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

[jsfisher]

doronshadmi, if you are going to use the Axiom of Infinity, quote it correctly. This shows your level of credibility. It has been provided to you several times on this page alone.

∃I (∅∈I ∧ ∀x∈I [(x∪{x})∈I])

Doronshadmi thinks he is being clever by not using "for all", which, as you point out, is what the Axiom actually says. It is part of his never-ending effort to disprove definitions, all the while being unable to provide his own.

 

[jsfisher]

By the way, Doronshadmi, were {{∅}} included in the set you are having so much difficulty defining, then you might imagine it thusly:

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

Two sequences of successors. Imagine that.

This set meets all the criteria you've specified for N; it just isn't the set for which you were hoping. Can to imagine what the set might be like if { {∅}, {{{∅}}} } were a member?

 

[doronshadmi]

By the way, Doronshadmi, were {{∅}} included in the set you are having so much difficulty defining, then you might imagine it thusly:

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

Two sequences of successors. Imagine that.

This set meets all the criteria you've specified for N; it just isn't the set for which you were hoping. Can to imagine what the set might be like if { {∅}, {{{∅}}} } were a member?

This is beautiful jsfisher thank you, N can have more than one base member, but it does not change the fact that given any base member, its successors are finite sets since they are based on all their finitely many predecessors (the term all is valid only for finitely many (distinct or non-distinct) objects).

Furthermore, there is a bijection between the bases and their successors, so we are closed under two proper subsets of N, that no one of them has its largest successor (where (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 the considered proper subsets of N:

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

definition 1: N is a non-finite set iff at least one of its base members is infinite.

It is easy to see that this is a circular definition, which is based on the infinite in order to define the infinite.

Doronshadmi thinks he is being clever by not using "for all", which, as you point out, is what the Axiom actually says. It is part of his never-ending effort to disprove definitions, all the while being unable to provide his own.

So jsfisher, by ∃N (∅ ∈ N ∧ ∀x ∈ N ( (x∪{x}) ∈ N)) N is infinite iff at least one x (a base member) is infinite (which is a circular reasoning), therefore ∀x is valid only if any given x (a base member) is finite.

In other words jsfisher, ∃N (∅ ∈ N ∧ ∀x ∈ N ( (x∪{x}) ∈ N)) establishes N as an infinite set only in your circular imagination.

 

[jsfisher]

This is beautiful jsfisher thank you, N can have more than one base member...

Finally! Ok then, you have not yet defined this set N to which you refer.

...but it does not change the fact that given any base member, its successors are finite sets...

You haven't defined finite set, either, and why couldn't one of these so-called base members be an infinite set?

 

[doronshadmi]

Finally! Ok then, you have not yet defined this set N to which you refer.

You are right, N is not a specific set, by the axiom of infinity.

You haven't defined finite set, either

∃ V ( ∅ ∈ V as its one and only one base member ∧ for any given x ∈ V ( ( x ∪ { x } ) ∈ V ) )

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

}

Definition 1: Cardinality is the 'size' of D iff |D| is defined by V.

Definition 2: N is the set of finite cardinalities iff any given n ∈ N is defined by its corresponding V member.

For example:

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

Definition 3: D is called finite iff it is bijective with some particular V member, where |D| is the corresponding N member.

Definition 4: (x∪{x}) is called the largest successor in V iff (x∪{x}) does not have a successor in V.

Definition 5: D is called non-finite iff |D| is not any particular N member, since V does not have the largest successor.

and why couldn't one of these so-called base members be an infinite set?

Because defining N as an infinite set by an infinite set is a circular reasoning that its result is an infinite regression (which is not exactly the property of something that is defined as a base for other things).

 

[jsfisher]

You are write, N is not a specific set, by the axiom of infinity



∃ V ( ∅ ∈ V as its one and only one base member ∧ for any given x ∈ V ( ( x ∪ { x } ) ∈ V ) )

What you have written is more like a theorem (or axiom). It is not a specification for a set. (You can start by losing the ∃V in exchange for something else.)

How do you propose to express "as its one and only base member" in the predicate calculus of set theory? (Keep in mind, too, 'base member' has no formal definition, either.)

How do you propose to express "for any given" in the predicate calculus of set theory?


At some point, too, you'll need to show that your set V is actually a set according to the set theory axioms.

 

[jsfisher]

and why couldn't one of these so-called base members be an infinite set?

Because defining N as an infinite set by an infinite set is a circular reasoning that its result is an infinite regression (which is not exactly the property of something that is defined as a base for other things).

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.

 

[RolandRat]

12 years and about 600 pages, impressive thread. What's it all about? I tried to read the 2008 opening post and my brain farted and died.

 

[jsfisher]

12 years and about 600 pages, impressive thread. What's it all about? I tried to read the 2008 opening post and my brain farted and died.

This thread started off with some excitement by Doronshadmi for some property he'd discovered about prime numbers. Turns out, the discovery was that prime numbers have no positive integer factors besides themselves and one. The topic gradually spiraled back to Doronshadmi's pet peeve with Mathematics: It doesn't handle infinity properly.

Doronshadmi doesn't like the idea that there could be a set of all of the integers, for example.

He takes some basis for this from Philosophy and an historic distinction between "actual" and "potential infinities", but Mathematics is not bound to any particular philosophic underpinning; it can accommodate more than one view. The "standard view" tends to be the one that is most useful. The "standard view" isn't to Doronshadmi's liking, so he argues that (all of) Mathematics is wrong.

 

[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.

Recurring decimals are another point where doronshadmi’s maths differ from the conventional approach. 1 / 3 * 3 does not equal 1 in his view, but .999...

 

[Little 10 Toes]

For the third time I notice doronshadmi dodged at least two direct questions from jsfisher posted here (http://www.internationalskeptics.com/forums/showthread.php?p=13187786#post13187786):
1) You are offering substitute words without conveying any meaning. What does "most complex member" mean?
and
2) For that matter, what bearing does having or not having a largest member have on this discussion? If there were a set which was both infinite and did have a "largest member" (by some appropriate definition of largest member), then what? What if there were set that had no largest member but was finite, then what?


Is there a reason you can't/won't answer these questions doronshadmi, yet again?

 

[Little 10 Toes]

And generally speaking, from what I remember, there needs to be philosophy when doing math. Basically, there is something "bad" if you're using numbers for "evil".

Quote from doronshadmi:

"In my opinion the current dichotomy between our Ethical skills and our Logical\Technological skills leading us very quickly to one of the dead ends to the Evolution."

http://www.internationalskeptics.com/forums/showpost.php?p=7682481&postcount=8


Here's a post that can clear things up:

As you have already demonstrated repeatedly, you don't answer questions, Doron. You just repeat, post links to repeats, shift, evade, then eventually end up with some flavor of "you just can't get it".

Give it up. You got nothing.

http://www.internationalskeptics.com/forums/showpost.php?p=7673019&postcount=16601

 

[doronshadmi]

What you have written is more like a theorem (or axiom). It is not a specification for a set. (You can start by losing the ∃V in exchange for something else.)

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

How do you propose to express "as its one and only base member" in the predicate calculus of set theory? (Keep in mind, too, 'base member' has no formal definition, either.)

Please keep in mind that, for example, Set or Cardinality do not have formal definitions in the predicate calculus of set theory.

But I'll try to figure out how to define it in the predicate calculus of set theory.

How do you propose to express "for any given" in the predicate calculus of set theory?

It does not have to be defined in the predicate calculus of set theory, but I'll try to figure out how to define it in the predicate calculus of set theory.

At some point, too, you'll need to show that your set V is actually a set according to the set theory axioms.

According to the algorithm above V is actually a set according to the set theory axioms, since any given member of it is a distinct and valid set.

If you disagree with me, then please show some invalid set, as defined by my algorithm above.

 

[doronshadmi]

You couldn't imagine the set possibly containing {{{}}} as a member -- that was your error.

It was indeed my error, but it is recovered by http://www.internationalskeptics.com/forums/showpost.php?p=13191898&postcount=203 and you actually ignored the most of it.

Now you cannot imagine it containing an infinite set as a member -- again, your error.

It is not about imagine, but about circular reasoning, as shown both in http://www.internationalskeptics.com/forums/showpost.php?p=13191898&postcount=203 and http://www.internationalskeptics.com/forums/showpost.php?p=13192433&postcount=205.

 

[caveman1917]

There has been some travelling down by-ways, too.

This thread has actually been quite productive for me, or at least the part about Godel's Incompleteness Theorem was. I had never been able to really wrap my head around GIT, I could follow the math but it never really "clicked" in my head to get an intuitive grasp of it. Until in this thread I realized the connection between GIT and the halting problem and I started looking at GIT through the lens of computer science and it finally "clicked" in my head. So thanks for this thread doronshadmi, it ended up being quite helpful for me.

 

[jsfisher]

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

That's an attempt to describe a process. It is not a definition of a set.

You need something like: V = { x : <logical expression> } where the logical expression provides the restrictions for what is and is not in the set. The logical expression must stay within the bounds of the set axioms (in particular, the Axiom of Restricted Comprehension).

 

[doronshadmi]

That's an attempt to describe a process. It is not a definition of a set.

jsfisher, there is no definition of a set by classical logic (what you call "the predicate calculus of set theory").

Since this is the case, my algorithm of V set is valid not less than your way to define V, and again this algorithm can be taken at once (in parallel) exactly as V members can be taken at once (in parallel).

You insist to take it as a process because of the physical restrictions of the current computers, but that is your restriction, not mine.

You need something like ...

Well, I do not need to use classical logic in order to define what are and what are not the members of V.

But I can try (if I wish) to do it by using classical logic.

Also it will be nice to define "base member" and "any given" in terms of classical logic.

 

[jsfisher]

jsfisher, there is no definition of a set by classical logic (what you call "the predicate calculus of set theory").

So, you "prove" Mathematics is wrong by simply rejecting its rules. Not very convincing.

Since this is the case, my algorithm of V set is valid not less than your way to define V, and again this algorithm can be taken at once (in parallel) exactly as V members can be taken at once (in parallel).

A set definition must definitively answer the question "Is x a member of the set?" for all x. That is a yes/no question. Your "algorithm" does not answer it.

You insist to take it as a process because of the physical restrictions of the current computers, but that is your restriction, not mine.

You present it as a process. It is all on you.

 

[doronshadmi]

This thread has actually been quite productive for me, or at least the part about Godel's Incompleteness Theorem was. I had never been able to really wrap my head around GIT, I could follow the math but it never really "clicked" in my head to get an intuitive grasp of it. Until in this thread I realized the connection between GIT and the halting problem and I started looking at GIT through the lens of computer science and it finally "clicked" in my head. So thanks for this thread doronshadmi, it ended up being quite helpful for me.

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.

 

[doronshadmi]

So, you "prove" Mathematics is wrong by simply rejecting its rules. Not very convincing.

Not Mathematics, but only what is known as classical mathematics in case that it uses the term "all" on endless things.

A set definition must definitively answer the question "Is x a member of the set?" for all x.

"all" is valid only for finitely many things.

That is a yes/no question. Your "algorithm" does not answer it.

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

You present it as a process. It is all on you.

It is presented in terms of the physical restrictions of the current computers, but its notion (that can be taken in parallel) is not limited by its presentation.