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

[jsfisher]

It does not need a yes/no question, since its answers can't be but yes.

You might want to reconsider that response. If the answer were always "yes", then everything would be a member of your set V. For {{{}}}, don't you want the answer to be "no"?

 

[caveman1917]

Dear caveman1917, it is nice to know that some thread that I started 12 years ago helped you to get GIT, but I am only one poster of it and in the case of GIT I think that jsfisher also helped you a lot to get GIT.

Well yes, it was jsfisher's comments which helped it, not yours - your comments are pseudo-mathematics. But you should still be credited for creating the thread and ultimately causing the whole discussion to take place.

 

[zooterkin]

There has been some travelling down by-ways, too. For example, doronshadmi appears to believe you cannot simply define a set by specifying a property its members have (all positive numbers, for example), you need to enumerate them.
...

Case in point:

It has the needed properties in order to define V members as follows:

{} is the one and only one base member of set V.

x={} ∈ V 
Do loop x forever
  (x∪{x}) ∈ V
   x=(x∪{x})               
Next

...

 

[RolandRat]

@JS and Zoot, thanks for explaining. I'll let you guys get back to it and check back in a few years to see if any progress has been made

 

[doronshadmi]

Case in point:

Take for example set V:

x={} ∈ V 
Do loop x forever
  (x∪{x}) ∈ V
   x=(x∪{x})               
Next

The basic properties of V are as follows:

1) {} is its one and only one base member.

2) Any member that is not {} is of the form (x∪{x}).

3) For any given x in V, (x∪{x}) is in V.

Definition: (x∪{x}) is called the largest successor in V iff (x∪{x}) does not have a successor in V.

4) No (x∪{x}) is the largest successor in V.

5) No enumeration is involved in properties 1 to 4.

There has been some travelling down by-ways, too. For example, doronshadmi appears to believe you cannot simply define a set by specifying a property its members have (all positive numbers, for example), you need to enumerate them.

Wrong, according to property (5).

 

[jsfisher]

Take for example set V:

x={} ∈ V 
Do loop x forever
  (x∪{x}) ∈ V
   x=(x∪{x})               
Next

Still not a description of a set. Still lacks a membership function.

 

[doronshadmi]

You might want to reconsider that response. If the answer were always "yes", then everything would be a member of your set V. For {{{}}}, don't you want the answer to be "no"?

The Axiom of Infinity does not define a set nor does it assert it as infinite. It simply asserts the existence of a set that satisfies two properties.

There is nothing in the Axiom itself to rule out an infinite set as a member. You couldn't imagine the set possibly containing {{{}}} as a member -- that was your error. Now you cannot imagine it containing an infinite set as a member -- again, your error.

Here it is, where N is not a specific set and also there are base members which are infinite sets.

∃ N ( ∅ ∈ N ∧ for any given x ∈ N ( ( x ∪ { x } ) ∈ N ) )


Definition: (x∪{x}) is called the largest successor in N iff (x∪{x}) does not have a successor in N.

Here is the bijective map between proper subsets of N, where for any given proper subset there is a distinct base member, and its other distinct members are defined according to (x∪{x}), such that no (x∪{x}) is the largest successor in any given proper subset):

N = {
∅,                    ---> {{∅}}, ---> ...   <-- base members
{∅},                  ---> {{∅}, {{∅}} }, ---> ...
{∅, {∅}},             ---> {{∅}, {{∅}}, {{∅}, {{∅}}}}, ---> ...
{∅, {∅}, {∅, {∅}}},  ---> {{∅}, {{∅}}, {{∅}, {{∅}}}, {{∅}, {{∅}}, {{∅}, {{∅}}}}}, ---> ...
...                         ...
}

Moreover, each proper subset has its own unique distinct members, since they are derived from their unique base member even though (x∪{x}) is a common property among N's proper subsets.

So N has actually two different properties that define any given proper subset of N, which are:

1) Distinct base member (where also infinite sets are base members).

2) A common form ( (x∪{x}) in case of this axiom ).

Since no proper subset of N has its largest successor, so is the case with the proper subsets of any given infinite member of N, which is one of its base members (the definition of the largest successor and properties (1) and (2) hold for any given set).

Moreover, since each proper subset of N has its own unique members (because they are derived by (x∪{x}) form from a distinct base member), we are able to define a diagonal set, which its members do not follow after the unique form of the members of any given proper subset of N.

But let's not forget that also this set (which is supposed to be a base member of N) even though it is not a member of any given proper subset of N, is indirectly defined by the definition of the largest successor and properties (1) and (2).

So, being closed under succession, is actually being closed under endless construction, which prevents |N|<=|N| as if N has all of its members, or ω as an ordinal number after all { |{}|, |{{}}|, |{{},{{}}}|, ... }

 

[jsfisher]

You might want to reconsider that response. If the answer were always "yes", then everything would be a member of your set V. For {{{}}}, don't you want the answer to be "no"?

You quote me, but don't respond to the remark? You need a membership function (i.e. m(z) if and only if z is a member of your set V). You don't have one.

The Axiom of Infinity does not define a set nor does it assert it as infinite. It simply asserts the existence of a set that satisfies two properties.

There is nothing in the Axiom itself to rule out an infinite set as a member. You couldn't imagine the set possibly containing {{{}}} as a member -- that was your error. Now you cannot imagine it containing an infinite set as a member -- again, your error.

Here it is, where N is not a specific set and also there are base members which are infinite sets.[/quote]

Whether there are so-called base members other than the empty set is undetermined, but at least you now accept that there is nothing in the Axiom of Infinity to exclude them, including infinite sets.

I guess you've made progress.

∃ N ( ∅ ∈ N ∧ for any given all x ∈ N ( ( x ∪ { x } ) ∈ N ) )

Stick with the axiom as written.

Definition: (x∪{x}) is called the largest successor in N...

Not sure what relevance you think all that has, but whatever.

...
So N has actually two different properties that define any given proper subset of N, which are:

1) Distinct base member (where also infinite sets are base members).

2) A common form ( (x∪{x}) in case of this axiom ).

Nonsense. {{}, {{}}}, for example, is a proper subset and satisfies neither property.

 

[doronshadmi]

You quote me, but don't respond to the remark? You need a membership function (i.e. m(z) if and only if z is a member of your set V). You don't have one.

Since only {} is the one and only one base member of set V, {{{}}} can't be defined as its successor by (x∪{x}) etc.

Stick with the axiom as written.

In predicate logic, a universal quantification is a type of quantifier, a logical constant which is interpreted as "given any" or "for all".

https://en.wikipedia.org/wiki/Universal_quantification

Not sure what relevance you think all that has, but whatever.

Then please ask, your "but whatever" is a sign that you don't wish to understand it.

Nonsense. {{}, {{}}}, for example, is a proper subset and satisfies neither property.

Well {{}, {{}}} is satisfied by both properties, since:

1) {} is its base member.

2) (x∪{x}) is the form that establishes {{}, {{}}} from {{}}∪{{{}}}.

Here it is:
{
∅ (the base member),
∅∪{∅}={∅} ,
{∅}∪{{∅}}={ ∅, {∅} }={{}, {{}}},
...
}

EDIT:

This part has to be corrected:

Since no proper subset of N has its largest successor, so is the case with the proper subsets of any given infinite member of N, which is one of its base members (the definition of the largest successor and properties (1) and (2) hold for any given set).

Into:

"No proper subset with infinitely many finite members of N has its largest successor, and so is the case with the proper subsets of any given infinite member of N, which is one of its base members (the definition of the largest successor and properties (1) and (2) hold for any given set)."

 

[jsfisher]

Since only {} is the one and only one base member of set V, {{{}}} can't be defined as its successor by (x∪{x}) etc.

Your post has nothing to do with the post to which you are responding. You have yet to define the set V. You need a membership function.

Where is your membership function that specifies whether or not some set is a member of V?

 

[doronshadmi]

Your post has nothing to do with the post to which you are responding. You have yet to define the set V. You need a membership function.

Where is your membership function that specifies whether or not some set is a member of V?

Open program
  Call function M(V,{})
Close

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

Function M(set_name,x)
x ∈ set_name 
  Do x forever
    (x∪{x}) ∈ set_name
     x=(x∪{x})               
  No-end
End

As you can see, my membership function gets a set name and a base member.

In this case it gets set name V and base member {}, and we have the following infinite set:

V={
    
       ∅, 
       {∅}, 
       { ∅, {∅} }, 
       { ∅, {∅} , {∅, {∅}} }, 
       { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, 
...  

}

which is infinite exactly because it does not have its largest successor.

You take my function in terms of a process because you get it in terms of the current physical limitations of computers, which work step by step.

But my membership function is not restricted to these limitations, since it can be taken in one step, a parallel one.

Please be aware that your membership function (i.e. m(z) if and only if z is a member of set V) can also be taken step by step, where in each step m(z) decides by some logical rule if z is a member of V, or not.

But I do not do that since not the number of steps is important, but the fact that in both cases V does not have its largest successor.

 

[jsfisher]

Open program
  Call function M(V,{})
Close

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

Function M(set_name,x)
x ∈ set_name 
  Do x forever
    (x∪{x}) ∈ set_name
     x=(x∪{x})               
  No-end
End

As you can see, my membership function gets a set name and a base member

You function needs to indicate if some set is a member of your set V or not. That would be a true/false proposition.

Your set V remains undefined.

 

[doronshadmi]

jsfisher, you opened the door into a paradise called N (here: http://www.internationalskeptics.com/forums/showpost.php?p=13191454&postcount=202), which you are unaware of it, exactly because you are focused on some gate keeper that decides if some set is a member of N, or not.

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

This time please be focused very carefully on N by being aware that it includes distinct base members, where each distinct base member has its own unique successors ("descendants") (we have unique proper subsets of N that do not have their largest successors ("descendants") as follows):

Definition: (x∪{x}) is called the largest successor in N iff (x∪{x}) does not have a successor in N.

∃ N ( ∅ ∈ N ∧ for any given x ∈ N ( ( x ∪ { x } ) ∈ N ) )


Here is a bijective map between such proper subsets of N, where for any given proper subset there is a distinct base member, and its other distinct members are defined according to (x∪{x}), such that no (x∪{x}) is the largest successor in any such given proper subset):

N = {
∅,                    ---> {{∅}}, ---> ...   <-- base members
{∅},                  ---> {{∅}, {{∅}} }, ---> ...
{∅, {∅}},             ---> {{∅}, {{∅}}, {{∅}, {{∅}}}}, ---> ...
{∅, {∅}, {∅, {∅}}},  ---> {{∅}, {{∅}}, {{∅}, {{∅}}}, {{∅}, {{∅}}, {{∅}, {{∅}}}}}, ---> ...
...                         ...
}

There is no gate keeper for N, since any given set that is not a successor of some base member, is a base member in N.

 

[Little 10 Toes]

... and how does this help you define cardinality?

 

[doronshadmi]

... and how does this help you define cardinality?

V is exactly the non-finite proper subset of N, which its base member is {}.

Open program
  Call function M(V,{})
Close

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

Function M(set_name,x)
x ∈ set_name 
  Do x forever
    (x∪{x}) ∈ set_name
     x=(x∪{x})               
  No-end
End

As you can see, my membership function gets a set name and a base member.

In this case it gets set name V and base member {}, and we have the following infinite set:

V={
    
       ∅, 
       {∅}, 
       { ∅, {∅} }, 
       { ∅, {∅} , {∅, {∅}} }, 
       { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, 
...  

}

Definition 1: (x∪{x}) is called the largest successor in V iff (x∪{x}) does not have its successor in V.

D is a placeholder for any given set.

Definition 2: Cardinality is the 'size' of D iff |D| is defined by V.

Definition 3: K is the set of finite cardinalities iff any given k ∈ K is defined by its corresponding V member.

For example:

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

Definition 4: D is called finite iff it is bijective with some particular V member, where |D| is the corresponding K member.

Definition 5: D is called non-finite iff |D| is not any particular K member, since V does not have its largest successor, by definition 1.

 

[jsfisher]

Nonsense. {{}, {{}}}, for example, is a proper subset and satisfies neither property.

Well {{}, {{}}} is satisfied by both properties....

I've learned something very valuable: Never post while in the middle of a video conference call.

I think I know what I had meant, but what I posted wasn't it. Still, since it is really an aside at the main current point and an excuse for Doronshadmi not not address the main point, I'll leave it for later, if it comes up again.

 

[jsfisher]

jsfisher, you opened the door into a paradise called N (here: http://www.internationalskeptics.com/forums/showpost.php?p=13191454&postcount=202), which you are unaware of it, exactly because you are focused on some gate keeper that decides if some set is a member of N, or not.

No, you were repeatedly stating grossly incorrect things about the Axiom of Infinity. I simply provided you an example of something not denied by the Axiom using the very presentation format you had been insisting on repeating in post after post.

Apparently it was effective, because you then were accepting that the Axiom didn't exclude things you had been insisting it did. Alas, though, you continued to cling to the also incorrect idea that the Axiom excluded infinite sets as these so-called base members.

Now, although the Axiom of Infinity does not exclude the possibility the set it asserts has some members you were convince it couldn't have, it doesn't require any of them, either. The Axiom does now and always has simply asserted the existence of a set for which two properties hold.

This time please be focused very carefully on N by being aware that it includes distinct base members...

"Base member" was an informal term I introduced for the sole purpose of conveying a concept. If you intend to use it in any formal sense, you will need to define it.

...where each distinct base member has its own unique successors...

Like there.

...
∃ N ( ∅ ∈ N ∧ for any given x ∈ N ( ( x ∪ { x } ) ∈ N ) )

You continue to recite the Axiom of Infinity while continuing to miss that the Axiom does not define the set, N.

If you want a certain set N, you need to define it, and you need to define it in a way acceptable to the language of set theory.

 

[jsfisher]

As you can see, my membership function gets a set name and a base member.

What I see is that your "membership function" isn't a membership function.

 

[doronshadmi]

The Axiom does now and always has simply asserted the existence of a set for which two properties hold.

The two properties are:

1) One property establishes sets, which are not of the form (x∪{x}) (where ∅ is an example of such set).

2) The other property establishes sets, which are of the form (x∪{x}).

"Base member" was an informal term I introduced for the sole purpose of conveying a concept. If you intend to use it in any formal sense, you will need to define it.

Definition 1: Given set N, any given member which is not of the form (x∪{x}), is called a base member in N.

Definition 2: Given set N, any given member which is of the form (x∪{x}), is called a successor member in N.

By definition 2 any given successor member in N is defined by (x∪{x}) as a common property among successors.

By definition 1 any given base member in N is defined by ~(x∪{x}) as a common property among bases.

So N members are for the form (x∪{x}) (successors) OR not of the form (x∪{x}) (bases), which is a tautology.

It means that the gate keeper of N paradise (seen in http://www.internationalskeptics.com/forums/showpost.php?p=13197067&postcount=233) is out of job.

You continue to recite the Axiom of Infinity while continuing to miss that the Axiom does not define the set, N.

If you want a certain set N, you need to define it, and you need to define it in a way acceptable to the language of set theory.

∃ N ( ∅ [where ∅ is an example of a member, which is not of the form (x∪{x})] ∈ N ∧ for any given x ∈ N ( ( x ∪ { x } ) ∈ N ) )

Definition 3: (x∪{x}) is called the largest successor in N iff (x∪{x}) does not have its successor in N.

N is infinite iff at least one of its successors does not have its largest successor, or in other words, if N is infinite, the term "all" is not satisfied (N paradise is closed under construction and no gate keeper is needed).

 

[doronshadmi]

What I see is that your "membership function" isn't a membership function.

It is a membership function that does not need a gate (yes/no) keeper.