Solve the paradox of set theory V7.4.1

In summary, the paradox of set theory V7.5, also known as Russell's paradox, is a paradox that arises when considering all power sets of the real number set R. According to Russell's paradox, all sets that do not contain themselves must contain all power sets of R, making the concept of a set containing itself meaningless. The solution to this paradox lies in axiomatic set theory, but some argue that the theory has strayed onto a different path. LiJunYu proposes several theorems to address this paradox, including the concept of the generalized continuum hypothesis and the idea that a set is meaningless if its cardinality is a limit. These theorems ultimately show that the cardinality of all sets is the limit of
  • #1
e271828
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Solve the paradox of set theory V7.5
by LiJunYu 2010.12.25 email: myvbvc@tom.com or 165442523@qq.com
Brief:All power sets of real number set R: P(R),P(P(R)),P(P(P(R))),...,Pn(R),...Because all Pn(R) does not contain its own,in Russell's paradox,"all sets which does not contain its
own" must contain all Pn(R),that is to say,it contains all of the cardinality of generalized continuum hypothesis {X0,X1,...Xn...},so became meaningless.
Although I know that axiomatic set theory is to solve the paradox arising from, but I think the axiom of set theory in the detours, and even strayed into the manifold road. If the

theory does not contain the following, I think <<set theory>> is incomplete.
Generalized continuum hypothesis: the cardinality of an infinite set must be one of X0, X1, ... Xn ...
Where X is the Greece character aleph, because I can not find the character, so the letter X was expressed in English.
Meaningless axioms: If the cardinality of an infinite set is limit limXn (n-->infinite), then this set is meaningless.
I can not find the sign expressed infinite in english computer,so I use the character "infinite".
This axiom is my introduction, I have not seen elsewhere.
This axiom is easy to understand, it is equivalent to axiomatic set theory in the concept of the true class, but after the introduction of the concept of this class of axiomatic set ,the

theory straying into the manifold road, at least I think so.
LiJunYu first theorem: If a set contains all of the cardinality of generalized continuum hypothesis , that is the set {X0,X1,...Xn...},the cardinality of this set is limXn (n-->infinite)
This theorem is obvious, by reduction to absurdity is not difficult to prove.
LiJunYu second theorem: If an infinite set also contains its own power set, that is the set A={...,P(A)),then the cardinality of this set A is limXn (n-->infinite)
Proof: Let the cardinality of infinite sets A is Xn, n be fixed. Because they contain an infinite set A subset or all of its power set, while the power set of the cardinality is X (n +1) = 2

^ Xn, Therefore, the cardinality becomes X (n +1), which assumes an infinite set A, the original cardinality is Xn, n is a constant contradiction,antinomy, so the cardinality of infinite

sets A is limXn (n-->infinite).
LiJunYu third theorem: If a set contains all power set of an infinite set , that is the set B={P(A),P(P(A)),P(P(P(A))),...,Pn(A),...} ,then the cardinality of this set B is limXn (n--

>infinite),for example,the real number set R,B={P(R),P1(R),P2(R),...,Pn(R),...},then the cardinality of this set B is limXn (n-->infinite).
All the power set, assuming infinite set A, then the power set P (A), power set of the power set P (P (A)), the power set of the power set of the power set P (P (P (A))), ... Pn

(A )... as all of its power set.
Because all power set of infinite set is the cardinality of the generalized continuum hypothesis in all of the cardinality, so by the LiJunYu first theorem know this theorem.
LiJunYu fourth Theorem: If the set P'n(A) have the same cardinality with the power set Pn(A), that is, then the set P'n(A) equivalent to the power set Pn(A) for LiJunYu

second and LiJunYu third theorem.
****I. cardinality Paradox
Theorem 1: The cardinality of all sets is limXn (n-->infinite).
The obvious, that in the "<set theory>> has long been, here repeat it. Because the set of all sets contains its own power set,by the LiJunYu second theorem the cardinality is

limXn (n-->infinite). So this set is referred to as the true class of axiomatic set theory.
****II. Russell's paradox
LiJunYu Fifth Theorem: If the set A does not contain its own,then the power set of A is P(A),it is also not contain its own.
Proof: by contradiction. To assume that anyone does not contain its own set of is A, assume that any child sets of set A is B,B is the set that contains itself, then there are

elements in the subset B, B is the element that contains a set of its own, and B is an element of set A , the set A does not contain all the elements of A's own set, so the element

of B is not included its own set, contradictions. So subset of A does not contain its own . The same reason can be used as the power set proof. Assuming the power set of set A is

a set of C, assume that C is a set that contains its own set, the set has an element of C , C is the element that contains its own set, but a set of elements of C is a subset of A,

according to the above ,The process has been proved by contradiction known any subset of A is the set does not contain itself, the element C is not included its own set ,

contradictions, so the power set is a set does not contain itself.
This is understandable, for example, the set {1,2,3} does not contain itself, then its power set and all subsets, also does not contain itself, it is very easy to understand, but

extended to an infinite set to it. Then real number set R does not contain its own , then any child set and power set of R does not contain itself.
Theorem 2: If the set B is all sets which does not contain its own,the cardinality of B is limXn (n-->infinite).
Proof: because the real numbers set R is not contain its own ,by LiJunYu Fifth theorems ,so all power set of R is not contain its own, that is, the power set of R is R1, the

power set of the power set of R is R2, the power set of the power set of the power set of R is R3. . . . ALL the power set Rn does not contain itself, then All the set does not

contain its own ,that is the set B,containing all the power set of R, by the theorem of LiJunYu third,the cardinality so is limXn (n-->infinite).That is set B contain {P(R),P1(R),P2

(R),...,Pn(R),...}.
So Russell's paradox in "All sets which do not contain its own set ",the cardinality of this set is limXn (n-->infinite), in axiomatic set theory call as the true class.
****III. Ordinal number paradox
Theorem 3: Any ordinal number set has a minimum order, so any ordinal number set on less than or equal relations are well-ordered set.
Theorem 3 is the <<Set Theory>> there's theorem, so there need not be proved.
LiJunYu sixth Theorem: Any set of ordinals ,its power set is also ordinal number .
Proof: for any ordinal number of set subset is ordinal set, so by Theorem 3 knowing subset is also a well-ordering set ,so the subset is an ordinal number, then all subsets of the

power set is ordinal number of the set, by Theorem 3 know that this power set is well-ordered set, so this power set is a ordinal number.
Theorem 4: The cardinality of the set of all ordinals is limXn (n-->infinite).
Proof: Let all order number of the set named A, by the LiJunYu sixth theorem, this set A power set is ordinal, it should also be included in the set A, then the set A contains its

own power set ,by LiJunYu second theorem ,The cardinality of this set is limXn (n-->infinite).

The problem of the cardinality paradox is "a set of all sets", the problem of Ordinal number paradox is "the set of all ordinals," the problem of Russell's paradox is "All the set

does not contain its own." Because according to the above shows that this the cardinality of the three sets are limXn (n-->infinite). then this is meaningless three sets, so the

paradoxes of set theory did not shake the existing science cardinality.
Axiomatic set theory that the introduction of the concept of class is correct, but then the issue is to complicate the simple, I solve the paradoxes of set theory with the most

simple language to understand them, abandoned the scientific axiom of set theory , meaning is very important.
****IV. The following in-depth discussion of the nature of some of the set
Proposition I: All the set which does not contain its own power set is true class? Yes.
Because all power set of the real numbers R does not contain its own power set . So all the set which does not contain its own power set must contain all power set of real

numbers R , by the third theorem of LiJunYu knowing it is true class. Why do all power set of the real numbers R does not contain its own power set, because assumption any

power set Rn contains its own power set , by knowing LiJunYu second theorem it is true class, which have a fixed Xn Rn, contradictory.
Proposition II: all the set which do not contain number 1 is true class? Yes.
Because the power set which do not contain number 1 is also a set which do not contain number 1, it is not difficult to prove by contradiction, because it does not element 1, so

its power set and can not contain element 1. So all of its power set does not contain elements 1. Assuming the real number set R after removing a number of set is named r, then

the power set of all r is r1, r2, ... rn, ... all does not contain element 1, the third by the theorem of LiJunYu knowing it is true class.
Proposition III: all the set which do contain number 1 is true class? Yes.
Because the power set of real number set R must contain elements of {1}, after removing the brackets is the element 1, the power set which remove parentheses is one by one

corresponding the power set of the original, only {1} into 1, so the brackets removed power set is same cardinality with the original set , which is the same cardinality, the same

token, all the power set of real numbers R, exists corresponding same cardinality power set,which contains element 1, by the theorem of LiJunYu fourth and third theorems know

it is true class .
Then all the set which do not contain number 1 really meaningless it? Not. This is the problem of the complete works . If the set is a set of a fixed cardinality Xn, all within this

subset of complete works and then discuss all the set does not contain 1, which makes sense, is not really class. If the set is a set of true class of all sets, the cardinality is limXn (n-

->infinite), then the will be true class . That is, any of "the set of all sets" made into a limited number count of subset , there must be one subset l is the true class. On Russell's

Paradox, "a set of all do not contain themselves", also because it's complete works is a set of all sets, will be meaningless, if it is within a set which has fixed cardinality Xn, then

meaningful carry on.
 
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  • #2
e271828 said:
Solve the paradox of set theory V7.5
by LiJunYu 2010.12.25 email: myvbvc@tom.com or 165442523@qq.com
Brief:All power sets of real number set R: P(R),P(P(R)),P(P(P(R))),...,Pn(R),...Because all Pn(R) does not contain its own,in Russell's paradox,"all sets which does not contain its
own" must contain all Pn(R),that is to say,it contains all of the cardinality of generalized continuum hypothesis {X0,X1,...Xn...},so became meaningless.

Please explain this, because I don't understand what you mean. What does the Russell paradox have to do with [tex]\mathcal{P}_n(\mathbb{R})[/tex]? It is easy to construct a set which contains all [tex]\mathcal{P}_n(\mathbb{R})[/tex], namely

[tex]\{[tex]\mathcal{P}_1(\mathbb{R}),\mathcal{P}_2(\mathbb{R}),...,\mathcal{P}_n(\mathbb{R}),...\}[/tex]

This set exists by the replacement axiom.
I do not understand what this has to do with the Russell paradox. This paradox states that [tex]A=\{x~\vert~x\notin x\}[/tex] cannot be a class.

Although I know that axiomatic set theory is to solve the paradox arising from, but I think the axiom of set theory in the detours, and even strayed into the manifold road. If the

theory does not contain the following, I think <<set theory>> is incomplete.

Yes, set theory is incomplete. This is basically Godels theorem. In fact, any possible theory is incomplete. So you won't be able to save anything. The only axiom system which is not influenced by Godels theorem is falso: http://estatis.coders.fm/falso/ [Broken]

Generalized continuum hypothesis: the cardinality of an infinite set must be one of X0, X1, ... Xn ...

This is not what the generalized continuum hypothesis states. It states: "If you have a set X of cardinality [tex]\alaph_\alpha[/tex], then [tex]\mathcal{P}(X)[/tex] has cardinality [tex]\aleph_{\alpha+1}[/tex]."

[QUOTE
Meaningless axioms: If the cardinality of an infinite set is limit limXn (n-->infinite), then this set is meaningless.
[/QUOTE]

What exactly do you mean with the notation [tex]\lim_{n\rightarrow \infty}{\aleph_n}[/tex]. This is not a standard notation in set theory.

Do you mean [tex]\aleph_\omega[/tex]? If yes, why is a set meaningless if it has cardinality [tex]\aleph_\omega[/tex]?

LiJunYu first theorem: If a set contains all of the cardinality of generalized continuum hypothesis , that is the set {X0,X1,...Xn...},the cardinality of this set is limXn (n-->infinite)

I don't think this is obvious. Could you first explain your notations and then supply a proof?

LiJunYu second theorem: If an infinite set also contains its own power set, that is the set A={...,P(A)),then the cardinality of this set A is limXn (n-->infinite)

The axiom of regularity explicitly forbids a set to contain it's own power set. So this theorem is void.

LiJunYu third theorem: If a set contains all power set of an infinite set , that is the set B={P(A),P(P(A)),P(P(P(A))),...,Pn(A),...} ,then the cardinality of this set B is limXn (n--

>infinite),for example,the real number set R,B={P(R),P1(R),P2(R),...,Pn(R),...},then the cardinality of this set B is limXn (n-->infinite).

The cardinality of all these sets is [tex]\aleph_0[/tex]. The set B is countable...

LiJunYu fourth Theorem: If the set P'n(A) have the same cardinality with the power set Pn(A), that is, then the set P'n(A) equivalent to the power set Pn(A) for LiJunYu

second and LiJunYu third theorem.

What do you mean with [tex]\mathcal{P}^\prime_n(A)[/tex]. And what do you mean with "equivalent"?

****I. cardinality Paradox
Theorem 1: The cardinality of all sets is limXn (n-->infinite).
The obvious, that in the "<set theory>> has long been, here repeat it. Because the set of all sets contains its own power set,by the LiJunYu second theorem the cardinality is

limXn (n-->infinite). So this set is referred to as the true class of axiomatic set theory.

Well, since LiJunYu's theorems are all false, and since you did not explain the notation [tex]\lim_{n\rightarrow \infty}{\aleph_n}[/tex], I do not think that this is a paradox...


****II. Russell's paradox
LiJunYu Fifth Theorem: If the set A does not contain its own,then the power set of A is P(A),it is also not contain its own.

A set cannot contain its own. This is the regularity axiom... So this theorem is void...
And since this theorem is void, all the rest of II is also void...

****III. Ordinal number paradox
Theorem 3: Any ordinal number set has a minimum order, so any ordinal number set on less than or equal relations are well-ordered set.

Wow, you're losing me here. What do you mean with a "minimum order"? And could you explain the thing after the comma. It makes no sense to me...

Theorem 3 is the <<Set Theory>> there's theorem, so there need not be proved.

OK, if it is a theorem of set theory, could you provide a reference?

LiJunYu sixth Theorem: Any set of ordinals ,its power set is also ordinal number .

Could you explain theorem 3 first. Then we'll have a look at your sixth theorem...


Theorem 4: The cardinality of the set of all ordinals is limXn (n-->infinite).
The set of all ordinals do not exist. It is a proper class. So therefore it can not have a cardinality.

Proposition II: all the set which do not contain number 1 is true class? Yes.

This is correct. The sets which do not contain 1 form a proper class...

Proposition III: all the set which do contain number 1 is true class? Yes.

This is also correct.



You know a lot about set theory, but it seems that you still have a few misconceptions about it. I would suggest reading "Introduction to set theory" by Hrbacek and Jech. It's a very fun book!
 
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  • #3
i am a chinese ,my english is not every good.i want to say:all Pn(R) does not contain its own,and the cardinality of R is X1,the cardinality of P(R) is X2,the cardinality of P(P(R)) is X3...the cardinality of Pn(R) is Xn+1...
 
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  • #4
limit limXn (n-->infinite) ,it is equivalent to the concept of the proper class in axiomatic set theory
 
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  • #5
[QUOTE
A set cannot contain its own. This is the regularity axiom... So this theorem is void...
And since this theorem is void, all the rest of II is also void...

[/QUOTE]

I say :If the set A does not contain its own,then the power set of A is P(A),it is also not contain its own.NOT "the set A does contain its own".
what you say is "A set cannot contain its own",i am the same meaning,why you say "is void"?
 
  • #6
"A set can contain its own",such as continuum fraction 2+1/(2+1/(2+1/...)),it is corresponding to set {2,{2,{2,{...}}}},it mean a real number.you can x=2+1/x,you can
solv it.
 
  • #7
Meaningless axioms: If the cardinality of an infinite set is limit limXn (n-->infinite),or limit limXn (n-->X0) or limit limXn (n-->omega),then this set is meaningless.
 
  • #8
You say:
LiJunYu third theorem: If a set contains all power set of an infinite set , that is the set B={P(A),P(P(A)),P(P(P(A))),...,Pn(A),...} ,then the cardinality of this set B is limXn (n--
>infinite),for example,the real number set R,B={P(R),P1(R),P2(R),...,Pn(R),...},then the cardinality of this set B is limXn (n-->infinite).

The cardinality of all these sets is X0. The set B is countable...
===================
you must wrong,the cardinality of real number set R is X1,why all powers set of R is X0?
 
  • #9
do you know what is power set Xn+1=2^Xn ?
 

1. What is the paradox of set theory V7.4.1?

The paradox of set theory V7.4.1 is a contradiction that arises in set theory when trying to define a set that contains all sets that do not contain themselves. This set cannot logically exist, leading to a paradox.

2. How did the paradox of set theory V7.4.1 come about?

The paradox was first introduced by Bertrand Russell in 1901 as an extension of his earlier work on the foundations of mathematics. It was later explored in more detail by mathematician and philosopher, Georg Cantor.

3. Why is the paradox of set theory V7.4.1 important?

The paradox highlights the limitations of set theory and the challenges of defining infinite sets. It also raises questions about the consistency of set theory and the foundations of mathematics.

4. What attempts have been made to solve the paradox of set theory V7.4.1?

Several solutions have been proposed, including the axiomatic approach, where the paradox is avoided by restricting the formation of sets, and the iterative approach, where sets are built up gradually. However, the paradox remains a topic of debate and has not been definitively solved.

5. How does the paradox of set theory V7.4.1 impact other areas of mathematics and science?

The paradox has implications for other branches of mathematics, such as logic and set theory, as it challenges the fundamental concepts and principles of these fields. It also has philosophical implications, as it raises questions about the nature of infinity and the foundations of mathematics.

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