Solve the paradox of set theory V7.2
by LiJunYu 2010.12.22 email: myvbvc@tom.com or 165442523@qq.com
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 base must be one of the infinite set X0, X1, ... Xn ....
Where X is the cardinality of aleph, because I can not find the characters, so the letter X was expressed in English.
Meaningless axioms: an infinite set of the base is limXn (n-->infinite), then this collection is meaningless.
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 collection contains all of the generalized continuum hypothesis in the base, that is the set {X0,X1,...Xn...},the base of this collection is limXn (n-->infinite)
This theorem is obvious, by reduction to absurdity is not difficult to prove.
LiJunYu second theorem: If an infinite collection also contains its own power set, then the base of this collection 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 base is X (n +1) = 2 ^ Xn, Therefore, the cardinality becomes X (n +1), which assumes an infinite set A, the original base 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 an infinite set of all power set, then the base of this collection 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 base, so by the LiJunYu first theorem know this theorem.
LiJunYu fourth Theorem: The set of same potential power set, etc., that is, with the power set of the same base set, also equivalent to the power set for LiJunYu second and LiJunYu third theorem.
****I. Base Paradox
Theorem 1: The set of all sets of the base 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 base is limXn (n-->infinite). So this collection is referred to as the true class of axiomatic set theory.
****II. Russell's paradox
LiJunYu Fifth Theorem: Any set does not contain its own,the child's set and power set of this set is also 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 collection 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 collection of its own, and B is an element of set A , the set A does not contain all the elements of A's own collection, 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 collection of C, assume that C is a collection that contains its own collection, the collection has an element of C , C is the element that contains its own collection, but a collection 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 collection 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 collection 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: All the set does not contain its own,its base is limXn (n-->infinite).
Proof: because the real numbers set R is not included its own collection ,by LiJunYu Fifth theorems ,so all power set of R is not included its own, that is, R of the power set is R1, R the power set of the power set is R2, R the power set of the power set of the power set is R3. . . . ALL the power set Rn does not contain itself, then All the set does not contain its own containing all the power set of R, by the theorem of LiJunYu third,the base so is limXn (n-->infinite).
So Russell's paradox in "All do not contain its own set ",the base of this collection 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 collection 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 collection 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 base of this collection is limXn (n-->infinite).
The problem of the base paradox is "a collection 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 base of the three collections are limXn (n-->infinite). then this is meaningless three sets, so the paradoxes of set theory did not shake the existing science base.
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 collection
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 potential with the original set , which is the same base, the same token, all the power set of real numbers R, exists corresponding same potential 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 Collection is a collection of a fixed base 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 Collection is a collection of true class of all sets, the base 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 collection of all do not contain themselves", also because it's complete works is a collection of all sets, will be meaningless, if it is within a set which has fixed base Xn, then meaningful carry on.