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Deeper than primes

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That would be the construction method that doesn't actually exist. You have failed to show otherwise. And that would be the mapping you have failed to show exists.
Wrong jsfisher, your X^2 reasoning is simply blind to http://www.internationalskeptics.com/forums/showpost.php?p=6925404&postcount=14419 .

Please try again, though. Please show us your bijective mapping between the natural numbers and the power set of {A}. Then go ahead and show us where you imagined "Cantor's construction method" to be.
Already done at http://www.internationalskeptics.com/forums/showpost.php?p=6928575&postcount=14425 .

Please try to get out of your arbitrary X^2 box.
 
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Very simple, any bijective degree between X^2 and 2^X.
If it's so simple, then there is no obstacle to define such a set.

S = { ? }

Go ahead, unless you fell in deep love with that "permanent question mark" from your previous adventure with math fantasies.

Btw1, "any bijective degree" is a term of a very peculiar meaning, coz the prefix bi means two and not any. Examples: binary number (combination of two digits) or bicycle (two wheels). In this case, bijection follows the suit; its a link between two and only two elements.

A <---> B

So your "any degree" needs some defining.

Btw2, when God found out about The New Testament to join the first edition now called The Old Testament, he decided to call the whole enchilada Bible. You know -- Bible -- like two testaments. Even he abides by the meaning of the prefixes, but you must have always something extra.
:rolleyes:
 
If it's so simple, then there is no obstacle to define such a set.
It is in front of your blind mind right at http://www.internationalskeptics.com/forums/showpost.php?p=6925404&postcount=14419 (I made some notation's mistakes that, so let us correct them:

1) P({}, ... , {1,2,3,...}) has to be P({1,2,3,...})

2) |P({}, ... , {1,2,3,...})| has to be |P({1,2,3,...})|

3) ...P(P({1,2,3}))... has to be ...P(P({1,2,3,...}))...

)

Take for example 3 and 8, all you have is to define any desirable bijective degree between 3 to 8 pairs (3 to 8 bis).
 
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It is in front of your blind mind right at http://www.internationalskeptics.com/forums/showpost.php?p=6925404&postcount=14419 (I made some notation's mistakes that, so let us correct them:

1) P({}, ... , {1,2,3,...}) has to be P({1,2,3,...})

2) |P({}, ... , {1,2,3,...})| has to be |P({1,2,3,...})|

3) ...P(P({1,2,3}))... has to be ...P(P({1,2,3,...}))...

)

Take for example 3 and 8, all you have is to define any desirable bijective degree between 3 to 8 pairs (3 to 8 bis).
Even the simple task of completing S = { ? } poses a great deal of difficulty to you. Just replace the friggin' question mark and then complete |S| = ? in terms of alephs, if you say that the continuum hypothesis is false.
 
Even the simple task of completing S = { ? } poses a great deal of difficulty to you. Just replace the friggin' question mark and then complete |S| = ? in terms of alephs, if you say that the continuum hypothesis is false.
You are still off topic and blind to http://www.internationalskeptics.com/forums/showpost.php?p=6925404&postcount=14419 .

For example:

S={1,2,3}

P(S)={{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}

and some bijective degree between S and P(S) is {{},{1},{2},{3},{1,2}}

Where:

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} |
-----------------------
 
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That would be the construction method that doesn't actually exist. You have failed to show otherwise. And that would be the mapping you have failed to show exists.
Wrong jsfisher, your X^2 reasoning is simply blind to http://www.internationalskeptics.com/forums/showpost.php?p=6925404&postcount=14419 .

It is unlikely we'd see something that simply is not there. Here's the first question again, in the hopes you'll read and comprehend it this time:
Where in Cantor's Theorem (or in the proof to Cantor's Theorem) is this construction method? Don't keep giving us what you imagine to be an example of its use. Point to the construction method directly.​

Please try again, though. Please show us your bijective mapping between the natural numbers and the power set of {A}. Then go ahead and show us where you imagined "Cantor's construction method" to be.
Already done at http://www.internationalskeptics.com/forums/showpost.php?p=6928575&postcount=14425 .

Please try to get out of your arbitrary X^2 box.

That would be the post where you provided two separate mappings from S to P(S). The first was A->{A}; the second was A->{}. Neither mapping is a bijection between the members of {A} and P({A}).

By way of vigorous hand-waving, you also generated a trivial mapping between {1,2} and {{},{A}}. This impressed nobody except you. Be that as it may, it is also not a bijective mapping between {A} and P({A}) or between the natural numbers and P({A}).

Care to try again?
 
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Take for example 3 and 8, all you have is to define any desirable bijective degree between 3 to 8 pairs (3 to 8 bis).


Bijective degree? Is this your new cover, Doron? Since you keep showing examples of mappings that are not bijective, are you now shifting the terminology?

This is very boring.

Simple statement of fact: There is no bijective mapping between a set of 3 things and a set of 8.

Multiple rounds and the introduction of integers doesn't alter that simple fact.
 
Wrong The Man, the invariance property of being natural number is common (invariant) for the set of natural numbers, yet (each natural number is different than any other natural number) AND (the largest natural number is not satisfied) (which is the variant property of the set of natural numbers).

Wrong about what? Do try to be specific if you can.

Yes all natural numbers are natural numbers which is exactly what makes them all members of the set of all natural numbers. Who has said anything about “the largest natural number” other than just you? Doron it isn’t a “variant property of the set of” all “natural numbers”. There is no “largest natural number”, if you think not then please show us what natural number is the largest. That fact is not variant or “increasing”. Again you are still deliberately trying to confuse a list with a set and have yet to show what is increasing (and thus changing) in your ‘ever increasing’ set.


The Man, the set of all natural numbers is an unsatisfied fantasy, and you can't comprehend it exactly because you do not understand the inseparability of the invariance with the variance, exactly as shown in the case (the invariant pi) AND (the variant curvature) among any collection (finite or not) of circles.

No Doron you assert yourself above that “the invariance property of being natural number is common (invariant) for the set of natural numbers” so all natural numbers are natural numbers and thus members of the set of all natural numbers. You just don’t like the term “all” for whatever reason that evidently has absolutely nothing to do with sets or your trivial assertion that Pi doesn’t vary because the diameter and circumference of different circles do vary to the same proportion.


Again…

By all means please explain to us the difference between increasing and decreasing with “no past (before) and no future (after)”?
 
You are still off topic and blind to http://www.internationalskeptics.com/forums/showpost.php?p=6925404&postcount=14419 .

For example:

S={1,2,3}

P(S)={{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}

and some bijective degree between S and P(S) is {{},{1},{2},{3},{1,2}}
It's obvious once again that when you say that Cantor's continuum hypothesis is false, you don't have the slightest idea what you are talking about.

The elements of the above S are all in Z -- they are integers and the set is finite. Now read again Cantor's hypothesis:

There is no set whose cardinality is strictly between that of the integers and that of the real numbers.[/i]

If you think that your S={1,2,3} -- the set S={ ? } I asked for -- is the set that Cantor's hypothesized of not existing, then just keep believing it is so.
 
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You just don’t like the term “all” for whatever reason that evidently has absolutely nothing to do with sets or your trivial assertion that Pi doesn’t vary because the diameter and circumference of different circles do vary to the same proportion.
Doron got nothing against "all"; it's just the circumstance when the word sneaks upon his mind and makes him worried whenever he opens his fridge. Is that ALL I got left for the weekend? I must double up the next time.
 
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It is unlikely we'd see something that simply is not there. Here's the first question again, in the hopes you'll read and comprehend it this time:
Where in Cantor's Theorem (or in the proof to Cantor's Theorem) is this construction method? Don't keep giving us what you imagine to be an example of its use. Point to the construction method directly.​



That would be the post where you provided two separate mappings from S to P(S). The first was A->{A}; the second was A->{}. Neither mapping is a bijection between the members of {A} and P({A}).

By way of vigorous hand-waving, you also generated a trivial mapping between {1,2} and {{},{A}}. This impressed nobody except you. Be that as it may, it is also not a bijective mapping between {A} and P({A}) or between the natural numbers and P({A}).

Care to try again?

The hand-waving is a direct result of you closed reasoning under X^2.

Your reasoning is a boring fantasy.

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

Again (in this post a I correct some expressions):

jsfisher, it is not P(S) unless you account for every P(S) member.

Actually you can do it infinity many times and still you will not get a complete collection of different objects in both sides of the 1-to-1 mapping.

Be aware that "infinity many times" is a present continuous parallel reasoning's expression, which is not limited to your step-by-step serial only reasoning.

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

By using a cross-contexts reasoning, let us use Cantor's construction method in order to define a 1-to-1 correspondence between natural numbers and any set of different objects.

First, let us demonstrate it on some finite case, for example, the powerset of set {1,2,3}.

P({1,2,3}) = ({},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}

By using Cantor's construction method with the members of {1,2,3} and the same amount of members taken from P({1,2,3}), one enables to explicitly define each P({1,2,3}) member and put it with 1-to-1 correspondence with a natural number, such that there is no natural number that is not mapped with some P({1,2,3}) member.

Cantor's construction method constructs an explicit P({1,2,3}) member as follows:

1) The defined explicit P({1,2,3}) member is the result of a 1-to-1 correspondence between {1,2,3} members and the same amount of members taken from P({1,2,3}).

2) The explicit constructed P({1,2,3}) member is the result of some 1-to-1 correspondence, such that it includes a {1,2,3} member only if this member does not exist as one of the members of the P({1,2,3}) member that is in a 1-to-1 correspondence with it.

By using this construction method several times, one enables to define a 1-to-1 correspondence between natural numbers and P({1,2,3}) members, as follows:

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

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

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

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

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

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

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

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

The same 1-to-1 correspondence construction method works also in the case of infinitely many objects, where in this case {{}, ... , {1,2,3,...}} is the power set of {1,2,3,...}, and also in this case we are able to define a 1-to-1 correspondence between 1 to {}, 2 to {1,2,3,...}, and any natural number > 2 with any {{}, ... , {1,2,3,...}} member between {} and {1,2,3,...}.

It is done by using Cantor's construction method |{{}, ... , {1,2,3,...}}| times (again, it has to be taken in parallel, such that there is no process but simply a permanent incompleteness that can't be comprehended by using only step-by step reasoning).

Since the set of all powersets does not exist (because ...P(P({1,2,3}}}... size is unsatisfied) and since we are generally able (by using Cantor's construction method) to construct P(S) members by mapping S members with |S| P(S) members |P(S)| times, and also to define a 1-to-1 correspondence between the natural numbers and any P(S) degree, then also the size of the set of natural numbers is unsatisfied.

In other words, we discover that the concept of collection of different objects is essentially incomplete, or in other words, the whole is greater than the amount of any collection of different objects, where the whole is the cross-contexts existence among this infinity many context-dependent objects, which their amount is unsatisfied (they can't be summed into a whole).
 
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By using a cross-contexts reasoning, let us use Cantor's construction method in order to define a 1-to-1 correspondence between natural numbers and any set of different objects.

First, let us demonstrate it on some finite case...

count_sheep.gif


1, 2, 3, 4, 5, 6, 7, . . .

The set of all integers is often denoted by a boldface Z, which stands for Zahlen (German for numbers).

GUTE NACHT = {ZZZZZZZ...ZZZZZZ...ZZZZZZZ...}
 
Yes all natural numbers are natural numbers which is exactly what makes them all members of the set of all natural numbers.
Again your limited reasoning airs its view.

The set of natural numbers, is some particular case of a collection of different objects.

Given any collection of different objects, succession (in terms of present continuous, where also parallel reasoning is used) is its essential property, whether it is finite (and in this case the succession is external w.r.t the collection) or infinite (and in this case the succession is internal w.r.t the collection).

You still do not understand the meaning of succession +1 in terms of cardinality as a present continuous, because parallel reasoning is beyond your only step-by-step serial reasoning.
 
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It's obvious once again that when you say that Cantor's continuum hypothesis is false, you don't have the slightest idea what you are talking about.

The elements of the above S are all in Z -- they are integers and the set is finite. Now read again Cantor's hypothesis:



If you think that your S={1,2,3} -- the set S={ ? } I asked for -- is the set that Cantor's hypothesized of not existing, then just keep believing it is so.
epix, you simply unable to do the needed generalization about finite or infinite S;P(S).

In both cases there is any wished bijective degree between S and P(S).

Once again you are using only "in the box" reasoning.
 
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Simple statement of fact: There is no bijective mapping between a set of 3 things and a set of 8.

Multiple rounds and the introduction of integers doesn't alter that simple fact.
This fact is the reflection of your limited reasoning, which can't comprehend Cantor's construction method in terms of the simultaneity of present continuous, that actually defines any wished bijective degree between S and P(S), which demonstrates Cantor's GCH fallacy.
 
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This fact is the reflection of your limited reasoning, which can't comprehend Cantor's construction method in terms of the simultaneity of present continuous, that actually defines any wished bijective degree between S and P(S), which demonstrates Cantor's GCH fallacy.

The cognitive limitations have been and continue to be your own, Doron. You see things that aren't there. This "Cantor's construction method" is just something you invented to fill your abyss of misunderstanding.

There is no bijective mapping between the members of any set and its power set. You have not and cannot demonstrate otherwise.
 
That would be the post where you provided two separate mappings from S to P(S). The first was A->{A}; the second was A->{}. Neither mapping is a bijection between the members of {A} and P({A}).
That is again your reasoning, which simply can't get the simple fact that natural numbers are not limited to any particular collection and therefore, there is a 1-to-1 correspondence between the natural numbers and any given collection of different objects, whether the given collection has finite or infinite amount of objects.

In other words, by using Cantor's construction method on {{},{A}} case, there is a bijedction between the natural numbers and {{},{A}} as follows:

Case 1:
-----------------------
A ↔ {A} | Provides 1 ↔ {}
-----------------------

Case 2:
-----------------------
A ↔ { } | Provides 2 ↔ {A}
-----------------------

and this is one and only one mapping.
 
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There is no bijective mapping between the members of any set and its power set. You have not and cannot demonstrate otherwise.
Only if your understanding of collections of different objects is closed by your artificial forcing reasoning.

Again you demonstrate the limitations of your own paradigm on your mind, as if it is some universal truth.

You are identified with your blind mind, jsfisher, and can't see beyond your imaginary box.
 
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