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

[jsfisher]

My point is straightforward: Traditional mathematics does not consistently and rigorously establish a non-finite set.

Great! Now, how do you support this point of view?

 

[doronshadmi]

Great! Now, how do you support this point of view?

1) There is nothing in your definition of relative cardinality (|A| <= |B| if and only if there exists an injection from A to B) which establishes a non-finite set.

2) The ZF(C) axiom of infinity establishes the existence of (at least one) set, I that has those two properties:

Property 1: ∅ ∈ I

Proeprty 2: ∀ x ∈ I ( ( x ∪ { x } ) ∈ I )

This axiom by itself does not establish a non-finite set, no matter what name is given to it.

3) No axiom schema can help here since it is by itself a set of axioms that can't be non-finite as long as a non-finite set is not consistently and rigorously established.

4) You can use any relations that you like among ZF axioms, but they can't establish a non-finite set in a way that is not arbitrary, exactly because cardinality is not directly defined by it.

Subtlety and nuance escape you. I did not define cardinality, at least not directly. Euclidean Geometry does not define point (directly), nor does ZF Set Theory define set (directly).

Instead, ZF Set Theory provides a set of axioms and axiom schema that characterize set properties; Euclidean Geometry characterizes points by way of its postulates; and I defined a relationship sufficient for comparing cardinalities of sets.

Guess what that does.

I do not have to guess, as long as your system does not define Cardinality directly, you can't consistently and rigorously establish a non-finite set.

 

[jsfisher]

1) There is nothing in your definition of relative cardinality (|A| <= |B| if and only if there exists an injection from A to B) which establishes a non-finite set.

Yes, very true. My approach to cardinality applies to all sets, not just some.

2) The ZF(C) axiom of infinity establishes the existence of (at least one) set, I that has those two properties:
...
This axiom by itself does not establish a non-finite set, no matter what name is given to it.

Another true statement.

Then again, what's missing from all this? What is it that you, Doronshadmi, continually overlook because you continue to apply an elementary school perspective to all of Mathematics?

Answer: A meaning for 'non-finite set'.

And how could that be remedied? A simple definition, perhaps, of what it means for a set to be non-finite? Maybe something like:

A set, X, is non-finite if and only if |N| <= |X|, where N is the von Neumann ordinal.​

Other definitions are possible, but this one will do.

I do not have to guess, as long as your system does not define Cardinality directly, you can't consistently and rigorously establish a non-finite set.

And yet reality would disagree with you.

 

[doronshadmi]

A simple definition, perhaps, of what it means for a set to be non-finite? Maybe something like:

A set, X, is non-finite if and only if |N| <= |X|, where N is the von Neumann ordinal.​

Other definitions are possible, but this one will do.

|N| <= |X| can be written as (|N|<|X|) OR (|N|=|X|).

In that case you consistently and rigorously need to show that N is a non-finite set (after all we discuss about mathematics, isn't it?)

Please provide another definition or consistently and rigorously show that N is a non-finite set.

And yet reality would disagree with you.

Please define reality.

 

[jsfisher]

|N| <= |X| can be written as (|N|<|X|) OR (|N|=|X|).

Other than your insistence on adding unnecessary things, why would we need to do that?

In that case you constantly and rigorously need to show that N is a non-finite set.

Since |N| <= |N|, done!

 

[doronshadmi]

Since |N| <= |N|, done!

Not done as long as you consistently and rigorously show that N is a non-finite set (after all we discuss about mathematics, isn't it?)

Please provide another definition or consistently and rigorously show that N is a non-finite set.

 

[jsfisher]

Not done as long as you consistently and rigorously show that N is a non-finite set (after all we discuss about mathematics, isn't it?)

You don't understand how definitions work, do you?

The term, non-finite set, has been defined. N is non-finite by that defnition.

 

[doronshadmi]

A Wikipedia quote (https://en.wikipedia.org/wiki/Ordinal_number#Von_Neumann_definition_of_ordinals):

First few von Neumann ordinals

0 	= { } 	         = ∅
1 	= { 0 } 	 = {∅}
2 	= { 0, 1 } 	 = { ∅, {∅} }
3 	= { 0, 1, 2 } 	 = { ∅, {∅} , {∅, {∅}} }
4 	= { 0, 1, 2, 3 } = { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }
...

A Wikipedia quote:

Perhaps a clearer intuition of ordinals can be formed by examining a first few of them: as mentioned above, they start with the natural numbers, 0, 1, 2, 3, 4, 5, … After all natural numbers comes the first infinite ordinal, ω,...

In this case one first has to consistently and rigorously show that the non-finite set of all natural numbers, exists.

 

[doronshadmi]

You don't understand how definitions work, do you?

The term, non-finite set, has been defined. N is non-finite by that defnition.

Sorry but your link does not work.

If you mean to "A set, X, is non-finite if and only if |N| <= |X|, where N is the von Neumann ordinal.", then, as very simply shown, this definition does not satisfy http://www.internationalskeptics.com/forums/showpost.php?p=13163749&postcount=88.

Your "definition" is a good example of how mathematics is arbitrarily done.

 

[Little 10 Toes]

Please define "non-finite set". Because you do not use the term "infinite set", your term must have a special meaning to you.

 

[jsfisher]

In this case one first has to consistently and rigorously show that the non-finite set of all natural numbers, exists.

No, one doesn't. One would need to show only that the von Neumann ordinal exists within set theory, which it does, but I'll leave that bit of trivia to you to show. (Hint: Axioms of Infinity and of Restricted Comprehension.) The definition I have provided for "non-finite set" is sufficient to show the von Neumann ordinal as a non-finite set.

If you want to label the members of the von Neumann ordinal with natural numbers, go right ahead, but that's your doing, not a requirement of mine.

 

[jsfisher]

I apologize. It seems I have introduced some confusion I did not intend by using the term "von Neumann ordinal" in a non-standard way.

The Axiom of Infinity establishes the existence of a set with certain properties. The set contains the empty set and what I will call other base elements. The set contains the successors of all of its elements (which would be its base elements, and their successors, and their successors, and ...).

A set which can be considered the minimal set satisfying the axiom's conditions is the set I will call N. It includes only the empty set as a base element. The empty set is the only element of N that is not the successor another element.

So, a set X is a non-finite set if and only if |N| <= |X|, where N is the minimal set established by the Axiom of Infinity.

 

[doronshadmi]

So, a set X is a non-finite set if and only if |N| <= |X|, where N is the minimal set established by the Axiom of Infinity.

ZF Axiom of Infinity is a good example of how Traditional Mathematics arbitrarily forces N to actually be non-finite by simply ignoring the fact that the largest successor does not exist as one of N's members.

As a result |N| (N size) can't be defined since N is closed under this impossible existence.

So to say that "a set X is a non-finite set if and only if |N| <= |X|" is simply nonsense.

Here is how this nonsense is done:

The Axiom of Infinity establishes the existence of a set with certain properties. The set contains the empty set and what I will call other base elements. The set contains the successors of all of its elements (which would be its base elements, and their successors, and their successors, and ...).

...and their successors, and their successors, and ..., which is a fundamental property that can't able The Axiom of Infinity to establish N.

Since N is not actually established, anything that is based on it, is not established.

It has to be clear that non-finite collection of members is taken in terms of actual infinity by traditional mathematicians, in order to establish N.

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

The claim that |N| exists as an extension of N members, without directly define Cardinality, has no consistent and rigor foundations.

 

[doronshadmi]

Please define "non-finite set". Because you do not use the term "infinite set", your term must have a special meaning to you.

Please read very carefully http://www.internationalskeptics.com/forums/showpost.php?p=13161559&postcount=73 including its link.

 

[jsfisher]

ZF Axiom of Infinity is a good example of how Traditional Mathematics arbitrarily forces N to actually be non-finite by simply ignoring the fact that the largest successor does not exist as one of N's members.

As a result |N| (N size) can't be defined since N is closed under this impossible existence.

You are arguing in circles.

You cannot disprove definitions. And you certainly cannot arbitrarily apply your own definitions to do so.

But you did.

So, how do you define what it means for a set to be non-finite.

 

[jsfisher]

Please read very carefully http://www.internationalskeptics.com/forums/showpost.php?p=13161559&postcount=73 including its link.

Nothing was defined there. Quite the opposite. There you insisted you didn't need to define anything because it was intuitive. This cycles back to your infamous proof by direct perception -- if you, Doronshadmi and only you, really, really believe something in Mathematics to be true, well, then, it must be true, because that's what you were taught on or about the third grade.

That's a great self-serving philosophy, I suppose, but not particularly useful nor convincing.

 

[doronshadmi]

Nothing was defined there.

Because you ignore the link there.

 

[doronshadmi]

Here is an improved version of my definitions (which, unlike you, I do not take a definition as something that can't be changed by better notions).

Let A be any arbitrary set.

Let N be the set of natural numbers.

Definition 1: Cardinality is the "size" of A (notated as |A|) iff it is related to the members of N, where 0 is a member of N.


Let < be less than.

Let > be more than.

Let = be equal to.


A relative measure of "sizes" of two sets does not have to be done by specific directions (the terms "from A to B" or "from B to A" are irrelevant).


Definition 2: The cardinality between the two sets A and B is a relative measure of their "sizes", where ( ((|A| < |B|) iff (|B| > |A|)) OR ((|A| = |B|) iff (|B| = |A|)) ).

(visually in "((|A| < |B|) iff (|B| > |A|))" there are two different symbols ("<" and ">"), where in "((|A| = |B|) iff (|B| = |A|))" there is the same symbol ("="), which intuitively reinforces the meaning of inequality or equality).

These definitions say nothing about being finite or non-finite sets.

So:

You are arguing in circles.
how do you define what it means for a set to be non-finite.

Definition 3: Set A is called finite iff given any n in N, |A| is any particular n

Definition 4: Set A is called non-finite iff given any n in N, |A| is not any particular n


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

By the standard notion "given any" is the same as "for all" ( as seen in https://en.wikipedia.org/wiki/Universal_quantification ) but not in my framework, where "given any" holds for both finite and non-finite sets, where "for all" holds only for finite sets.


Let ≤ be less than or equal to.

Let ≥ be greater than or equal to.

Definition 5: The cardinality between sets A and B is called non-strict inequality iff ((|A| ≤ |B|) iff (|B| ≥ |A|)).

 

[jsfisher]

Let N be the set of natural numbers.

Definition 1: Cardinality is the "size" of A (notated as |A|) iff it is related to the members of N, where 0 is a member of N.

I doubt that is what you meant. "If and only if" is a boolean relationship with operands that are either true or false. When used as part of a definition, the if-and-only-if is taken as a constraint requiring the left-hand side (the thing being defined) bear the same truth value as the right-hand side.

You are claiming that the left-had side, "Cardinality is the 'size' of A", is a statement that can be either true or false.

On the other hand, the right-hand side, "[set A] is related to the members of N", appears to be a true/false valued proposition as would be required, but what does "related to" mean?


Your claimed Definition 1 isn't a definition, and it introduces another undefined term.

What is your definition for "non-finite set"?

 

[doronshadmi]

I doubt that is what you meant. "If and only if" is a boolean relationship with operands that are either true or false. When used as part of a definition, the if-and-only-if is taken as a constraint requiring the left-hand side (the thing being defined) bear the same truth value as the right-hand side.

You are claiming that the left-had side, "Cardinality is the 'size' of A", is a statement that can be either true or false.

On the other hand, the right-hand side, "[set A] is related to the members of N", appears to be a true/false valued proposition as would be required, but what does "related to" mean?


Your claimed Definition 1 isn't a definition, and it introduces another undefined term.

What is your definition for "non-finite set"?

Thank you.

Being a member of N is not determined only by some specific property of that member, where "related to" is general enough in order to capture both the common or the specific properties of being a member of N.

For example: Not any given N member = 0 (common property), but any given element = 0 is N member (specific property).