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Old 19th September 2020, 01:01 PM   #521
jsfisher
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Are you unable to tell us what you mean? Use your words. Examples are not definitions. Dodges are not definitions.
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Old 19th September 2020, 01:42 PM   #522
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Oh, what the heck. Let's dissect your latest heralded post to see just how much correctness it contains.

Originally Posted by doronshadmi View Post
V = { v : (v is not non-finite member of von Neumann ordinals) } = The set of all natural numbers in terms of sets.
This is not a set description within the rules of ZF or ZFC.

Quote:
v(v∪{v} ∈ V ...
This part of the expression insures the entire expression evaluates to false, no matter how you define V.

Quote:
...∧ |v∪{v}| < |V|)...
This part does not evaluate at all since cardinality lacks meaning. You need to define your terms.

Quote:
which means that |V| is not the cardinality of set V.
There is no basis for this conclusion. Undefined terms guarantees it, as does the lack of facility with predicate calculus, but since we have already been forewarned of what to expect, we know the conclusion is embedded in the definition. Anything between the definition and this conclusion is unnecessary.

Quote:
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|).
Maybe in your fantastical world, but if you stay within the bounds of my definitions, everything works out fine.

Quote:
One asks: Is there a bijection among all V members?
What does that even mean? A bijection is a mapping with certain properties from one set to another. A bijection among sets? 'Tis a mystery until you explain your usage.

Quote:
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.
Sets are not bijective with anything. Mappings can be bijective; sets, not at all.

Quote:
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|).
Care to define what you mean by "no-bijection"? By the way, again, this attempt at another expression would also evaluate to false.

Quote:
Here it is:...
Examples aren't proof.

Quote:
which means that < (no-bijection) and = (bijection) are not gathered into <= in order to define |V| <= |X|, |V| <= |V|, in the first place.
"Gathered into"?? Be that as it may, if you are referring here to my cardinality definition, it does not require any deconstruction by you to ascertain the meaning of the relational operator I used. It is fully contained in the definition. All you have done is attempted to reach some bogus conclusion by altering the original definition.

Quote:
Since |V| is not strictly established, the Cantorian notion of strict cardinality for set V, does not hold water.
You haven't come anywhere near establishing anything. You haven't defined what you mean by cardinality, yet.

Quote:
So once again, thank you
At least until you change your mind, again.


So, let's see. Every part, every single part of your post, Doronshadmi, is riddled with mistakes, omissions, and just plain nonsense. There is nothing there for you to point to and be proud. Please stop linking back to it as if it merited high praise. It is line-by-line garbage.

And stop pretending.
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Old 19th September 2020, 03:12 PM   #523
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Originally Posted by jsfisher View Post
There is no set of the von Neumann ordinals simply because it is too big to be a set.
V = { v : (v is not non-finite member of von Neumann ordinals) } = The set of all natural numbers in terms of sets.

Originally Posted by jsfisher View Post
This is not a set description within the rules of ZF or ZFC.
In that case the set of all natural numbers in terms of sets, is undefined within the rules of ZF or ZFC.

Originally Posted by jsfisher View Post
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.
In other words, you are dishonest with yourself.

|V| is not the cardinality of V simply because 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|.

So, exactly as the von Neumann set of ordinals it is too big to be a set, so is the case of all natural numbers in terms of sets, it is too small to be a set.

No wonder that λ was invented out of nowhere (as seen in http://www.internationalskeptics.com...&postcount=451) in order to cover that for all v∪{v} in V, |v∪{v}| is too small in order to be valued as |V|.

Without the ad hoc invention out of nowhere of |V|=λ as a weak limit cardinal (such that λ is neither a successor cardinal nor zero) ZF(C) does not hold water.
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For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video.

Last edited by doronshadmi; 19th September 2020 at 03:47 PM.
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Old 19th September 2020, 04:10 PM   #524
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Originally Posted by doronshadmi View Post
V = { v : (v is not non-finite member of von Neumann ordinals) } = The set of all natural numbers in terms of sets.
This is not a valid definition for your set, V. It violates the Axiom of Restricted Comprehension, for example.

Quote:
In that case the set of all natural numbers in terms of sets, is undefined within the rules of ZF or ZFC.
That does not follow. Just because you are unable to express, within the bounds of set theory, a definition of the minimal set that satisfies the Axiom of Infinity does not mean it cannot be done. From there, the natural numbers are not hard to reach.

Getting to all the ordinals that von Neumann had in mind, as you insist on doing, is an extremely difficult and long route. It is also working backwards, as you are prone to do. And it also requires accepting as given all the things you are trying to dismiss.



Why are you still dodging defining cardinality as you use the term?
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Old Yesterday, 06:29 AM   #525
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Originally Posted by jsfisher View Post
This is not a valid definition for your set, V. It violates the Axiom of Restricted Comprehension, for example.
It is valid because of the following reasons:

1) All v in V and all v∪{v} in V are not non-finite sets, in terms of "von Neumann's treatment of the natural numbers".

2) For all v in V and for all v∪{v} in V, |v| is a strict cardinality of v, and |v∪{v}| is strict cardinality of v∪{v}.

3) V does not have strict cardinality since for all v∪{v} in V, |v∪{v}| < |V|, which means that |V| can't be a measurement value of V.

4) Without strict cardinality, V is not a set in terms of ZF(C), and therefore can't be a member of itself, in the first place.
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That is also over the matrix, is aware of the matrix.

That is under the matrix, is unaware of the matrix.

For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video.

Last edited by doronshadmi; Yesterday at 06:58 AM.
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Old Yesterday, 07:01 AM   #526
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Originally Posted by doronshadmi View Post
It is valid because of the following reasons...
No part of your bogus argument matters until you define your terms.

Cardinality is the big one you continue to dodge. Please define what you mean by cardinality.

ETA: You will need to define "strict cardinality", too.
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Old Yesterday, 01:45 PM   #527
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Originally Posted by jsfisher
You can, in fact, use those meanings to show that the expression |A| <= |B| is identical to (|A| < |B|) OR (|A| = |B|)

Definition 1: |A| = |B| iff there is bijection from A to B (where in case of bijection, B can be replaced by A and we get bijection from A to itself).

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

v or v∪{v} are not non-finite sets, where the cardinality of v (notated as |v|) or the cardinality of v∪{v} (notated as |v∪{v}|) is defined (by definition 1) by bijection form a given domain to itself.


Definition 2: |A| < |B| iff (there is injection no surjection from A to B) OR (A=∅ ∧ B~=∅).

v ∈ V(v∪{v} ∈ V ∧ |v| < |v∪{v}|) by definition 2, which means that the expression |A| <= |B| (injection from A to B) is not satisfied from v to v∪{v} in V.

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|)


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.
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That is also over the matrix, is aware of the matrix.

That is under the matrix, is unaware of the matrix.

For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video.

Last edited by doronshadmi; Yesterday at 01:54 PM.
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Old Yesterday, 01:52 PM   #528
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Originally Posted by doronshadmi View Post
Definition 1: |A| = |B| iff there is bijection from A to B (where in case of bijection, B can be replaced by A and we get bijection from A to itself).

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

v or v∪{v} are not non-finite sets, where the cardinality of v (notated as |v|) or the cardinality of v∪{v} (notated as |v∪{v}|) is defined (by definition 1) by bijection form a given domain to itself.

Definition 2: |A| < |B| iff (there is injection no surjection from A to B) OR (A=∅ ∧ B~=∅).

v ∈ V(v∪{v} ∈ V ∧ |v| < |v∪{v}|) by definition 2, which means that the expression |A| <= |B| (injection from A to B) is not satisfied from v to v∪{v} in V.

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|)


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.

doronshadmi can't even get bijection right.

Last edited by Little 10 Toes; Yesterday at 01:55 PM. Reason: added original post before doronshadmi starts editing it. again.
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Old Yesterday, 02:29 PM   #529
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Originally Posted by doronshadmi View Post
Definition 1: |A| = |B| iff there is bijection from A to B
Great! except your definition doesn't exclude anything that would support your definition for non-finite set. 'Tis a problem, no?

Quote:
(where in case of bijection, B can be replaced by A and we get bijection from A to itself)
You have a compulsion to overstate the obvious and irrelevant, don't you?

Quote:
V = { v : (v is not non-finite member of von Neumann ordinals) } = The set of all natural numbers in terms of sets.
Not a valid definition. Plus, your "non-finite member" is in jeopardy of being an undefined term.

Quote:
Definition 2: |A| < |B| iff (there is injection no surjection from A to B) OR (A=∅ ∧ B~=∅).
Ok, then. Still no support for your "non-finite set" definition. You'll need to try again defining that term.


Kinda funny, too, you end up with substantially the same definition for cardinality as I, just with more steps.




So, what is your definition for non-finite set? You broke the one you provided before.
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Old Yesterday, 03:04 PM   #530
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Doronshadmi, exactly what post of mine are you quoting in the following?


Originally Posted by doronshadmi View Post
Originally Posted by jsfisher
You can, in fact, use those meanings to show that the expression |A| <= |B| is identical to (|A| < |B|) OR (|A| = |B|)
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Old Yesterday, 10:40 PM   #531
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Originally Posted by jsfisher View Post
Doronshadmi, exactly what post of mine are you quoting in the following?
http://www.internationalskeptics.com...97&postcount=2
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Old Today, 12:48 AM   #532
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Originally Posted by jsfisher View Post
Not a valid definition. Plus, your "non-finite member" is in jeopardy of being an undefined term.
I start by using your definition of non-finite set, here it is:
Originally Posted by jsfisher View Post
That would be back when I defined a set X to be non-finite when |N| <= |X| (where N is the minimal set satisfying the requirements of the Axiom of Infinity). Since finite and non-finite are complementary terms, defining one automatically defines the other
So not non-finite set is a finite set by your terms.

Originally Posted by jsfisher
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.
Great, so by using von Neumann's treatment of the natural numbers, we define V as follows:

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

Originally Posted by jsfisher View Post
Kinda funny, too, you end up with substantially the same definition for cardinality as I, just with more steps.
Kinda funny, that you don't understand that your definition of non-finite set holds, only if |N| holds.

Originally Posted by jsfisher View Post
A bijection is a mapping with certain properties from one set to another.
Unlike your relative approach that relies on set A and another set B, I first directly define the cardinality of not non-finite set by using the same set, as follows:

Definition 1: |A| = |B| iff there is bijection from A to B (where in case of bijection, B can be replaced by A and we get bijection from A to itself)

So first I am focused on what happens inside set V, among its members, and it is done in order to check the validity of |V| (will be done later).

v or v∪{v} are not non-finite sets, where the cardinality of v (notated as |v|) or the cardinality of v∪{v} (notated as |v∪{v}|) is defined (by definition 1) by bijection form a given domain to itself (by direct definition of cardinality).

So first |v| or |v∪{v}| (which are the not non-finite members of von Neumann's treatment of the natural numbers) are defined directly.

Definition 2: |A| < |B| iff (there is injection no surjection from A to B) OR (A=∅ ∧ B~=∅).

v ∈ V(v∪{v} ∈ V ∧ |v| < |v∪{v}|) by definition 2, which means that the expression |A| <= |B| (injection from A to B) is not satisfied from v to v∪{v} in V.

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|)


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 = {

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

}

Order is not irrelevant since all we is that v ∈ V(v∪{v} ∈ 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.
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That is also over the matrix, is aware of the matrix.

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For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video.

Last edited by doronshadmi; Today at 02:42 AM.
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Old Today, 03:10 AM   #533
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Some correction of the previous post:

The sentence at the end of the diagram has to be replaced by:

Order is irrelevant since all we care is that v ∈ V(v∪{v} ∈ V) no matter where they are in V.
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That is also over the matrix, is aware of the matrix.

That is under the matrix, is unaware of the matrix.

For more details, please carefully observe Prof. Edward Frenkel's video from https://youtu.be/PFkZGpN4wmM?t=697 until the end of the video.

Last edited by doronshadmi; Today at 03:12 AM.
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Old Today, 04:39 AM   #534
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Originally Posted by doronshadmi View Post
I start by using your definition of non-finite set, here it is:

So not non-finite set is a finite set by your terms.
Is this your final answer, then? No more changes?

Since your latest definition for "finite set" depends on your definition(s) for cardinality, you'll need to save it until later. The term is undefined until you've adequately defined cardinality.

Quote:
Great, so by using von Neumann's treatment of the natural numbers, we define V as follows:

V = { v : (v is not non-finite member of von Neumann's treatment of the natural numbers) } = The set of all natural numbers in terms of sets.
Wrong on two counts (still). The term "non-finite member" is not (yet) defined, and your set-builder formulation falls outside what is allowed.

Quote:
Kinda funny, that you don't understand that your definition of non-finite set holds, only if |N| holds.
Definitions don't "hold". They simply define.

Quote:
Unlike your relative approach that relies on set A and another set B, I first directly define the cardinality of not non-finite set by using the same set, as follows:

Definition 1: |A| = |B| iff there is bijection from A to B (where in case of bijection, B can be replaced by A and we get bijection from A to itself)
That's still a "relative approach", and you have provided nothing to restrict it to non-finite sets.

Quote:
So first I am focused on what happens inside set V...
Congratulations. Your Definition 1 establishes that |V| = |V| and that for all v in V, |v| = |v|.

Quote:
Definition 2: |A| < |B| iff (there is injection no surjection from A to B) OR (A=∅ ∧ B~=∅).
I doubt this is what you really mean. By this definition |V| < |V|. You need more words for the left side of the OR (and you don't need the rest of the OR at all).

Quote:
v ∈ V(v∪{v} ∈ V ∧ |v| < |v∪{v}|) by definition 2, which means that the expression |A| <= |B| (injection from A to B) is not satisfied from v to v∪{v} in V.
Why not? If |v| < |v ∪ {v}| (your definition) then there must be an injection from v to V. If there is an injection from v to V, then |v| <= |v ∪ {v}| (my definition).

Quote:
Now let's look if |A| <= |B| (injection from A to B) can be used in order to establish |V|.
Same deal.
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Old Today, 07:00 AM   #535
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jsfisher, you jumped all over my previous post.


My previous post has to be read in the following order, in order to be understood:

1) First I use your definitions, which define the complement terms "non-finite" and "finite" (which I write as "not non-finite")

2) Then I use von Neumann's treatment of the natural numbers, which is right in line with your point, in order to define V not non-finite members, which are v and v∪{v}.

3) Then I directly define the cardinality for all v in V, by bijection (=) from v to itself.

4) Than I indirectly define the cardinality for all v and v∪{v, by injection non-surjection (<) from v to v∪{v or from ∅ to ~∅, in V.

5) (3) and (4) are done within V, at this stage.

6) Only than I use (4) in order to show that |V| is too big in order to be valued as the cardinality of set V, simply because
v ∈ V(v∪{v} ∈ V ∧ |v| < |v∪{v}| < |V|).

Originally Posted by jsfisher View Post
Since your latest definition for "finite set" depends on your definition(s) for cardinality, you'll need to save it until later. The term is undefined until you've adequately defined cardinality.
Wrong, jsfisher, in case that you have missed it, I start by using your definition of "finite set".

Originally Posted by jsfisher View Post
Wrong on two counts (still). The term "non-finite member" is not (yet) defined, and your set-builder formulation falls outside what is allowed.
Wrong again jsfisher, at this stage I still use your definition of the term "non-finite member".


Originally Posted by jsfisher View Post
Definitions don't "hold". They simply define.
Yes I already aware of your religious approach about definitions.


Originally Posted by jsfisher View Post
That's still a "relative approach", and you have provided nothing to restrict it to non-finite sets.
It is a "direct approach" since the bijection is done from A to itself, where v is not non-finite member (still by your definition of non-finite) of von Neumann's treatment of the natural numbers, which is right in line with your point, isn't it jsfisher?

Originally Posted by jsfisher View Post
Congratulations. Your Definition 1 establishes that |V| = |V| and that for all v in V, |v| = |v|
Wrong, definition 1 directly defines the cardinality of a not non-finite member v (in set V) by bijection from v to itself, where |V| is not considered at all at this stage.


Originally Posted by jsfisher View Post
I doubt this is what you really mean. By this definition |V| < |V|. You need more words for the left side of the OR (and you don't need the rest of the OR at all).
Again, at this stage we argue only about what is inside set V, such that v is not non-finite member (still by your definition of non-finite) of von Neumann's treatment of the natural numbers, which is right in line with your point.


Originally Posted by jsfisher View Post
Why not? If |v| < |v ∪ {v}| (your definition) then there must be an injection from v to V.
Wrong again, bijection (=) is from v to itself in V, where injection non-surjection (<) is from |v| to |v ∪ {v}| or form ∅ to ~∅ in V (anything about V is not considered yet, but only about what within it.)

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

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 = {

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

}

Order is not irrelevant since all we is that v ∈ V(v∪{v} ∈ 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 http://www.internationalskeptics.com...&postcount=523.
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Last edited by doronshadmi; Today at 08:00 AM.
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Old Today, 08:24 AM   #536
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Originally Posted by doronshadmi View Post
1) First I use your definitions, which define the complement terms "non-finite" and "finite" (which I write as "not non-finite")
That definition requires an already existing definition for cardinality. You cannot use mine because you reject it. You can't use yours because you haven't introduced it/them yet. So you cannot use my definition for finite/non-finite set.

Quote:
2) Then I use von Neumann's treatment of the natural numbers, which is right in line with your point, in order to define V not non-finite members, which are v and v∪{v}.
Except you didn't for reasons I have stated multiple times already.

Quote:
3) Then I directly define the cardinality for all v in V, by bijection (=) from v to itself.
Except you didn't. Your definition does not restrict itself to just your set V (which you have failed to define). In case you have forgotten:
Definition 1: |A| = |B| iff there is bijection from A to B
Quote:
4) Than I indirectly define the cardinality for all v and v∪{v, by injection non-surjection (<) from v to v∪{v or from ∅ to ~∅, in V.
Except you didn't. Your definition does not restrict itself to just your set V, nor does it give you your anticipated result.

Quote:
5) (3) and (4) are done within V, at this stage.
Except they weren't. If that is what you meant, then you need to state that in your definitions, explicitly. You didn't. You will also need to fix your inability to introduce your set V, too, before you can refer to it.

Quote:
6) Only than I use (4) in order to show that |V| is too big in order to be valued as the cardinality of set V, simply because
v ∈ V(v∪{v} ∈ V ∧ |v| < |v∪{v}| < |V|).
Except you didn't....
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Old Today, 04:42 PM   #537
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Originally Posted by doronshadmi View Post
snip

1) First I use your definitions, which define the complement terms "non-finite" and "finite" (which I write as "not non-finite")
There's the right way, the wrong way, and the let's-make-things-more-complicated-by-adding-junk-and-making-up-things way.

Guess which way you use doronshadmi?

You've already been using finite/non-finite.

You only started using this nonsense term in the last 4 days. Specifically post 503.

In fact, jsfisher was trying (that's a recurring theme) to correct you in using the right terms before you "write 'not non-finite". Finite and non-finite. They are complimentary. Not finite set and non-finite set, but just finite and non-finite.

Let's stick with the correct terms. Stop making up words.
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