Deeper than primes

[doronshadmi]

If you are trying to talk about the cardinality of what looks like you’re trying to represent as sets, you are still wrong. The cardinality of a set is not itself a set.

Again if the members of X are not sets then they do not have cardinality. You have previously said that the members of your set X, N and K sets are cardinal numbers which means that your required members are specifically not sets and thus do not themselves have cardinality.

No.

What I show is that if any member is a set, then any member has cardinality, and this cardinality is comparable to |X|.

For example:

N={ {{}}, {{},{{}}}}, {{},{{}},{{},{{}}}}, ... }

and:

|N| > |{{}}|
|N| > |{{},{{}}}}|
|N| > |{{},{{}},{{},{{}}}}|
...

This is the smallest transfinite caedinal because each comparable cardinality is finite.

This is not the case with K={ { {{}}, {{},{{}}}}, {{},{{}},{{},{{}}}}, ... }, {{}} , {{},{{}}} , {{},{{}},{{},{{}}}} , ... }

because K has a member that its cardinality is a transfinite cardinal, etc ...

 

[doronshadmi]

And where did this forcing take place? Please be specific.

He claims that a non-finite set of localities can be non-locality (the parts (the localities) have the same size as the whole (the non-local)).

So you are saying there are fewer even numbers that integers. Interesting point of view you have, there. Totally unsupported by the facts, but interesting nonetheless.

No.

I say that when two non-finite collections of localities are mapped, this mapping has nothing to do with size, it simply a non-finite collection of 1-1 mappings between pairs of distinct objects.

 

[doronshadmi]

A couple of technical points:



The more common convention is for the set of natural numbers to include zero.

1) You are wrong.

2) Even if {} is a member of N, my argument holds.

I see we have added more words trying to correct previous bogus statements, but this is still not a definition. It's a tautology. In fact, that whole middle part is unnecessary. It can be reduced to:

If X is an infinite set, then |X| is a transfinite cardinal​

I see that you are playing the prefect God of deduction, and this is exactly where your reasoning fails (by your reasoning no improvement or correction is allowed since you are perfect).


This is wrong:

If X is an infinite set, then |X| is a transfinite cardinal​

because this is arbitrary.


This is right:

If |X| > the cardinality of any member of X, then |X| is a transfinite cardinal​

because this is not arbitrary.

 

[zooterkin]

Your are wrong:

1={{}}

2={{},{{}}}

3={{},{{}},{{},{{}}}}

...

N={1,2,3,...}

K={N,1,2,3,...}

X is a non-finite collection.

General definition:

If |X| > the cardinality of any member of X, then |X| is a transfinite cardinal.

The smallest transfinite cardinal:

If |X| > the cardinality of any member of X and the cardinality of any member of X is finite, then |X| is the smallest transfinite cardinal.

So, if X is the set of real numbers between 1 and 2, then 3 is a transfinite number?

A couple of technical points:



The more common convention is for the set of natural numbers to include zero.

1) You are wrong.

2) Even if {} is a member of N, my argument holds.

zero is not the same as the empty set.

 

[doronshadmi]

So, if X is the set of real numbers between 1 and 2, then 3 is a transfinite number?

We are talking only about members that have cardinality, and members that have cardinality must be sets.

So in the case of R we deal with the power set of N (where each member of this power set is itself a set and therefore it has cardinality)

zero is not the same as the empty set.

In our case all we care is the cardinality of some member, where the member is itself a set.

In other words:

0 = |{}| , where {} is a member of N={{}, {{}}, {{},{{}}}}, {{},{{}},{{},{{}}}}, ... } if N={0,1,2,3,…}

By the way I made a mistake, by writing:

1={{}}

2={{},{{}}}

3={{},{{}},{{},{{}}}}
…

It has to be:

1=|{{}}|

2=|{{},{{}}}|

3=|{{},{{}},{{},{{}}}}|
…

and as you see N={0,1,2,3,…} is actually a set that represents the cardinality of the members of N.

 

[zooterkin]

We are talking only about members that have cardinality, and members that have cardinality must be sets.

Then your definition needs to specify that, whatever it means.

 

[doronshadmi]

Then your definition needs to specify that, whatever it means.

This is the standard definition in any set theory that is based on sets, any member is a set and any set has cardinality.


Here is my definition:

If |X|(the cardinality of X) > the cardinality of any member of X and X is an infinite set, then |X| is a transfinite cardinal.

We can do even better:

1) X is a set and any member of X (if exists) is a set.

2) If |X|(the cardinality of X) > the cardinality of any member of X , then |X| is a transfinite cardinal.

It is clealry understood that (2) is true only of X is an infinite set.

 

[zooterkin]

This is the standard definition in any set theory that is based on sets, any member is a set and any set has cardinality.


Here is my definition:

If |X| > the cardinality of any member of X and X is an infinite set, then |X| is a transfinite cardinal.

Where does that say that the members of X are sets?

 

[doronshadmi]

Where does that say that the members of X are sets?

This is the standard agreement about any standard set theory, and I am talking here about set theory that is based on the standard agreement.


Again:

1) X is a set and any member of X (if exists) is a set.

2) If |X|(the cardinality of X) > the cardinality of any member of X , then |X| is a transfinite cardinal.

It is clealry understood that (2) is true only of X is an infinite set.

For concrete examples, please look at http://www.internationalskeptics.com/forums/showpost.php?p=4953267&postcount=5521 .


3) If |X|(the cardinality of X) > the cardinality of any member of X and the cardinality of any member of X is a finite cardinal ((2) is not satisfied in the case of finite cardinals) , then |X| is the smallest transfinite cardinal.



Now:

The set of natural numbers is N={ {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}, ... }

The set of even numbers is E={ {{},{{}}}, {{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}} ... }

and |E| = |N| because (according to (3)) both of them are the smallest transfinite cardinal.

 

[jsfisher]

He claims that a non-finite set of localities can be non-locality (the parts (the localities) have the same size as the whole (the non-local)).

Really? Cantor used your non-local concept? When and where, exactly did he do that?

I say that when two non-finite collections of localities are mapped, this mapping has nothing to do with size, it simply a non-finite collection of 1-1 mappings between pairs of distinct objects.

So cardinality, the measure of set size, has nothing to do with size.

 

[jsfisher]

Where does that say that the members of X are sets?

If he restricts his universe to ZFC set theory, then all members are sets. (Of course, he won't be able to do that, since his arguments ramble and wander so much.)

 

[jsfisher]

1) X is a set and any member of X (if exists) is a set.

2) If |X|(the cardinality of X) > the cardinality of any member of X , then |X| is a transfinite cardinal.

It is clealry understood that (2) is true only of X is an infinite set.

Nonsense. It is only clearly understood if it is clearly stated as a requirement. Without such a statement, it isn't a requirement, and the set X can be finite. Your "definition" then falls apart. ETA: E.g. X = {{}}.

On the other hand, if it is clearly stated, then the IF-part of (2) is unnecessary. |X| is a transfinite cardinal number in all cases.

So, either way, the "definition" you offer is completely bogus.

3) If |X|(the cardinality of X) > the cardinality of any member of X and the cardinality of any member of X is a finite cardinal ((2) is not satisfied in the case of finite cardinals), then |X| is the smallest transfinite cardinal.

This continues to be wrong for reasons given in a prevous post. The parenthetical remark is also wrong.

 

[doronshadmi]

Ok, you are right, thank you jsfisher.

Let us do it your way:

1) X is a set and any member of X (if exists) is a set.

2) If X is an infinite set, then |X| is a transfinite cardinal.

3) If |X| is a trafnitine cardinal, such that |X| > the cardinality of any member of X and any member of X is a finite set, then |X| is the smallest transfinite cardinal.

Now:

The set of natural numbers is N={ {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}, ... }

The set of even numbers is E={ {{},{{}}}, {{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}} ... }

and |E| = |N| because (according to (3)) both of them are the smallest transfinite cardinal.


Also:

K={ { {{}}, {{},{{}}}}, {{},{{}},{{},{{}}}}, ... }, {{}} , {{},{{}}} , {{},{{}},{{},{{}}}} , ... }

and according (3) |K| is not the smallest transfinite cardinal.

 

[jsfisher]

1) X is a set and any member of X (if exists) is a set.

Unnecessary. Your universe is ZFC set theory.

2) If X is an infinite set, then |X| is a transfinite cardinal.

This is a statement of fact, not a definition.

3) If |X| is a trafnitine cardinal, such that |X| > the cardinality of any member of X and any member of X is a finite set, then |X| is the smallest transfinite cardinal.

Nope. I already gave you a counterexample to this. (By the way, your any's really should be every's. All would work equally well.)

...
and according (3) |K| is not the smallest transfinite cardinal.

Nope, on two counts. Not only is (3) bogus from the outset, your reasoning is as well. You have committed the logical fallacy of Denying the Antecedent.

 

[doronshadmi]

Nope. I already gave you a counterexample to this.

No, this is a new (3) case (|X| is a transfinite cardinal in the new (3) case, which is something that was not a fact in the old (3) case). Please read it carefully again according to the changes that have been done at post http://www.internationalskeptics.com/forums/showpost.php?p=4953821&postcount=5533 .

Again:

The new case (your counterexample does not work):

1) X is a set and any member of X (if exists) is a set.

2) If X is an infinite set, then |X| is a transfinite cardinal.

3) If |X| is a transfnitie cardinal (which means that X is an infinite set (see 2)), such that |X| > the cardinality of any member of X AND any member of X is a finite set, then |X| is the smallest transfinite cardinal.

---------------------------------------
The old case (your counterexample works):

1) X is a set and any member of X (if exists) is a set.

2) If |X|(the cardinality of X) > the cardinality of any member of X , then |X| is a transfinite cardinal.

3) If |X|(the cardinality of X) > the cardinality of any member of X and the cardinality of any member of X is a finite cardinal ((2) is not satisfied in the case of finite cardinals) , then |X| is the smallest transfinite cardinal.

We are no longer in the old case.
---------------------------------------

So, according to the new case:

The set of natural numbers is N={ {{}}, {{},{{}}}, {{},{{}},{{},{{}}}}, ... }

The set of even numbers is E={ {{},{{}}}, {{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}} ... }

and |E| = |N| because (according to (3)) both of them are the smallest transfinite cardinal.


Now:

K={ { {{}}, {{},{{}}}}, {{},{{}},{{},{{}}}}, ... }, {{}} , {{},{{}}} , {{},{{}},{{},{{}}}} , ... }

and according (3) |K| is not the smallest transfinite cardinal because one of the members of K is an infinite set (which is not true in the case of N and E non-finite sets).

(By the way, your any's really should be every's. All would work equally well.)

All does not have any meaning in the case of non-finite sets, and this is exactly my argument against Cantor's reasoning (he takes a notion that has a meaning only in the case of finite sets, and forces it on non-finite sets).

Really? Cantor used your non-local concept? When and where, exactly did he do that?

He does not have to. Forcing All on a non-finite set is equivalent to the argument that non-finite localities are Non-locality.

 

[jsfisher]

No, this is a new (3) case (|X| is a transfinite cardinal in the new (3) case, which is something that was not a fact in the old (3) case). Please read it carefully again according to the changes that have been done at post http://www.internationalskeptics.com/forums/showpost.php?p=4953821&postcount=5533 .

Again:

The new case (your counterexample does not work):

1) X is a set and any member of X (if exists) is a set.

2) If X is an infinite set, then |X| is a transfinite cardinal.

3) If |X| is a transfnitie cardinal (which means that X is an infinite set (see 2)), such that |X| > the cardinality of any member of X AND any member of X is a finite set, then |X| is the smallest transfinite cardinal.

Ok, let's check that. Here is my counter-example, again:

Let I be the minimal set guaranteed by the Axiom of Infinity.
Let P be the power set of I.
Let X be the set { {a} : a ∈ P }.
​

Is X a set? Check!
Is every member of X a set? Check!
Is X an infinite set? Check!
Is every member of X a finite set? Check!
Is |X| > the cardinality of every member of X? Check!

So, all the conditions you laid out are satisfied, but |X| is not the smallest transfinite cardinal number.

Is you "definition" bogus? Check!

...
K={ { {{}}, {{},{{}}}}, {{},{{}},{{},{{}}}}, ... }, {{}} , {{},{{}}} , {{},{{}},{{},{{}}}} , ... }

and according (3) |K| is not the smallest transfinite cardinal because one of the members of K is an infinite set (which is not true in the case of N and E non-finite sets).

You are still guilty of the same logical fallacy as before: Denying the Antecedent. You argument is bogus.

 

[doronshadmi]

Is every member of X a finite set?

No.

One of the members of the powerset of I (which is a non-finite set called X, and it is not a non-finite set if I is not a non-finite set) is I (where I is a non-finite set).

Let I be the minimal set guaranteed by the Axiom of Infinity.

Just do not tell us that also finite sets are guaranteed by the Axiom of Infinity.

----------------------------------------------

I say that when two non-finite collections of localities are mapped, this mapping has nothing to do with size, it simply a non-finite collection of 1-1 mappings between pairs of distinct objects.

So cardinality, the measure of set size, has nothing to do with size.

No, you assume that the mapping technique that is used between finite sets, is valid also between non-finite sets.

But I show that this is a wrong assumption.

 

[jsfisher]

No.

Which one isn't?

By the very construction of X ( = {{a} : a ∈ P } ), every element is a singleton set. Every element of X is of the form {a}, and |{a}| is 1.

 

[doronshadmi]

Which one isn't?

By the very construction of X ( = {{a} : a ∈ P } ), every element is a singleton set. Every element of X is of the form {a}, and |{a}| is 1.

And how do you conclude that |X|>|N| (where |N| is an example of the smallest transfinite cardinal)?

For example: there is not problem to show that:

 X       I

{a} <--> a
{b} <--> b
{c} <--> c
...

 

[jsfisher]

And how do you conclude that |X|>|N| (where |N| is an example of the smallest transfinite cardinal)?

By the construction of set X, we know |X| = |P|. Since |P| > |I|, we know |X| > |I|. Since sets I and X are both infinite sets, and |X| > |I|, |X| cannot be the smallest transfinite cardinal.