Cont: Deeper than primes - Continuation 1/3*9

[doronshadmi]

delete ...

 

[jsfisher]

Since N is the minimal set satisfying the requirements of the Axiom of Infinity, does it mean that it is at least the set of all sets that are not non-finite sets, which are the members of the von Neumann set of ordinals?

"[T]he set of all sets that are not non-finite sets"!! Phew, that's convoluted.

No, the minimal set satisfying the requirements of the Axiom of Infinity, which I prefer to call N, is not the set of all finite sets. Informally, N has the empty set as a member, all successor that stem from the empty set as members, and nothing else.

And for the avoidance of doubt, this informal description cannot be easily prettied up to make it a formal definition of N.

 

[doronshadmi]

V = { v : (v is not non-finite member of von Neumann set of ordinals) }


Here are some of the members of V:

V = {
       	    ∅, 
            ↓  [B]No-bijection[/B]
     	   {∅},
            ↓  [B]No-bijection[/B] 
     	   { ∅, {∅} },
            ↓  [B]No-bijection[/B] 
     	   { ∅, {∅} , {∅, {∅}} },
            ↓  [B]No-bijection[/B] 
     	   { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} },
            ↓  [B]No-bijection[/B] 
...  

}

where ∀v ∈ V ( ( v ∪ { v } ) ∈ V )

There is no-bijection from ∀v∈V to ∀v∪{v} in V (where V order is irrelevant).

Since ∀ ( v ∪ { v } ) ∈ V |v∪{v}| < |V|, |V| is not the cardinality of V.

 

[doronshadmi]

In other words, Since |V| is not formally established, set X can't be non-finite since |V| <= |X| (where V is the minimal set satisfying the requirements of the Axiom of Infinity) is not well defined, thanks to ∀.

 

[jsfisher]

V = { v : (v is not non-finite member of von Neumann set of ordinals) }

(1) The ordinals defined by von Neumann do not form a set.
(2) You haven't defined what you mean by "non-finite" with regards to sets.
(3) If you choose to accept the ordinals into the discussion, then since they are heavily founded in cardinality and many other concepts you abhor, you are implicitly accepting all those things.

 

[jsfisher]

In other words, Since |V| is not formally established, set X can't be non-finite since |V| <= |X| (where V is the minimal set satisfying the requirements of the Axiom of Infinity) is not well defined, thanks to ∀.

In other words, you don't understand the definition I have provided for the comparisons of set cardinality.

|A| <= |B| if and only if there is an injection from A to B​

is entirely well-defined.

It's a definition, by the way. It is not subject to proof or disproof.

Moreover, despite your fascination with bijections (which are absent from my definition), you have not in any way shown there is no set X meeting the criterion for |V| <= |X|. In fact, if X is the set V, itself, it is trivial to show |V| <= |V|.

Also, if you are now back to shunning quantification, then your entire approach to an argument falls apart. It heavily relies on quantification.

 

[doronshadmi]

(1) The ordinals defined by von Neumann do not form a set.

V = { v : (v is not non-finite member of von Neumann set of ordinals) } = The set of all natural numbers in terms of sets.

Since ∀ ( v ∪ { v } ) ∈ V |v∪{v}| < |V|, |V| is not the cardinality of set V.

In other words, you don't have a basis for your X as a non-finite set, and |V| <= |V| is not exceptional exactly because ∀ ( v ∪ { v } ) ∈ V |v∪{v}| < |V|.

One asks: Is there a bijection among all V members?

The answer is yes: Given any arbitrary member x in V (where V is the set of all natural numbers in terms of sets) the cardinality of x is |x|, which means that x is bijective with itself.

But ∀ x ∈ V (where x=v or x=v∪{v}) there is no-bijection from v to v∪{v} ∧ ∀ (v ∪{v} ) ∈ V |v∪{v}| < |V|

 

[doronshadmi]

Please ignore post #507.

(1) The ordinals defined by von Neumann do not form a set.

V = { v : (v is not non-finite member of von Neumann set of ordinals) } = The set of all natural numbers in terms of sets.

Since ∀(v∪{ v}) ∈ V |v∪{v}| < |V|, |V| is not the cardinality of set V.

In other words, you don't have a basis for your X as a non-finite set, and |V| <= |V| is not exceptional exactly because ∀(v∪{v}) ∈ V |v∪{v}| < |V|.

One asks: Is there a bijection among all V members?

The answer is yes: Given any arbitrary member x in V (where V is the set of all natural numbers in terms of sets) the cardinality of x is |x|, which means that x is bijective with itself.

But ∀x ∈ V (where x=v or x=v∪{v}) there is no-bijection from v to v∪{v} ∧ ∀(v ∪{v}) ∈ V |v∪{v}| < |V|

Here it is:

V = {

  ----------------------------------> [B]Bijection[/B], order is irrelevant among all x members (where x=[I]v[/I] or x=[I]v[/I]∪{[I]v[/I]}).
  |   0  =   |∅|, 
  |          ↓  [B]No-bijection[/B]
  |   1  = |{ ∅ }|,
  |          ↓  [B]No-bijection[/B] 
  |   2  = |{ ∅, {∅} }|,
  |          ↓  [B]No-bijection[/B] 
  |   3  = |{ ∅, {∅} , {∅, {∅}} }|,
  |          ↓  [B]No-bijection[/B] 
  |   4  = |{ ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }|,
  |          ↓  [B]No-bijection[/B] 
  |
  v
  [B]No-Bijection[/B], order is irrelevant among all x members (where x=[I]v[/I] or x=[I]v[/I]∪{[I]v[/I]}) and [B]∀[/B]([I]v[/I]∪{[I]v[/I]}) ∈ V |[I]v[/I]∪{[I]v[/I]}| < |V|
... 

}

which means that < (no-bijection) and = (bijection) are not gathered into <= in order to define |V| <= |X|, |V| <= |V|.

By the way, my definition for non-finite sets was based on a definition for relative cardinality. Are you now adopting my definition of cardinality, too?

No, I take a ride on it with you in order to find if it holds water.

Since |V| is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water.


So once again, thank you ∀.

 

[doronshadmi]

I like ∀

 

[jsfisher]

Please ignore post #507.


V = { v : (v is not non-finite member of von Neumann set of ordinals) } = The set of all natural numbers in terms of sets.

There is no such set as the von Neumann set of ordinals.
You still haven't defined what you mean by non-finite (or finite, either will do) set.

...∀(v∪{ v}) ∈ V...

Still don't understand quantification, do you. That is complete gibberish.

 

[doronshadmi]

There is no such set as the von Neumann set of ordinals.

V is not the von Neumann set of ordinals, bot only all of its members that are equivalent to all the natural numbers.

Still don't understand quantification, do you. That is complete gibberish.

Still don't understand that ∀(v∪{v}) ∈ V |v∪{v}| < |V|, which means that |V| is not the cardinality of set V (the set of all natural numbers).

If ∀(v∪{v}) ∈ V is a complete gibberish, it can be replaced by ∀v(v∪{v} ∈ V)

 

[doronshadmi]

V = { v : (v is not non-finite member of von Neumann set of ordinals) } = The set of all natural numbers in terms of sets.

∀v(v∪{v} ∈ V) ∧ |v∪{v}| < |V|, which means that |V| is not the cardinality of set V.

In other words, you don't have a basis for your X as a non-finite set, and |V| <= |V| is not exceptional exactly because ∀v(v∪{v} ∈ V) ∧ |v∪{v}| < |V|.

One asks: Is there a bijection among all V members?

The answer is yes: Given any arbitrary member x in V (where V is the set of all natural numbers in terms of sets) the cardinality of x is |x|, which means that x is bijective with itself.

But ∀x ∈ V (where x=v or x=v∪{v}) there is no-bijection from v to v∪{v} ∧ ∀v(v∪{v} ∈ V) ∧ |v∪{v}| < |V|.

Here it is:

V = {

  ----------------------------------> [B]Bijection[/B], order is irrelevant among all x members (where x=[I]v[/I] or x=[I]v[/I]∪{[I]v[/I]}).
  |   0  =   |∅|, 
  |          ↓  [B]No-bijection[/B]
  |   1  = |{ ∅ }|,
  |          ↓  [B]No-bijection[/B] 
  |   2  = |{ ∅, {∅} }|,
  |          ↓  [B]No-bijection[/B] 
  |   3  = |{ ∅, {∅} , {∅, {∅}} }|,
  |          ↓  [B]No-bijection[/B] 
  |   4  = |{ ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }|,
  |          ↓  [B]No-bijection[/B] 
  |
  v
  [B]No-Bijection[/B], order is irrelevant among all x members (where x=[I]v[/I] or x=[I]v[/I]∪{[I]v[/I]}) and [B]∀[/B][I]v[/I]([I]v[/I]∪{ [I]v[/I]} ∈ V) ∧|[I]v[/I]∪{v}| < |V|
... 

}

which means that < (no-bijection) and = (bijection) are not gathered into <= in order to define |V| <= |X|, |V| <= |V|, in the first place.




Since |V| is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water.


So once again, thank you ∀.

 

[jsfisher]

V is not the von Neumann set of ordinals, bot only all of its members that are equivalent to all the natural numbers.

There is no such thing as the von Neumann set of ordinals, so any reference to its members is meaningless.

If ∀(v∪{v}) ∈ V is a complete gibberish....

It is.


Please take some time out from your repetitious posts to define your terms. You need to explain what you mean by:

finite or infinite or non-finite set
cardinality
no-bijection​

Concise, simple definitions. No need to wander off towards some agenda you are trying to push, just the definitions, simple and concise.

And can we not agree that "|A|" is a notation that means "the cardinality of set A"? It is a definition and not subject to disproof. Statements like "|V| is not the cardinality of set V" are just dumb.

 

[doronshadmi]

There is no such thing as the von Neumann set of ordinals, so any reference to its members is meaningless.

In that case N, known as the set of all natural numbers (such that all natural numbers are defined in terms of sets) does not exist, and you don't have the minimal non-finite set to be based on.

And can we not agree that "|A|" is a notation that means "the cardinality of set A"?

Only if there are finitely many members (or no members at all) in A.

Since you don't get that yet, let's simplify it for you:

A is a finite set iff |A| is the cardinality of A.

Since the term "non-finite set" is the complement of the term "finite set", so are the following expressions:

"|A| is the cardinality of A" (in case of finite sets) is the complement of "|A| is not the cardinality of A" (in case of non-finite sets).

 

[jsfisher]

...define your terms. You need to explain what you mean by:

finite or infinite or non-finite set
cardinality
no-bijection​

Here, if these are of any use to you, are mine:

• Notation: |A| means the cardinality of set A
• Notation: N means the minimal set satisfying the requirements of the Axiom of Infinity. If the Axiom of Infinity is given by ∃I (Ø∈I ∧ ∀x (x∈I ⇒ x ⋃ {x} ∈ I)) then N = {x∈I : ∀y ((Ø∈y ∧ ∀z ((z∈y ⇒ z ⋃ {z} ∈ y))) ⇒ x∈y)}
• Definition: |A| ≤ |B| means there is an injection from A to B. For completeness:
  • |A| = |B| means |A| ≤ |B| ∧ |B| ≤ |A|
  • |A| < |B| means ¬(|B| ≤ |A|)
• Definition: A is an infinite set means |N| ≤ |A|
• Definition: No-bijection has no meaning of which I am aware.

 

[jsfisher]

In that case N, known as the set of all natural numbers (such that all natural numbers are defined in terms of sets) does not exist, and you don't have the minimal non-finite set to be based on.

Your conclusion does not follow. There is no set of the von Neumann ordinals simply because it is too big to be a set.

A is a finite set iff |A| is the cardinality of A.

So, by definition, you will be restricting cardinality to only finite sets. You are building in a restriction a priori in order to get your desired conclusion.

Rather self-serving, and not particularly useful.

Nonetheless, what then is your definition for cardinality that self-limits somehow (without any reference to finite sets, since that would be circular)?

 

[doronshadmi]

Syntactic correction:

V = { v : (v is not non-finite member of von Neumann ordinals) } = The set of all natural numbers in terms of sets.

∀v(v∪{v} ∈ V ∧ |v∪{v}| < |V|), which means that |V| is not the cardinality of set V.

In other words, you don't have a basis for your X as a non-finite set, and |V| <= |V| is not exceptional exactly because ∀v(v∪{v} ∈ V ∧ |v∪{v}| < |V|).

One asks: Is there a bijection among all V members?

The answer is yes: Given any arbitrary member x in V (where V is the set of all natural numbers in terms of sets) the cardinality of x is |x|, which means that x is bijective with itself.

But ∀x ∈ V (where x=v or x=v∪{v}) there is no-bijection from v to v∪{v} ∧ ∀v(v∪{v} ∈ V ∧ |v∪{v}| < |V|).

Here it is:

V = {

  ----------------------------------> [B]Bijection[/B], order is irrelevant among all x members (where x=[I]v[/I] or x=[I]v[/I]∪{[I]v[/I]}).
  |   0  =   |∅|, 
  |          ↓  [B]No-bijection[/B]
  |   1  = |{ ∅ }|,
  |          ↓  [B]No-bijection[/B] 
  |   2  = |{ ∅, {∅} }|,
  |          ↓  [B]No-bijection[/B] 
  |   3  = |{ ∅, {∅} , {∅, {∅}} }|,
  |          ↓  [B]No-bijection[/B] 
  |   4  = |{ ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }|,
  |          ↓  [B]No-bijection[/B] 
  |
  v
  [B]No-Bijection[/B], order is irrelevant among all x members (where x=[I]v[/I] or x=[I]v[/I]∪{[I]v[/I]}) and [B]∀[/B][I]v[/I]([I]v[/I]∪{[I]v[/I]} ∈ V ∧|[I]v[/I]∪{v}| < |V|).
... 

}

which means that < (no-bijection) and = (bijection) are not gathered into <= in order to define |V| <= |X|, |V| <= |V|, in the first place.


Since |V| is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water.


So once again, thank you ∀

 

[doronshadmi]

• Definition: No-bijection has no meaning of which I am aware.

You can't be aware of it as long as you reject N as the set of all natural numbers in terms of sets, as given in http://www.internationalskeptics.com/forums/showpost.php?p=13229226&postcount=517.

 

[jsfisher]

You can't be aware of it as long as you reject N as the set of all natural numbers in terms of sets, as given in http://www.internationalskeptics.com/forums/showpost.php?p=13229226&postcount=517.

No one can "be aware of it" until you tell us what you mean by "no-bijection". I know what it means to have a bijection between two sets. I know what it means for there to be no bijection between two sets. This "no-bijection" is a mystery.

Are you unable to tell us what you mean? Use your words. Examples are not definitions.

 

[doronshadmi]

No one can "be aware of it" until you tell us ...

You simply can't say "I am not aware of it".

As long as you need some us in order to be aware of it, you don't say to yourself "I am not aware of it".

Without this fundamental awareness, you actually unable to be aware of it.

So first please leave this us ghost out of the framework, and read http://www.internationalskeptics.com/forums/showpost.php?p=13229226&postcount=517 to yourself by using your awareness, without be depend on any us ghost, at least in this first stage.