Understanding Set Theory: Order and Intersection of Sets A and B

[QuantumClue]

All elements in \mathcal{A} match all elements of \mathcal{B}. The order of the information in \mathcal{B} is important to understand the dynamics of the information. \mathcal{A} does not have a logical order of information. However \mathcal{B} does have a logical order. Because there is no order in \mathcal{A} there is no proper intersection.

What I want to know is if there is any way to express the statement \mathcal{A} has no mathematical order of information in it's Set, but still has all \mathcal{M} \in \mathcal{B} where \mathcal{M} is to denote its members?

Thanks

 

[SW VandeCarr]

All elements in \mathcal{A} match all elements of \mathcal{B}. The order of the information in \mathcal{B} is important to understand the dynamics of the information. \mathcal{A} does not have a logical order of information. However \mathcal{B} does have a logical order. Because there is no order in \mathcal{A} there is no proper intersection.

What I want to know is if there is any way to express the statement \mathcal{A} has no mathematical order of information in it's Set, but still has all \mathcal{M} \in \mathcal{B} where \mathcal{M} is to denote its members?

Thanks

Think about it. Let's say set B is ordered by the natural numbers up to cardinality N_b. Because A has the same cardinality as B, there exists a one to one mapping between the sets. You say that A is "unordered". You also say that the elements of A are the 'same' as in set B. In what way are the same? If they are the same, the elements of set A will have the same order as B. If they are not the same, then anyone to one mapping will "order" set A according to set B.

If A and B are the same then A\cup B = A\cap B

 

[disregardthat]

You will have to explain yourself clearer than this. A bijection between sets means that they have the same cardinality. A bijection between sets which preserves a well-ordering means that they correspond to the same ordinal. The latter implies the former, but not vice versa. Was it this you were thinking of?

 

[QuantumClue]

You will have to explain yourself clearer than this. A bijection between sets means that they have the same cardinality. A bijection between sets which preserves a well-ordering means that they correspond to the same ordinal. The latter implies the former, but not vice versa. Was it this you were thinking of?

So would I state this as |\mathcal{A}|=|\mathcal{B}|?

 

[disregardthat]

So would I state this as |\mathcal{A}|=|\mathcal{B}|?

What are you trying to state? That a there is a well-ordering preserving function from A to B? A well-ordering of a set S is an ordering such that any subset of S has a least element. Is this what you want ?

I'm not familiar with any common notation for expressing that two sets corresponds to the same ordinal, but of course you could define |A| = |B| to mean exactly that. But mind that |A| = |B| commonly refers to that A and be have the same cardinality.

 

[QuantumClue]

Say \mathcal{B} is an ascending order of information (|\psi_1>,|\psi_2>,|\psi_3>... ... |\psi_n>) then is an order relation in set theory yes? Meaning I am supposing that \mathcal{A} does not have an ascending hill of information - the elements contained within the Set are completely random in the notation. But every element in \mathcal{A} can be corresponded to \mathcal{B} as all the relevant information is contained in \mathcal{A}.

 

[disregardthat]

If you are talking about finite sets cardinality and ordinal correspondence trivially coincide.

So you are trying to come up with a mathematical way to express that the elements a_1,a_2,...,a_n are ordered in B and corresponding to elements not ordered in A? If so, you could let A be the disjoint union of the sets \{a_1\}, \{a_2\},...,\{a_n\}, i.e. A = \{a_1\} \sqcup \{a_2 \} \sqcup ... \sqcup \{ a_n \} = \bigsqcup_{k=1}^n \{a_k\} and B = \{(a_1,1),...,(a_n,n)\}. Then an element x in A would not remember its position in the order, but an element (x,k) in B would. The reason we are taking disjoint union is so that the set A would not forget that you consider a_i and a_j different even though they are equal (in, say, value if you are talking about numbers).

 

[QuantumClue]

If you are talking about finite sets cardinality and ordinal correspondence trivially coincide.

So you are trying to come up with a mathematical way to express that the elements a_1,a_2,...,a_n are ordered in B and corresponding to elements not ordered in A? If so, you could let A be the disjoint union of the sets \{a_1\}, \{a_2\},...,\{a_n\}, i.e. A = \{a_1\} \sqcup \{a_2 \} \sqcup ... \sqcup \{ a_n \} = \bigsqcup_{k=1}^n \{a_k\} and B = \{(a_1,1),...,(a_n,n)\}. Then an element x in A would not remember its position in the order, but an element (x,k) in B would. The reason we are taking disjoint union is so that the set A would not forget that you consider a_i and a_j different even though they are equal (in, say, value if you are talking about numbers).

I think I have finished the statement now. I will let you look over it. I am still trying to learn how to use logical set theory properly.

The difference of \mathcal{A} - \mathcal{B} = \emptyset gives us an empty set, as all elements contained in both \mathcal{A} and \mathcal{B} are the same in magnitude.

Using doxastic logic,

\Box \mathcal{B} \vdash \forall \mathcal{E} \in \mathcal{A}

''it is necessary that'' \mathcal{B} derives from all elements in contained in \mathcal{A}. Call this a new Group \mathfrak{G}.

Now it is interesting to note that \mathcal{A} does not need to imply \mathcal{B}. So

\lnot \mathcal{A} \vdash \mathcal{B}

which reads \mathcal{A} does not imply the existence of \mathcal{B}. So

\mathcal{A} \notin \mathfrak{G}

If \mathfrak{G} was a group of a certain symmetry. The members \mathcal{M} \in \mathcal{A} are unordered, but match all elements ordered in \mathcal{B}, then we can use Jarles solution

\mathcal{A} = \{a_1\} \sqcup \{a_2 \} \sqcup ... \sqcup \{ a_n \} = \bigsqcup_{k=1}^n \{a_k\}

and

\mathcal{B} = \{(a_1,1),...,(a_n,n)\}.

 

[SW VandeCarr]

The members \mathcal{M} \in \mathcal{A} are unordered, but match all elements ordered in \mathcal{B}, then we can use Jarles solution

I don't understand the precise meaning of "match" in this case. I think you need to define it in a logical way, not empirically. If the elements of A have some property i that uniquely "matches" a property of an element in a totally ordered finite set B of the same cardinality, then A is already totally ordered far as I understand it. That is, a unique bijection exists whereby every ordered element in B has a unique mapping with an element in A. Therefore the statement: "The members \mathcal{M} \in A are unordered." can't be true or you have a contradiction.

EDIT: Note that your empirical definition of set A has a least element and the fact that any finite total order is a well order.

 

[QuantumClue]

I don't understand the precise meaning of "match" in this case. I think you need to define it in a logical way, not empirically. If the elements of A have some property i that uniquely "matches" a property of an element in a totally ordered finite set B of the same cardinality, then A is already totally ordered far as I understand it. That is, a unique bijection exists whereby every ordered element in B has a unique mapping with an element in A. Therefore the statement: "The members \mathcal{M} \in A are unordered." can't be true or you have a contradiction.

EDIT: Note that your empirical definition of set A has a least element and the fact that any finite total order is a well order.

If |\psi_1>, |\psi_2>... |\psi_n> is an ordered information on the set \mathcal{B} then \mathcal{A} has members which match individually the members of \mathcal{B} but does not posses its order theory.

What makes members ordered? Exactly? I assume the layout of information exists to create order, am I wrong in this assumption?

 

[QuantumClue]

I think I have finished the statement now. I will let you look over it. I am still trying to learn how to use logical set theory properly.

The difference of \mathcal{A} - \mathcal{B} = \emptyset gives us an empty set, as all elements contained in both \mathcal{A} and \mathcal{B} are the same in magnitude.

Using doxastic logic,

\Box \mathcal{B} \vdash \forall \mathcal{E} \in \mathcal{A}

''it is necessary that'' \mathcal{B} derives from all elements in contained in \mathcal{A}. Call this a new Group \mathfrak{G}.

Now it is interesting to note that \mathcal{A} does not need to imply \mathcal{B}. So

\lnot \mathcal{A} \vdash \mathcal{B}

which reads \mathcal{A} does not imply the existence of \mathcal{B}. So

\mathcal{A} \notin \mathfrak{G}

If \mathfrak{G} was a group of a certain symmetry. The members \mathcal{M} \in \mathcal{A} are unordered, but match all elements ordered in \mathcal{B}, then we can use Jarles solution

\mathcal{A} = \{a_1\} \sqcup \{a_2 \} \sqcup ... \sqcup \{ a_n \} = \bigsqcup_{k=1}^n \{a_k\}

and

\mathcal{B} = \{(a_1,1),...,(a_n,n)\}.

Just to add, all this purports to a physical existence. It is an applied logic on the beginning of time. All the symbols I have used, relates to a description of our universe, by stating that before time exists something existed, instead of nothing. The inscription \Box \mathcal{B} \vdash \forall \mathcal{E} \in \mathcal{A} is very important to understand you cannot get anything from nothing. You can only imply the existence of our universe and its collection of information (given as) \mathcal{B} if you are willing to imply the existence of \mathcal{A}. And since \mathcal{A} represents the existence of ''something'' before the big bang, so there is a chronological order to the existence of \mathcal{B}. But interestingly as noted, for fundamental reasons, the something which existed before the big bang does not necesserily infer the existence of \mathcal{B}, because there was an infinite amount of possibilities which where allowed.

So for all practical purposes, one must understand that somehow an existence was chosen. I never specified that \mathcal{A} was infinite in its information, only that it has information which can be individually represented by the members of the universal set. And the only reason you can find the difference of the sets to be an empty set, is because you cannot infer the universal set \mathcal{B} as a reality, and expect any other possibilities to exist. This is another way to state that there is no real valid evidence to infer on the existences of other possible universes, when \mathcal{B} encapsulates our definition of ''everything''.

 

[SW VandeCarr]

If |\psi_1>, |\psi_2>... |\psi_n> is an ordered information on the set \mathcal{B} then \mathcal{A} has members which match individually the members of \mathcal{B} but does not posses its order theory.

What makes members ordered? Exactly? I assume the layout of information exists to create order, am I wrong in this assumption?

I'm ignoring your immediately preceding post since it's inappropriate for this (or any other) forum according to PF rules. Regarding your previous question, if you state elements of A match specific elements of B, you have specified a unique bijection between A and B given both sets have the same cardinality N and B is totally ordered. If you didn't specify this, there would be N! possible bijections; one for every permutation of A. However you did state the elements of A match specific elements in B. This implies a unique bijection and therefore you cannot then say A is unordered since it follows the order of B.

EDIT: Since you didn't define "match" when I asked you, I'll suggest this:An element in A matches a unique element in B if and only if for every i, a natural number: (1,2,...,n-1, n) there exists an ith element a_i in A which maps to a the ith element b_i in B..