I will offer a few definitions. A function f from A to B is a subset of AxB such that each element of A occurs at most once in f. A function f is 1:1 if and only if for every A, there is a corresponding B (=f(A)) and if distinct domain elements correspond to distinct range elements (i.e. f(A) = f(A') -> A = A'). A function is onto if for every B, there is an A such that f(A) = B. If I were doing this as a class lecture, I would give lots of examples at this point. Ask if you need them.

Two sets have the same cardinality if there is a function F between the sets that is both 1:1 and onto. If there is no possible 1:1 and onto map between A and B, but there is a 1:1 (but not onto) map from A to B, then we say (with some abuse of notation) that A < B or B>A.

Now, given a set X and the set Power(X), we let D be an arbitrary mapping between them. First, I will show (by constructing a specific D*) that D can be 1:1. Let D* map each element x \in X to the set {x}. {x} is clearly a member of Power(X), and two distinct elements x and y will generate two distincts mapped sets {x} and {y}. Hence X can be mapped 1:1 (but not onto) into the set Power(X)>

Now, I will show that D is never an onto map.

Consider an arbitary element x and its mapping y (i.e. (x,y) \in D). There are two possibilities here -- either x \in y or x \not\in y. Consider the set Z of all x (\in X) for which x \not\in y. (This is clearly a set by the Axiom of separation, and clearly a subset of X,
and hence a member of the power set of X. I claim that for no value of z (\in X) is (z,Z) a member of D. (if there were such an element, then it would introduce a contradiction). Hence D does not include the subset Z in its range and is not onto.

Since D cannot be onto, but can be 1:1, we conclude that X < Power(X).
Click to expand...

Thank you for repeating on Cantor's proof by contradiction.

Now hold your horses, and by using what you already know about Cantor's theorem, please read very carefully my first post.

Thank you.
 
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doronshadmi said:
Thank you for repeating on Cantor's proof by contradiction.
Click to expand...

Yes. It was my (obviously vain) hope that you would understand it if you saw a slightly different version.

Now hold your horses, and by using what you already know about Cantor's theorem, please read very carefully my first post.
Click to expand...

I read it very carefully. It consists, again, of words that you obviously do not understand. In this case, the words include "set," "power," "mapping," and "infinite" --- and a word that you did not use but that is key to the argument, "diagonalization." You also have no idea what > means when applied to sets.

I'm a teacher by profession. I read badly-written student cruft for a living. One thing that I can assure you of --- the reason that no one can extract any meaning from your gibberish is not through lack of reading skills on their part. No amount of reading skill can extract what isn't there. You go beyond badly written student cruft to complete word salad that indicates a complete lack of understanding, not only of the terminology you wish to study, but of the basic ideas and methods of the field.
 
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Whoa, Doronshadmi is still making new posts...

If you are so sure that you are right about these things, why not try getting published for a little peer review?

My guess is that somewhere deep down inside you know that you are looney tunes.
 
Ok, lemme see if I understand enough to rewrite this.

Given normal definitions of a set and the power set of a set, is the cardinality of a power set always larger than the set itself? Pow(X) > X?


Given an arbitrary mapping between a set and its power set, call it M (assume it is 1:1 and onto, this we seek to contradict) define another set, called D, where this definition holds (note M and D may themselves be represented as sets, specifically of pairs, where the first, x, is an element of X, and the second, y, is an element of Pow(X)):

D is the subset of M {x -> y | x E X and y E Pow(X)} where x is not an element of set y.

His example for this is OK, assuming set X contains each of those symbols.

We know there is an x' in M that maps to D, D being an element of Pow(x).


So consider the element of M that is {x', D}. Is this itself in D?


Well, if x' is an element of one of the members of D, then it cannot be in D.

If x' is not an element of one of the members of D, then it must be in D.

Therefore (?) there is at least one element of Pow(X), namely D, a seemingly reasonably-defined element, that has no corresponding element in X.

Therefore M cannot be constructed. Therefore Pow(X) > X



Don't know if there's a flaw there offhand, but is that about it?
 
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doronshadmi said:
The general idea that some object is greater than some other object.
Click to expand...

... and how does claiming that an arbitrary object is greater than some other arbitrary object help prove 'Pow(X) > X'? For suitable definitions of 'greater than' we already know that some objects are greater than some other objects. Here's one such example:
my bike is greater than your bike
see if you can guess a suitable definitions of 'greater than' that makes that a well formed statement :)
 
drkitten said:
I'm a teacher by profession.
Click to expand...

Maybe, but you are not a real teacher.

Real teachers do not kill the souls of their students by telling them
that they do not understand what they say.

Real teachers are brave enough to be also students as a part of being real teachers, at least in order to understand their students in order to help them to learn by themselves better.

No me and not you and not anyone else on this planet has the authority on wisdom and on different points of view of already known knowledge.

If you have something to say about what I wrote, be brave enough to do it step by step and in details.

This is the only way to show if I am right or wrong, or maybe if there is something new and interesting in what I say, which do not follow the agreed point of view of this subject.

But this:
drkitten said:
read it very carefully. It consists, again, of words that you obviously do not understand. In this case, the words include "set," "power," "mapping," and "infinite" --- and a word that you did not use but that is key to the argument, "diagonalization." You also have no idea what > means when applied to sets.
Click to expand...

This is the way of a person which is definitely not a real teacher.
 
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That you do not understand what you say is not killing your soul. You're using words whose definitions you have misconstrued and are using them to come to a false conclusion.

As a mathematician in training, I find your notation and word usage to be sloppy and vague. Hence, your OP does not constitute anything resembling a good proof. Please, take Beerina's and drkitten's suggestions and attempt to rewrite it, thank you.

Feel free to use either naive set theory (which you might be ok with) or Zermelo's axiomatic set theory.
 
doronshadmi, you are not a real student. But if you practice being a real student instead of practicing being a bad teacher, you might get better.
 
doronshadmi said:
Real teachers do not kill the souls of their students by telling them
that they do not understand what they say.
Click to expand...

Are you drkitten's student? You certainly don't behave that way.

You make nonsensical posts with great confidence on a public educational forum. Part of the purpose of this place is to expose such garbage for what it is.

If you want to learn something, ask - don't declare. Or if you must state nonsense as fact don't whine when it gets labeled as such. If you can't handle it, go somewhere else.
 
Doron's a tough nut to crack in part because he's actualy not coming from standard Mathematics. Over the years he's devoloped his own language of "Monadic Mathematics" that borrows terms from a number of different sources, but doesn't use them in precisely the same manner as the place he got them. So it comes to a kind of language confusion.

He's really requiring that we learn his own idiolect (That's a lingusitic term for individual language, not an insult.)
I don't think he realizes how often there's a disconnect between what he thinks he's saying and what a professional mathematician is going to read based on the language.

Mathematics, of course, has numerous dialects and fringe systems. So, we'd have to judge Dorons's system on it own ground.

My curiosty had some time to read one of his .pdfs on his "Monadic Mathematics."
I had an interesting flashback to over 30 years ago when I was in college and tutoring reading and math during the summer.
We used these:
http://www.educationallearninggames.com/cuisenaire-rods.asp
Cuisenaire Rods.
I can well imagine that the manipulation of these is the ultimate source of Doron's mathematical concepts.
He could use these to illustrate for us what he's about.

Now they are excelent tools for teaching basic math in a concrete fashion.
But a Mathematics of Cuisenaire Rods or some pedological equivalent isn't going to reach many of the more abstract concepts contemporary Mathematics is built upon (So Doron's issues with Cantor).

It's clever though, to have formalized a pedological stage in math learning.

BTW, Doron. A teacher who tells you s/he can't understand what you are trying to communicate and demands that you communicate clearly is treating you with compassion and fairness.
Real teachers and real mathematicians are going to be frank with you for your own sake. If you cannot take that and benifit from it, you are your own worst enemy.

Doron, if you haven't seen Cuisenaire Rods in action before, look them up. You'd love them.
 
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doronshadmi said:
Maybe, but you are not a real teacher.

Real teachers do not kill the souls of their students by telling them
that they do not understand what they say.
Click to expand...

Did that happen to you? Did a teacher kill your soul? Is that what this is about?

Try to encompass the concept that people can't understand you because you're actually spouting gibberish. Don't ask me explain in detail why it's gibberish, because gibberish is not susceptible to such analysis. It cannot be atomised, it exists only as a whole.
 
At the risk of again failing to find meaning among the gibberish, I'll point out this:

doronshadmi said:
If –as a part of our proof– we define E and F in such a way that E > F, can we still claim that our proof holds?
Click to expand...


I think doron is trying to say he believes that X < PowerSet(X) is true by definition and therefore no proof was necessary.
 
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doronshadmi said:
Maybe, but you are not a real teacher.

Real teachers do not kill the souls of their students by telling them
that they do not understand what they say.

Real teachers are brave enough to be also students as a part of being real teachers, at least in order to understand their students in order to help them to learn by themselves better.

...

This is the way of a person which is definitely not a real teacher.
Click to expand...

Hey, fella, that's a really serious professional allegation you're making.

Perhaps you ought to consider a massive, overwhelming apology real soon now?
 
jsfisher said:
I think doron is trying to say he believes that X < PowerSet(X) is true by definition and therefore no proof was necessary.
Click to expand...

He's insinuating that there's a "Hidden Asumption" thingy here, he hopes we'll notice for a change, so that he'll finally have an opening we can take into his discussion. Unfortunately it's not one I get.
I think I need the Cuisenaire Rods. :):rolleyes:
 
Apathia said:
He's insinuating that there's a "Hidden Asumption" thingy here, he hopes we'll notice for a change, so that he'll finally have an opening we can take into his discussion.
Click to expand...


If there is such an assumption, it is well hidden. We shall see. At least doron didn't try to tell us the diagonal proof the reals are non-countable is bogus.
 
jsfisher said:
I think doron is trying to say he believes that X < PowerSet(X) is true by definition and therefore no proof was necessary.
Click to expand...

You are in the right direction.

If you know Cantor's proof by contradiction, then please read my first post (which is not a proof of anything but some questions and thoughts about the proof) and air your view about it.

Thank you.

About my question of the necessity of the axiom of the power set, drkitten is right if we are talking about some proof, because our initial terms is that Power(X) is a set, and this initial term is the axiom of the power set.

But Cantor actually gives us (as a part of proof about Power(X)) the exact "recipe" of how to define Power(X) in such a way that cannot be but > X.

This definition of power(X) is actually independent of any proof about power(X), and by using this definition we actually get the set power(X) out of set X, in such a way that cannot be but Power(X) > X, and all this is done (I think) independently of ZF axiom of the power set.
 
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doronshadmi - I am unwilling to give any of your posts serious consideration.

This is Vanishingly unlikely to change.

I don't think that your ideas have any merit.

Your delusion of a 'hidden assumption' is not even wrong.

I'll check into this thread occasionally to see what others are saying - your nonsense has attracted some good posters in opposition - but your ramblings are of no interest or import.

I wish you'd loose your delusions of grandeur and learn enough mathematics to be able to appreciate the magnficence of what has been accomplished.
 
doronshadmi said:
If you know Cantor's proof by contradiction, then please read my first post (which is not a proof of anything but some questions and thoughts about the proof) and air your view about it.
Click to expand...


I am, I have, it isn't, and I will.

Your opening post is rather confused about terminology and the specifics of Cantor's diagonal proof. However, this sentence jumps out at me:

doronshadmi said:
So Cantor's method actually defines Power(X)={{}, … , {a, b, c, d, …}} as a part of a proof that says something about Power(X) (and in this case: Power(X) > X).
Click to expand...

(1) It is not a method; it is a proof.
(2) Cantor did not define Power Set as part of his proof.
(3) Nothing in Power Set's definition assures P(S) > S without proof.
(4) The word, "so", implies the statement follows from earlier statements in the post. It does not.
 
doronshadmi said:
But Cantor actually gives us (as a part of proof about Power(X)) the exact "recipe" of how to define Power(X) in such a way that cannot be but > X.

This definition of power(X) is actually independent of any proof about power(X), and by using this definition we actually get the set power(X) out of set X, in such a way that cannot be but Power(X) > X, and all this is done (I think) independently of ZF axiom of the power set.
Click to expand...


The Axiom of Power Set defines power set, not Cantor. Here's the axiom:

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

Quite an elegant definition of P(A), don't you think?
 
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Or in other words, P(A) contains all the possible subsets of ? Am I reading that right?
E.g.
P(2) = P({ {{}}, {} }) = { { {{}}, {} }, { {{}} }, { {} }, {} } = { 2, 2\1, 1, 0 }, and so |P(2)| = 4

Nifty!
(edited to fix incorrect numeric equivalences)
 
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doronshadmi said:
If you have something to say about what I wrote, be brave enough to do it step by step and in details.
Click to expand...

I have done, previously in this thread. Just to start with, your failure to define < (or even to recognize that what is needed is a definition of <, not the introduction of two new undefined symbols) renders your work invalid.

if there is something new and interesting in what I say,
Click to expand...

There is not. Self-righteous ignorance is neither new nor interesting.
 
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jsfisher said:
(2) Cantor did not define Power Set as part of his proof.
Click to expand...

Yes he did. Power(X) members are actually D sets, and D sets are defined
by Cantor as a part of his proof about Power(X) > X.
 
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doronshadmi said:
1) We do not need a proof by contradiction in order to conclude that Power(X) > X,
because Cantor actually defines Power(X) (by D's) in such a way that the result cannot be but
Power(X) > X.

2) Is a proof is still a proof if we actually determine the result, as a pert of the proof?

3) Do we need the ZF axiom of the power set (after all Cantor's method actually defines it, without any need of this axiom)?
Click to expand...

IANAM but...
On #1, I think you're jumping the gun when you say that cantor defines Power(x). The power set is not D but the set of all unique Ds. In other words constructing D is a step in one algorithm for the power set. I think Cantor's definition is very closely related to the recursive algorithm for the power set(you can see it on the wikipedia page for power set). One(non-technical) way of thinking about this is that Cantor makes his proof by looking at the implicit steps in the algorithm for the Power Set and observing that the output must always be larger than the input. There is no problem with this. A careful examination and understanding of the assumptions and details of an operation is exactly what is required to make proofs about an operation.

On #2 he doesn't determine the result, he shows that the result follows from the definition. You wouldn't complain if someone defined a function and made an inductive argument using the function. Then noted that this argument from induction demonstrates a property of the function would you?

On #3 I'm pretty sure you still would need it. One way of looking at the power set axiom is not just that it defines the power set but also that it guarantees the existence and uniqueness of a power set. Making sure this is the case is crucial.
 
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doronshadmi said:
Yes he did. Power(X) members are actually D sets, and D sets are defined
by Cantor as a part of his proof about Power(X) > X.
Click to expand...


No, P(X) members are subsets of X. That is a matter of definition and not anything Cantor provided in his proof. The fact that Cantor constructed D in such a way to be a subset of X in no way means D defines P(X).

Here's the essence of Cantor's proof. Given any mapping f(x) from X to P(X), define

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

D must be a subset of X, but it can be shown that

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

and therefore no mapping from X to P(X) can be onto.

No where in there does a definition for P(X) nor its construction appear.
 
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jsfisher said:
No where in there does a definition for P(X) nor its construction appear.
Click to expand...
Wrong, please read the follow:
zosima said:
IANAM but...
On #1, I think you're jumping the gun when you say that cantor defines Power(x). The power set is not D but the set of all unique Ds. In other words constructing D is a step in one algorithm for the power set.
Click to expand...

If D was a "single step" (as you claim) where each step is disconnected of any other D "step" (as you claim), then Cantor was not able to use D in order to conclude anything as a part of his proof.

For example, for infinitely many mappings between X and its power set, there is, for example D={}.

Some example (which is no more than a one of infinitely many other examples) of D={} is:
Code: 
X    Power(X)

a <--> {a}
b <--> {b}
c <--> {c}
d <--> {d}
…
and it is clear that D={} cannot be mapped with any of set X members.


Now let us look at the opposite version of D={}, which is D={a,b,c,d,…}

Some example (which is no more than a one of infinitely many other examples) of D={a,b,c,d,…} is:
Code: 
X    Power(X)

a <--> {b}
b <--> {c}
c <--> {d}
d <--> {e}
…
and it is clear that D={a,b,c,d,…} cannot be mapped with any of set X members.


Between D={} and D={a,b,c,d,…} there are the infinitely many D versions that not one of them is D={} or D={a,b,c,d,…}, and these infinitely many D versions are exactly Power(X) members.

In general there is no D version (and any D version is definitely a member of Power(X)) that is mapped with some X member.

In other words, Cantor explicitly defines Power(X) as an inseparable part of a proof that concludes something about Power(X) (and in this case the conclusion is: |Power(X)|>|X|).

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.
 
Last edited:
doronshadmi said:
Wrong, please read the follow:

If D was a "single step" (as you claim) where each step is disconnected of any other D "step" (as you claim), then Cantor was not able to use D in order to conclude anything as a part og his proof.

For example, for infinitely many mappings between X and its power set, there is, for example D={}.

Some example (which is no more than a one of infinitely many other examples) of D={} is:
Code: 
X    Power(X)
 
a <--> {a}
b <--> {b}
c <--> {c}
d <--> {d}
…
and it is clear that D={} cannot be mapped with any of set X members.


Now let us look at the opposite version of D={}, which is D={a,b,c,d,…}

Some example (which is no more than a one of infinitely many other examples) of D={a,b,c,d,…} is:
Code: 
X    Power(X)
 
a <--> {b}
b <--> {c}
c <--> {d}
d <--> {e}
…
and it is clear that D={a,b,c,d,…} cannot be mapped with any of set X members.


Between D={} and D={a,b,c,d,…} there are the infinitely many D versions that not one of them is D={} or D={a,b,c,d,…}, and these infinitely many D versions are exactly Power(X) members.

In general there is no D version (and any D version is definitely a member of Power(X)) that is mapped with some X member.

In other words, Cantor explicitly defines Power(X) as an inseparable part of a proof that concludes something about Power(X) (and in this case the conclusion is: |Power(X)|>|X|).

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.
Click to expand...

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?
 
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