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

Yes they are at issue.

No, not my monkey; not my circus. You are the one with the overwhelming need for "von Neumann's treatment of the natural numbers" in order to define the cardinality and finiteness of sets.

(Make no mistake (|A| ≤ |B| ∧ |B| ≤ |A|) or ¬(|B| ≤ |A|) are not the same as (|A| ≤ |B| means there is an injection from A to B))

Gosh, I hope not. The expression |A| ≤ |B| ∧ |B| ≤ |A| and the expression ¬(|B| ≤ |A|) were meant to mean something entirely different from |A| ≤ |B|.
 
No, not my monkey; not my circus. You are the one with the overwhelming need for "von Neumann's treatment of the natural numbers" in order to define the cardinality and finiteness of sets.
jsfisher, there is no overwhelming need for "von Neumann's treatment of the natural numbers".

Without your set N as the minimal set satisfying the requirements of the Axiom of Infinity, you don't have Cardinality beyond the natural numbers.

So there is no choice but to very carefully check if |N| is the cardinality of all N members, which are exactly the von Neumann's treatment of the natural numbers , which is right in line with your point.

All is needed for this careful check, can be found in http://www.internationalskeptics.com/forums/showpost.php?p=13232039&postcount=540 including its links an the summery is:

=========================================================

Now let's look if |A| <= |B| (injection from A to B) can be used in order to establish |V|.

∀v ∈ V(v∪{v} ∈ V ∧ |v| < |v∪{v}| < |V|)


In simple words:

For all v∪{v} in V, |v∪{v}| is too small in order to be valued as |V|, which means that |V| can't be defined as the cardinality of all V members.

In other words: |A| <= |B| (and definitely |V| <= |V|) do not hold water.


As for the set of all natural numbers, |N| is not established.


Here it is:

Code:
V = {

  ----------------------------------> [B]Bijection[/B] according to [B]Definition 1[/B] 
  |                                   (order is irrelevant).
  |   0  =   ∅, 
  |          ↓  [B]No-bijection[/B] according to [B]Definition 2[/B] (order is irrelevant). 
  |   1  = { ∅ },
  |          ↓  [B]No-bijection[/B] according to [B]Definition 2[/B] (order is irrelevant).  
  |   2  = { ∅, {∅} },
  |          ↓  [B]No-bijection[/B] according to [B]Definition 2[/B] (order is irrelevant).  
  |   3  = { ∅, {∅} , {∅, {∅}} },
  |          ↓  [B]No-bijection[/B] according to [B]Definition 2[/B] (order is irrelevant).
  |   4  = { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} },
  |          ↓  [B]No-bijection[/B] according to [B]Definition 2[/B] (order is irrelevant).
  |
  v
  [B]No-Bijection[/B] according to [B]Definition 2[/B] (order is irrelevant). 
... 

}

Order is irrelevant since all we care is that [B]∀[/B][I]v[/I] ∈ V([I]v[/I]∪{[I]v[/I]} ∈ V)

which means that < (no-bijection (by definition 2)) and = (bijection (by definition 1)) are not gathered into <= in order to define |V| <= |X|, |V| <= |V|, right inside set V, as clearly seen in the diagram above.

Without |V|, there is no basis to the extension of Cardinality beyond the natural numbers.

================================================== =======

The trick of |V| as a weak limit cardinal, is no more no less than an ad hoc artificial trick out of nowhere, which its aim it to save ZF(C) from its big crash, as already given in www.internationalskeptics.com/forums/showpost.php?p=13229460&postcount=523.
 
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jsfisher, there is no overwhelming need for "von Neumann's treatment of the natural numbers".

Then why did you continually insist on including your set V before you could (attempt to) define cardinality and finite set?

Without your set N as the minimal set satisfying the requirements of the Axiom of Infinity, you don't have Cardinality beyond the natural numbers.

Throughout this current thread arc, I offered the set N, the minimal set meeting the requirements of the Axiom of Infinity, as a given. You are free to assume it exists without the need for you to define the set in any formal way. However, if you want to move beyond just the existence of that set, that is all on you.

So there is no choice but very carefully check if |N| is the cardinality of all N members...

No, |N| is the cardinality of N, not any or all of its members.
 
No, |N| is the cardinality of N, not any or all of its members.

N=V={ v : (v is a set of von Neumann's treatment of the natural numbers) } = The set of all natural numbers in terms of sets (which is right in line with your point).

|V| is the cardinality of V iff ∀v ∈ V(v∪{v} ∈ V ∧ |v∪{v}| ≤ |V|)

Since ∀v ∈ V(v∪{v} ∈ V ∧ |v∪{v}| < |V|) |V| is not the cardinality of V, simply because |v∪{v}| can't be valued as |V|, exactly because there is no injection from v∪{v} to V, where more details are already given in http://www.internationalskeptics.com/forums/showpost.php?p=13232039&postcount=540 and all of its links.

=======================

A non-finite set is any set that its cardinality can't be defined.

There is bijection from a non-finite set to its non-finite proper subset exactly because their cardinalities can't be defined.
 
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Ok.

...={ v : (v is a set of von Neumann's treatment of the natural numbers) } = The set of all natural numbers in terms of sets
Not ok. You'd need to present how that is done.

(which is right in line with your point).
I have no idea why you keep writing this. The natural numbers are rather distant from "point" regarding cardinality as a relative concept.

|V| is the cardinality of V iff ∀v ∈ V(v∪{v} ∈ V ∧ |v∪{v}| ≤ |V|)

You can stop right there. |V| is the cardinality of V (period). That is what the notation, |V|, means.

Perhaps you mean that the cardinality of V is undefined or something? The meaning of the notation, though, stands.

...can't be valued as |V|...
Valued? My definition for cardinality does not involve values, only an ordering relation.

If you want to insist on values, you need to build that into your definition, but don't expect it from mine.
 
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Not ok. You'd need to present how that is done.
It is done by induction, where the initial member is ∅.

I have no idea why you keep writing this. The natural numbers are rather distant from "point" regarding cardinality as a relative concept
Because of this:
And von Neumann's treatment of the natural numbers is right in line with my point: Just about everything in Mathematics is rooted in Set Theory.


You can stop right there. |V| is the cardinality of V (period). That is what the notation, |V|, means.
Wrong, it is a notion with nothing behind it exactly because there is no injection from v∪{v} to V.

Perhaps you mean that the cardinality of V is undefined or something? The meaning of the notation, though, stands.
Not in case of V, |V| is a notation with nothing behind it.

Valued? My definition for cardinality does not involve values, only an ordering relation.
Your definition of cardinality fails in case of von Neumann's treatment of the natural numbers in terms of sets.

If you want to insist on values, you need to build that into your definition, but don't expect it from mine.
Value is used here by its informal meaning as a general concept of one's ability to know something about a given thing.

But I agree with you that "|v∪{v}| can't be valued as |V|" may be replaced by "|v∪{v}| < |V| for all v∪{v} in V".
 
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There are ever-so many things wrong in the post, but I choose to focus on just this:

Wrong, it is a notion with nothing behind it exactly because there is no injection from v∪{v} to V.

Sure there is. (You could just use v, by the way. The successor of v isn't necessary and doesn't make it special in any way.) Every element of v ⋃ {v} is an element of V. The identity function is a perfectly adequate injection from v ⋃ {v} to V.

Therefore (under my definition): |v ⋃ {v}| ≤ |V|
 
It's like deja vu all over again.

Ok, so now what is your definition for cardinality this time?

Already given directly by definition 1 and indirectly by definition 2 in http://www.internationalskeptics.com/forums/showpost.php?p=13230807&postcount=532 where the definitions are restricted to the sets as done by von Neumann's treatment of the natural numbers.

In order to move beyond this restriction one actually goes beyond the notion of sets, or more generally beyond collections, as suggested in the link that is found at the end of post http://www.internationalskeptics.com/forums/showpost.php?p=13219833&postcount=451.
 
Sure there is. (You could just use v, by the way. The successor of v isn't necessary and doesn't make it special in any way.) Every element of v ⋃ {v} is an element of V. The identity function is a perfectly adequate injection from v ⋃ {v} to V.

Therefore (under my definition): |v ⋃ {v}| ≤ |V|
It is about time that you look very carefully at http://www.internationalskeptics.com/forums/showpost.php?p=13231956&postcount=538 and do not forget that we are talking about ∀v ∈ V(v∪{v} ∈ V) so there is no way that out of nowhere there is injection from v∪{v} to V.

You still force injection on a domain that does not have injection.

Moreover, by |v ⋃ {v}| ≤ |V| you actually claim the finite sets (all the members of V are finite sets) may have cardinality of a non-finite set.
 
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...[D]o not forget that we are talking about ∀v ∈ V(v∪{v} ∈ V) so there is no way that out of nowhere there is injection from v∪{v} to V.

Is this another example of your lack of understanding of quantification? ∀v∈V (v ⋃ {v} ∈ V) is a simple logic expression (that happens to evaluate to true for the case at hand). "For all v that are members of V, v ⋃ {v} is a member of V." Yep, a true statement.

What is it you imagine going on beyond that?

You still force injection on a domain that does not have injection.

I have no idea to what you are referring.

Moreover, by |v ⋃ {v}| ≤ |V| you actually claim the finite sets (all the members of V are finite sets) may have cardinality of a non-finite set.

If A < B, then certainly A ≤ B. Do you claim otherwise?
 
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It's like deja vu all over again.

Ok, so now what is your definition for cardinality this time?

Already given directly by definition 1 and indirectly by definition 2 in http://www.internationalskeptics.com/forums/showpost.php?p=13230807&postcount=532 where the definitions are restricted to the sets as done by von Neumann's treatment of the natural numbers.

For easy reference, here are those two definitions:
Definition 1: |A| = |B| iff there is bijection from A to B
Definition 2: |A| < |B| iff (there is injection no surjection from A to B) OR (A=∅ ∧ B~=∅).​
Neither is "direct" nor "indirect". They each define the meaning of a relation between the cardinality of two not necessarily distinct sets.
Neither is restricted to the members of any particular set.
Neither supports your current definition for infinite set.

Additionally, Definition 2 is problematic because "injection no surjection" is not a meaningful phrase, and if you meant "injection that is not a surjection" then it has other problems. Everything from the OR and on is unnecessary.
 
Is this another example of your lack of understanding of quantification? ∀v∈V (v ⋃ {v} ∈ V) is a simple logic expression (that happens to evaluate to true for the case at hand). "For all v that are members of V, v ⋃ {v} is a member of V." Yep, a true statement.

What is it you imagine going on beyond that?
All v and v⋃{v} in V are finite sets that define all the natural numbers (according to von Neumann's treatment of the natural numbers) and only them.

So no member of V has the cardinality of V.


I have no idea to what you are referring.
All you can find within V is finite sets, such that:

The cardinality of each one of them (notated as |A|) is defined only by bijection form A to itself (what I call direct definition of cardinality, since it is based only on mapping from A to itself).

OR

The cardinality of them (notated as |A| or |B|) is defined only by no mapping form A=∅ to B~=∅ or by injection non surjection from A to B (what I call indirect definition of cardinality, since it is based only on mapping from A to B, such that |A| can't be but < |B| (because in our case A=v and B=v⋃{v}, such that v or v⋃{v} are finite sets, according to von Neumann's treatment of the natural numbers).

If A < B, then certainly A ≤ B. Do you claim otherwise?
Again, we are restricted only to the finite sets v or v⋃{v} that are defined according to von Neumann's treatment of the natural numbers, where A=v and B=v⋃{v}. So |A| can't be but < |B|, which is certainly not A ≤ B.
 
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For easy reference, here are those two definitions:
Definition 1: |A| = |B| iff there is bijection from A to B
Definition 2: |A| < |B| iff (there is injection no surjection from A to B) OR (A=∅ ∧ B~=∅).​
Neither is "direct" nor "indirect". They each define the meaning of a relation between the cardinality of two not necessarily distinct sets.
Neither is restricted to the members of any particular set.
Neither supports your current definition for infinite set.

Additionally, Definition 2 is problematic because "injection no surjection" is not a meaningful phrase, and if you meant "injection that is not a surjection" then it has other problems. Everything from the OR and on is unnecessary.
jsfisher, in order to establish the minimal set satisfying the requirements of the Axiom of Infinity, one first has to show that all the natural numbers (which are defined by von Neumann's treatment of the natural numbers in terms of sets) are actually establish a set that satisfying the requirements of the Axiom of Infinity.

So at the first stage the following definitions are restricted only to all the members of this minimal set, such that the initial member is the empty set (v=∅) and the rest of the members are defined as v⋃{v}, which are all finite sets:

Definition 1: |v| = |v| iff there is bijection from v to v.

Definition 2: |v| < |v⋃{v}| iff (there is an injective non-surjective function from v to v⋃{v}) OR (v=∅ ∧ v⋃{v}(≠∅)).

Definitions 1 and 2 look trivial at first glance, but these are exactly the only two possible functions among the finite sets, which define all the natural numbers by von Neumann's treatment, where also (v=∅ ∧ v⋃{v}(≠∅)) must be added since there is no function from v=∅ to v⋃{v}(≠∅), but we wish to include also the inequality |∅| < |v⋃{v}(≠∅)|.

B.T.W, |∅| < |v⋃{v}(≠∅)| is not covered by your mathematical framework, which is entirely based on the injective function.

Actually, one does not need more than this uncovered case, in order to show that a mathematical framework (which is entirely based on the injective function) does not satisfy the requirements of the Axiom of Infinity.

Since ∀v ∈ V(v∪{v} ∈ V ∧ |v∪{v}| can't be but < |V|) |V| is not established, simply because it can't be defined as the cardinality of all V members, which are all finite sets by von Neumann's treatment of all the natural numbers.

=======================

A non-finite set is any set that its cardinality can't be defined.

There is bijection from a non-finite set to its non-finite proper subset exactly because their cardinalities can't be defined.

=======================


Classical mathematicians are using an ad hoc trick, and inconstantly establish out of nowhere λ :
( https://en.wikipedia.org/wiki/Limit_ordinal )
In set theory, a limit ordinal is an ordinal number that is neither zero nor a successor ordinal. Alternatively, an ordinal λ is a limit ordinal if there is an ordinal less than λ, and whenever β is an ordinal less than λ, then there exists an ordinal γ such that β < γ < λ. Every ordinal number is either zero, or a successor ordinal, or a limit ordinal.
( https://en.wikipedia.org/wiki/Limit_cardinal )
In mathematics, limit cardinals are certain cardinal numbers. A cardinal number λ is a weak limit cardinal if λ is neither a successor cardinal nor zero. This means that one cannot "reach" λ from another cardinal by repeated successor operations. These cardinals are sometimes called simply "limit cardinals" when the context is clear.
This means that one cannot "reach" λ from another cardinal by repeated successor operations.
This means that λ is beyond collections, as seen, for example, in http://www.internationalskeptics.com/forums/showpost.php?p=12584400&postcount=3285.
 
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All v and v⋃{v} in V are finite sets that define all the natural numbers (according to von Neumann's treatment of the natural numbers) and only them.

You are allowed the minimal set satisfying the Axiom of Infinity. Call it N, call it V, doesn't matter, but anything beyond just that derived from the axiom, you need to build.
 
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jsfisher, in order to establish the minimal set satisfying the requirements of the Axiom of Infinity, one first has to show that all the natural numbers (which are defined by von Neumann's treatment of the natural numbers in terms of sets) are actually establish a set that satisfying the requirements of the Axiom of Infinity.

No, that isn't a requirement. It is much simpler than that. Given the Axiom of Infinity, ∃I (Ø∈I ∧ ∀x (x∈I ⇒ x ⋃ {x} ∈ I)), I can define the set N as follows:
N = {x∈I : ∀y ((Ø∈y ∧ ∀z (z∈y ⇒ z ⋃ {z} ∈ y)) ⇒ x∈y)}​
That finds entirely within the bounds of the set theory axioms. And notice that there are no natural numbers to be found anywhere.

If you feel the approach you described is necessary, then, please show us how it is done.



B.T.W, |∅| < |v⋃{v}(≠∅)| is not covered by your mathematical framework, which is entirely based on the injective function.

Are you unaware that there is in fact an injective mapping from ∅ to any set of your choosing? The identify mapping, f(x) = x, will do nicely.
 
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No, that isn't a requirement. It is much simpler than that. Given the Axiom of Infinity, ∃I (Ø∈I ∧ ∀x (x∈I ⇒ x ⋃ {x} ∈ I)), I can define the set N as follows:
N = {x∈I : ∀y ((Ø∈y ∧ ∀z (z∈y ⇒ z ⋃ {z} ∈ y)) ⇒ x∈y)}​
That finds entirely within the bounds of the set theory axioms. And notice that there are no natural numbers to be found anywhere.
Once again you skip on all the finite sets, as defined by von Neumann's treatment of all the natural numbers.

This is how you are doing it.

N = {x∈I : ∀y ((Ø∈y ∧ ∀z (z∈y ⇒ z ⋃ {z} ∈ y)) ⇒ x∈y)}

N is defined by the following set builder:

{x in I as follows: For all (y Ø in y, AND for all z, z in y implies (the union of z with {z}) in y) implies x in y)}

Now, Ø and all z (where all z are, at first, finite sets according to von Neumann's treatment of all the natural numbers.




Are you unaware that there is in fact an injective mapping from ∅ to any set of your choosing? The identify mapping, f(x) = x, will do nicely.
Are you unaware that since there are no members in ∅, there can't be any function (let alone 1-to-1 function) from ∅ to any set of your choosing?
 
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Correction of my previous post.
If you feel the approach you described is necessary, then, please show us how it is done.
It is much simpler than that. Given the Axiom of Infinity, ∃I (Ø∈I ∧ ∀x (x∈I ⇒ x ⋃ {x} ∈ I)), you defined the set N as follows:

N = {x∈I : ∀y ((Ø∈y ∧ ∀z (z∈y ⇒ z ⋃ {z} ∈ y)) ⇒ x∈y)} (it is literally much simpler, since, unlike you, I do not skip on all the finite sets according to von Neumann's treatment of all the natural numbers, which is already satisfies the requirements of the Axiom of Infinity).

In words:

{x in I is as follows: For all y (Ø in y, AND for all z, z in y implies (the union of z with {z}) in y) implies x in y)}

Now, Ø and all z (where all z are, at first, finite sets according to von Neumann's treatment of all the natural numbers, so all is needed in order to satisfy the requirements of the Axiom of Infinity, is already found by the von Neumann's treatment of all the natural numbers.

Therefore one can't skip on it (as you do time after time, jsfisher).

Are you unaware that there is in fact an injective mapping from ∅ to any set of your choosing? The identify mapping, f(x) = x, will do nicely.
Ok vacuous true where p→q and p is false.
 
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Once again you skip on all the finite sets, as defined by von Neumann's treatment of all the natural numbers.

I skipped any treatment of the natural numbers because I don't need any. You are the one insisting it be there, I have shown it isn't required.

This is how you are doing it.

N = {x∈I : ∀y ((Ø∈y ∧ ∀z (z∈y ⇒ z ⋃ {z} ∈ y)) ⇒ x∈y)}

Yes, that is what I wrote and what you already quoted. Repeating it was unnecessary.

N is defined by the following set builder:

No, N is defined by the previous expression.

{x in I as follows: For all (y Ø in y, AND for all z, z in y implies (the union of z with {z}) in y) implies x in y)}

Restating it again does not add value. And you made an error in the process. Be that as it may, though, I am confident you have no idea how the expression actually achieves the goal. You recognize the individual "words" by you don't comprehend their collective meaning.

Now, Ø and all z (where all z are, at first, finite sets according to von Neumann's treatment of all the natural numbers.

Yep, proof you don't comprehend their collective meaning.



Are you unaware that since there are no members in ∅, there can't be any function (let alone 1-to-1 function) from ∅ to any set of your choosing?

Oh? Are you unfamiliar with the phrase, vacuously true? F : ∅ → X is an injective map from the empty set to any set X. A function F is an injective map from the not necessarily distinct sets A to B iff:
∀a∈A (∃b∈B (F(a) = b ∧ ∀c∈A (F(a) = F(c) ⇒ a = c)))​
Nothing there forbids A or B from being empty.
 

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