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Old 16th August 2020, 10:11 AM   #201
jsfisher
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Originally Posted by Little 10 Toes View Post
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.
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Old 16th August 2020, 10:55 AM   #202
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By the way, Doronshadmi, were {{∅}} included in the set you are having so much difficulty defining, then you might imagine it thusly:

Code:
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?
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Old 17th August 2020, 12:10 AM   #203
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Originally Posted by jsfisher View Post
By the way, Doronshadmi, were {{∅}} included in the set you are having so much difficulty defining, then you might imagine it thusly:

Code:
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:
Code:
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.

Originally Posted by jsfisher View Post
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.
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Last edited by doronshadmi; 17th August 2020 at 01:30 AM.
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Old 17th August 2020, 04:02 AM   #204
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Originally Posted by doronshadmi View Post
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.

Quote:
...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?
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Old 17th August 2020, 12:32 PM   #205
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Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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 ) )

Code:
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 nN is defined by its corresponding V member.

For example:
Code:
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.

Originally Posted by jsfisher View Post
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).
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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; 17th August 2020 at 12:46 PM.
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Old 17th August 2020, 12:52 PM   #206
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Originally Posted by doronshadmi View Post
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.
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Old 17th August 2020, 01:04 PM   #207
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Originally Posted by doronshadmi View Post
Originally Posted by jsfisher View Post
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.
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Old 17th August 2020, 01:25 PM   #208
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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.
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Old 17th August 2020, 02:27 PM   #209
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Originally Posted by RolandRat View Post
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.
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Last edited by jsfisher; 17th August 2020 at 02:28 PM.
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Old 17th August 2020, 02:37 PM   #210
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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...
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Old 17th August 2020, 04:09 PM   #211
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For the third time I notice doronshadmi dodged at least two direct questions from jsfisher posted here (http://www.internationalskeptics.com...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?
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Old 17th August 2020, 04:12 PM   #212
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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...81&postcount=8


Here's a post that can clear things up:
Quote:
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...ostcount=16601

Last edited by Little 10 Toes; 17th August 2020 at 04:22 PM.
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Old 17th August 2020, 11:49 PM   #213
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Originally Posted by jsfisher View Post
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.
Code:
x={} ∈ V 
Do loop x forever
  (x∪{x}) ∈ V
   x=(x∪{x})               
Next
Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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.
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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.
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Old 17th August 2020, 11:55 PM   #214
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Originally Posted by jsfisher View Post
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...&postcount=203 and you actually ignored the most of it.

Originally Posted by jsfisher View Post
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...&postcount=203 and http://www.internationalskeptics.com...&postcount=205.
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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.
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Old 18th August 2020, 03:56 AM   #215
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Originally Posted by zooterkin View Post
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.
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Old 18th August 2020, 04:08 AM   #216
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Originally Posted by doronshadmi View Post
It has the needed properties in order to define V members as follows:

{} is the one and only one base member of set V.
Code:
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).
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Old 18th August 2020, 05:23 AM   #217
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Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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.
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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.
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Old 18th August 2020, 05:31 AM   #218
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Originally Posted by doronshadmi View Post
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.

Quote:
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.

Quote:
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.
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Old 18th August 2020, 05:34 AM   #219
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Originally Posted by caveman1917 View Post
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.
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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.
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Old 18th August 2020, 06:06 AM   #220
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Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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.
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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; 18th August 2020 at 06:07 AM.
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Old 18th August 2020, 06:25 AM   #221
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Originally Posted by doronshadmi View Post
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"?
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Old 18th August 2020, 07:03 AM   #222
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Originally Posted by doronshadmi View Post
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.
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Old 18th August 2020, 10:27 AM   #223
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Originally Posted by zooterkin View Post
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:
Originally Posted by doronshadmi View Post
It has the needed properties in order to define V members as follows:

{} is the one and only one base member of set V.
Code:
x={} ∈ V 
Do loop x forever
  (x∪{x}) ∈ V
   x=(x∪{x})               
Next
...
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Old 18th August 2020, 10:46 AM   #224
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@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
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Old 20th August 2020, 12:19 AM   #225
doronshadmi
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Originally Posted by zooterkin View Post
Case in point:
Take for example set V:

Code:
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.

Originally Posted by 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.
Wrong, according to property (5).
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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; 20th August 2020 at 12:35 AM.
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Old 20th August 2020, 04:50 AM   #226
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Originally Posted by doronshadmi View Post
Take for example set V:

Code:
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.
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Old 20th August 2020, 04:57 AM   #227
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Originally Posted by jsfisher View Post
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"?
Originally Posted by jsfisher View Post
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):
Code:
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 { |{}|, |{{}}|, |{{},{{}}}|, ... }
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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; 20th August 2020 at 06:23 AM.
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Old 20th August 2020, 06:28 AM   #228
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Originally Posted by doronshadmi View Post
Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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.

Quote:
∃ N ( ∅ ∈ N ∧ for any given all x ∈ N ( ( x ∪ { x } ) ∈ N ) )
Stick with the axiom as written.

Quote:
Definition: (x∪{x}) is called the largest successor in N...
Not sure what relevance you think all that has, but whatever.

Quote:
...
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.
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Old 20th August 2020, 07:31 AM   #229
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Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
Stick with the axiom as written.
Quote:
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

Originally Posted by jsfisher View Post
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.

Originally Posted by jsfisher View Post
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:
Originally Posted by doronshadmi
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)."
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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; 20th August 2020 at 08:06 AM.
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Old 20th August 2020, 08:37 AM   #230
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Originally Posted by doronshadmi View Post
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?
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Old 20th August 2020, 01:01 PM   #231
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Originally Posted by jsfisher View Post
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?
Code:
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:

Code:
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.
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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; 20th August 2020 at 02:09 PM.
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Old 20th August 2020, 03:23 PM   #232
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Originally Posted by doronshadmi View Post
Code:
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.
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Old 21st August 2020, 06:23 AM   #233
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jsfisher, you opened the door into a paradise called N (here: http://www.internationalskeptics.com...&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):

Code:
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.
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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; 21st August 2020 at 06:39 AM.
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Old 21st August 2020, 06:44 AM   #234
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... and how does this help you define cardinality?
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Old 21st August 2020, 01:01 PM   #235
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Originally Posted by Little 10 Toes View Post
... and how does this help you define cardinality?
V is exactly the non-finite proper subset of N, which its base member is {}.

Code:
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:

Code:
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 kK is defined by its corresponding V member.

For example:
Code:
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.
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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; 21st August 2020 at 01:04 PM.
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Old 21st August 2020, 05:55 PM   #236
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Originally Posted by doronshadmi View Post
Originally Posted by jsfisher View Post
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.
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Old 21st August 2020, 06:16 PM   #237
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Originally Posted by doronshadmi View Post
jsfisher, you opened the door into a paradise called N (here: http://www.internationalskeptics.com...&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.

Quote:
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.

Quote:
...where each distinct base member has its own unique successors...
Like there.

Quote:
...
∃ 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.
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Old 21st August 2020, 06:19 PM   #238
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Originally Posted by doronshadmi View Post
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.
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Old 22nd August 2020, 04:27 AM   #239
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Originally Posted by jsfisher View Post
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}).

Originally Posted by jsfisher View Post
"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...&postcount=233) is out of job.

Originally Posted by jsfisher View Post
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).
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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; 22nd August 2020 at 05:23 AM.
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Old 22nd August 2020, 05:28 AM   #240
doronshadmi
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Originally Posted by jsfisher View Post
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.
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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.
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