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**e271828****Member**- Registered: 2010-12-24
- Posts: 5

Solve the paradox of set theory V7.5.1

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),or limit limXn (n-->X0) or limit limXn (n-->omega),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 proper 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 proper 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 B=P(A),then set B also does not contain its own.

Proof: by contradiction. To assume that any one 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 proper 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 proper 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 proper 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 proper class, which have a fixed Xn Rn, contradictory.

Proposition II: all the set which do not contain number 1 is proper 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 proper class.

Proposition III: all the set which do contain number 1 is proper 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 proper 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 proper class. If the set is a set of proper class of all sets, the cardinality is limXn

(n-->infinite), then the will be proper 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 proper 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.

*Last edited by e271828 (2010-12-28 15:26:58)*

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**DrSteve****Member**- Registered: 2010-11-15
- Posts: 10

If I'm understanding correctly, then what you're saying is that the limit of P_n(R) is a proper class.

This simply isn't true. The limit of a countable sequence of cardinals is a cardinal. In particular, this limit is a set (it's just a countable union of sets which is a set by the union axiom).

You can continue to get much larger sets than this by continuing to take power sets.

If you're going to be taking the SAT, check out my book:

http://thesatmathprep.com/SAT_Sales_Page.html

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**e271828****Member**- Registered: 2010-12-24
- Posts: 5

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......

*Last edited by e271828 (2010-12-27 15:29:18)*

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**e271828****Member**- Registered: 2010-12-24
- Posts: 5

limit limXn (n-->infinite) ,it is equivalent to the concept of the proper class in axiomatic set theory ,is that right?

*Last edited by e271828 (2010-12-27 15:29:53)*

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**e271828****Member**- Registered: 2010-12-24
- Posts: 5

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.

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**e271828****Member**- Registered: 2010-12-24
- Posts: 5

i do not want to say: limit of P_n(R) is a proper class.

i want to say : limit lim P_n(R) (n-->omega) is a proper class.

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