Deeper than primes

[doronshadmi]

Using Cantor's theorem exactly once proves you wrong.

The basic mathematical fact remains, |S| < |P(S)| for any set S.

Wrong, it simply the basic fact of being closed under X^2 (under the matrix) and compare it only once to 2^X.

 

[doronshadmi]

No rounds anywhere.

Wrong. There is at least one round, which stays closed under X^2.

Please this time read carefully http://www.internationalskeptics.com/forums/showpost.php?p=6919720&postcount=14376 in order to realize how it is possible to use a reasoning, which is not closed under the one round of X^2.


By the way, I have a typo mistake in round 7 of http://www.internationalskeptics.com/forums/showpost.php?p=6919720&postcount=14376.

It has to be something like:

Round 7 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {3} | Provides {2,3} ↔ 7
3 ↔ {1,2} |
-----------------------

So in this case we get the following bijection:

Round 1 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {2} | Provides {} ↔ 1
3 ↔ {3} |
-----------------------

Round 2 particular case:
-----------------------
1 ↔ { } |
2 ↔ {2} | Provides {1} ↔ 2
3 ↔ {3} |
-----------------------

Round 3 particular case:
-----------------------
1 ↔ {1} |
2 ↔ { } | Provides {2} ↔ 3
3 ↔ {3} |
-----------------------

Round 4 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {2} | Provides {3} ↔ 4
3 ↔ { } |
-----------------------

Round 5 particular case:
-----------------------
1 ↔ { } |
2 ↔ {3} | Provides {1,2} ↔ 5
3 ↔ {1,3} |
-----------------------

Round 6 particular case:
-----------------------
1 ↔ { } |
2 ↔ {2} | Provides {1,3} ↔ 6
3 ↔ {1,2} |
-----------------------

Round 7 particular case:
-----------------------
1 ↔ {1} |
2 ↔ {3} | Provides {2,3} ↔ 7
3 ↔ {1,2} |
-----------------------

Round 8 particular case:
-----------------------
1 ↔ { } |
2 ↔ {1} | Provides {1,2,3} ↔ 8
3 ↔ {2} |
-----------------------

 

[doronshadmi]

"Increasing"? Again what natural number was not a natural number before you 'increased' your set of natural numbers? You keep deliberately portending that a set is a list.

A set is an ever increasing mathematical object.

This is exactly the reason of why the set of all powersets does not exist.

An ever increasing development is a present continuous state, and has nothing to do with before or after.

Your only step-by-step reasoning does not help to get it.

 

[The Man]

A set is an ever increasing mathematical object.

No it isn't . What is “ever increasing” about the set {1} or the empty set?

This is exactly the reason of why the set of all powersets does not exist.

What reason exactly would that be? That your claim “A set is an ever increasing mathematical object.” is demonstrably wrong?

An ever increasing development is a present continuous state, and has nothing to do with before (past) or after (future).

Yes it does, it was less “before (past)” and ‘increased’ after (future). Again you remain the staunchest opponent of your own notions. So we can add “increasing” to the list of words you just don’t want to understand.

Your only step-by-step reasoning does not help to get it.

Stop trying to simply posit aspect of your own failed reasoning on to others.

 

[epix]

Originally Posted by jsfisher
Using Cantor's theorem exactly once proves you wrong.

The basic mathematical fact remains, |S| < |P(S)| for any set S.

Wrong, it simply the basic fact of being closed under X^2 (under the matrix) and compare it only once to 2^X.

Can't you understand a simple thing? Before there was a power set, there was a question whether you can put two subsets of naturals on 1:1 correspondence with the counting numbers. Suppose you have two subsets even and odds:

{{2, 4, 6, 8, ...},{1, 3, 5, 7, ...}}

If you start with the evens

2, 4, 6, 8, ...
1, 2, 3, 4, ...

you never get the chance to hit the odds, coz the evens progress without a bound, so do the odds in case you would like to try your luck the other way. Since the power set of naturals include the naturals as its subset, you never get the chance to count other members of the subsets of the naturals, like prime numbers, Fibonacci numbers, Lucas numbers, triangular numbers, even numbers and so on. If you start with prime numbers, the story is the same, coz primes run without a bound as well. That's why |P(N)|>|N|. But that's not the proof by itself -- you need to arrive at a logical contradiction.

LOL. I'm just kidding.

LOL. What if i'm not kidding?

P(N) = {{1}, {2}, {3}, {4}, ... {1,2}, {1,3}, {1,4}, ...}
________1___2____3___4 ...

 

[jsfisher]

Using Cantor's theorem exactly once proves you wrong.

The basic mathematical fact remains, |S| < |P(S)| for any set S.

Wrong, it simply the basic fact of being closed under X^2 (under the matrix) and compare it only once to 2^X.

Great! Then it should be no problem for you to provide a simple counter example to disprove Cantor's theorem. Here, please show show us a bijection of the elements of {A, B} and its power set, {{}, {A}. {B}, {A, B}}. I'll even start you out with a table. You need only fill in the right-hand side:

  S      P(S)
------+-------

  A  <->  ?
  B  <->  ?

 

[jsfisher]

Wrong. There is at least one round, which stays closed under X^2.

Define "round". Like so many words, you seem to have your own novel meaning for it that completely transcends common usage.

Please this time read carefully http://www.internationalskeptics.com/forums/showpost.php?p=6919720&postcount=14376 in order to realize how it is possible to use a reasoning, which is not closed under the one round of X^2.

Perhaps, with your superior out of context reasoning, you could point out exactly where in that post you establish any sort of bijection between S and P(S). All I see are a sequence of different mappings, none of which are a bijection.

 

[jsfisher]

A set is an ever increasing mathematical object.

That a curious place this Doronetics is. Everything is so busy; action verbs everywhere.

I cannot help but wonder, at what rate is, say, the set of integers growing? How much did it increase from yesterday to today? What are some of the new numbers recently added to the set?

 

[zooterkin]

A set is an ever increasing mathematical object.

This is exactly the reason of why the set of all powersets does not exist.

An ever increasing development is a present continuous state, and has nothing to do with before or after.

Your only step-by-step reasoning does not help to get it.

You're the one who seems to be applying a step-by-step approach, adding members to a set one by one. When the set is defined, the members are in it, they don't have to be looked at one at a time to decide if they are in or out.

 

[doronshadmi]

Yes it does, it was less “before (past)” and ‘increased’ after (future). Again you remain the staunchest opponent of your own notions. So we can add “increasing” to the list of words you just don’t want to understand.

Any given collection is under the laws of probability.

EDIT:

Please try to understand the permanent question mark (which is a present continuous state) at the top of the following diagram:

 

[doronshadmi]

Define "round".

Any form that is equivalent to X^2.

Perhaps, with your superior out of context reasoning, you could point out exactly where in that post you establish any sort of bijection between S and P(S). All I see are a sequence of different mappings, none of which are a bijection.

Because you are looking at each mapping separately, and as a result you are closed under X^2 matrix.

 

[jsfisher]

Yes it does, it was less “before (past)” and ‘increased’ after (future). Again you remain the staunchest opponent of your own notions. So we can add “increasing” to the list of words you just don’t want to understand.

Any given collection is under the laws of probability.

As often happens, the reply has absolutely nothing to do with the quoted post.

 

[doronshadmi]

Define "round".

Any form that is equivalent to X^2.

Perhaps, with your superior out of context reasoning, you could point out exactly where in that post you establish any sort of bijection between S and P(S). All I see are a sequence of different mappings, none of which are a bijection.

Because you are looking at each mapping separately, and as a result you are closed under X^2 matrix.

Is there some reasoning that prevent from you to get the explicit bijection between some natural number and P(S) (the bijection is represented by bolded symbols), as explicitly shown in http://www.internationalskeptics.com/forums/showpost.php?p=6920517&postcount=14382 ?

 

[doronshadmi]

As often happens, the reply has absolutely nothing to do with the quoted post.

jsfisher, the permanent question mark at the top of the considerd diagram is timeless and therefore has nothing to do with before (past) and after (future).

The Man and you still do not grasp the accurate meaning of present continuous.

 

[jsfisher]

Any form that is equivalent to X^2.

That is a very odd and unworkable definition. Care to try again?

...
Isthere some reasoning that prevent from you to get the explicit bijection between some natural number and P(S) (the bijection is represented by bolded symbols), as explicitly shown in http://www.internationalskeptics.com/forums/showpost.php?p=6920517&postcount=14382 ?

Well, you haven't provided one, yet. Your claimed example includes "8" for example. That's not a member of S, so if that is your example, you again FAIL.

 

[jsfisher]

jsfisher, the permanent question mark at the top of the considerd diagram is timeless and therefore has nothing to do with before (past) and future (after).

The Man and you still do not grasp the accurate meaning of present continuous.

Do tell!!! What is the accurate meaning of "present continuous"? Does it involve gerunds?

 

[doronshadmi]

What is “ever increasing” about the set {1} or the empty set?

The powerset of them, and as I explicitly show in

http://www.internationalskeptics.com/forums/showpost.php?p=6916077&postcount=14361

http://www.internationalskeptics.com/forums/showpost.php?p=6918253&postcount=14366

http://www.internationalskeptics.com/forums/showpost.php?p=6920517&postcount=14382

there is a bijection between natural numbers and any P(S), where the set of all powersets does not exist.

 

[doronshadmi]

Well, you haven't provided one, yet. Your claimed example includes "8" for example. That's not a member of S, so if that is your example, you again FAIL.

It is not about (S (X^2)) < (P(S) (2^X)).

It is about the bijection between any given natural number and any given P(S) member , and as I explicitly show in

http://www.internationalskeptics.com/forums/showpost.php?p=6916077&postcount=14361

http://www.internationalskeptics.com/forums/showpost.php?p=6918253&postcount=14366

http://www.internationalskeptics.com/forums/showpost.php?p=6920517&postcount=14382

there is a bijection between natural numbers and any P(S), where the set of all powersets does not exist.

 

[doronshadmi]

Do tell!!! What is the accurate meaning of "present continuous"? Does it involve gerunds?

It is a "permanently out of the (X^2) box" reasoning.

EDIT:

Please try to understand the permanent question mark (which is a present continuous state) at the top of the following diagram:

 

[zooterkin]

Any form that is equivalent to X^2.

So squares are round?