Cont: Deeper than primes - Continuation 2

[jsfisher]

If you want to disprove Cantor's Theorem then work from the proof of Cantor's Theorem, not some confusion you think might be somewhat like the proof.

 

[doronshadmi]

If you want to disprove Cantor's Theorem then work from the proof of Cantor's Theorem, not some confusion you think might be somewhat like the proof.

Please support your claim that what is written in http://www.internationalskeptics.com/forums/showpost.php?p=10944628&postcount=728 is some confusion that might be somewhat like the proof, by your detailed reply to the content of http://www.internationalskeptics.com/forums/showpost.php?p=10944628&postcount=728.

 

[doronshadmi]

Ok, let's continue.

By using logic it has been shown that a given member of P(S) can't be mapped with any member of S, whether S is finite or an infinite set.

In both cases (the finite and the infinite) |S| and |S|+1 are involved, where in the finite case, it is trivially clear that |S| ≠ |S|+1.

By using the same logic also in the infinite case (as used by Cantor in order to define a member of P(S) that can't be mapped with any member of S) we also get |S| ≠ |S|+1, for example:

If S={a,b,c,...} then the result of the mapping

a --> {a}
b --> {b}
c --> {c}
...

is {} (which is a member of P(S) that logically can't be mapped with any member of S) where the existence of {} is guaranteed (if ZF is considered) by the axiom of replacement.

So by this logic we get |S| ≠ |S|+1 (where |S| is a transfinite number).

But according to the transfinite number system |S| = |S|+1.

In other words, according to the transfinite number system |S| = |S|+1 and |S| ≠ |S|+1 hold.

So, there are at least two options:

1) (|S| = |S|+1) AND (|S| ≠ |S|+1) and in this case (if ZF is considered) the transfinite system is inconsistent.

2) (|S| = |S|+1) OR (|S| ≠ |S|+1) and in this case (if ZF is considered) the transfinite system must be developed further in order to logically distinguish between |S| = |S|+1 and |S| ≠ |S|+1.

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

So actually what I show is the need to re-search the transfinite system.

 

[jsfisher]

Ok, let's continue.

By using logic it has been shown that a given member of P(S) can't be mapped with any member of S, whether S is finite or an infinite set.

In both cases (the finite and the infinite) |S| and |S|+1 are involved, where in the finite case, it is trivially clear that |S| ≠ |S|+1.

By using the same logic also in the infinite case (as used by Cantor in order to define a member of P(S) that can't be mapped with any member of S) we also get |S| ≠ |S|+1, for example:

If S={a,b,c,...} then the result of the mapping

a --> {a}
b --> {b}
c --> {c}
...

is {} (which is a member of P(S) that logically can't be mapped with any member of S) where the existence of {} is guaranteed (if ZF is considered) by the axiom of replacement.

So by this logic we get |S| ≠ |S|+1 (where |S| is a transfinite number).

But according to the transfinite number system |S| = |S|+1.

In other words, according to the transfinite number system |S| = |S|+1 and |S| ≠ |S|+1 hold.

So, there are at least two options:

1) (|S| = |S|+1) AND (|S| ≠ |S|+1) and in this case (if ZF is considered) the transfinite system is inconsistent.

2) (|S| = |S|+1) OR (|S| ≠ |S|+1) and in this case (if ZF is considered) the transfinite system must be developed further in order to logically distinguish between |S| = |S|+1 and |S| ≠ |S|+1.

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

So actually what I show is the need to re-search the transfinite system.

What does any of this have to do with reducing Cantor's Theorem to a conjecture?

 

[Little 10 Toes]

Ok, let's continue.

By using logic it has been shown that a given member of P(S) can't be mapped with any member of S, whether S is finite or an infinite set.

Really?

From Power set^WP

If S is the set {x, y, z }, then the subsets of S are:

{} (also denoted the empty set)
{x}
{y}
{z}
{x, y }
{x, z }
{y, z }
{x, y, z }
and hence the power set of S is {{}, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}}

If I remember my notation of elements, I think I can map {x} as a given member of set S to {x} as a given member of the power set of set S (a.k.a. P(S)).

I think you need to stop using doron-logic.

 

[jsfisher]

Really?

From Power set^WP

If S is the set {x, y, z }, then the subsets of S are:

{} (also denoted the empty set)
{x}
{y}
{z}
{x, y }
{x, z }
{y, z }
{x, y, z }
and hence the power set of S is {{}, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}}

If I remember my notation of elements, I think I can map {x} as a given member of set S to {x} as a given member of the power set of set S (a.k.a. P(S)).

I think you need to stop using doron-logic.

You can make it much simpler.

Consider S = {a} and its power set P = { {}, {a} }
Ok, Doron, which member of P cannot possibly be mapped from any member of S?

 

[doronshadmi]

You can make it much simpler.

Consider S = {a} and its power set P = { {}, {a} }
Ok, Doron, which member of P cannot possibly be mapped from any member of S?

You are missing my argument, which is:

For distinct |P(S)| mappings (where each mapping is between all |S| members of S and |S| members of P(S)) there is exactly one distinct member of P that is not in the range of each mapping, or in other words, it is always the case (according to the logic that is used by Cantor) that |S| ≠ |S|+1, whether S is finite, or not.

But according to the transfinite number system |S| = |S|+1, so two possible results that do not "agree" with each other, are both parts of the transfinite number system.

 

[doronshadmi]

I wish to be clearer about |S|+1.

|S|+1 does not mean that a given member of set P(S) has to be added to set S in order to conclude that |S|+|{some distinct member of P(S)}| ≠ |S|.

So |S|+|{some distinct member of P(S)}| ≠ |S| holds whether S is finite, or not.

But according to the transfinite number system |S|+|{some distinct member of P(S)}| = |S|, so two possible results that do not "agree" with each other are both parts of the transfinite number system.

In other words, the transfinite system has to be re-searched.

More details are given in http://www.internationalskeptics.com/forums/showthread.php?p=10947308#post10947308, where (|S|+1) = (|S|+|{some distinct member of P(S)}|).

Also I wish to correct some mistake in http://www.internationalskeptics.com/forums/showpost.php?p=10944628&postcount=728:

The example |{a,b,c}| < |{{},{a},(b},{c}}| has to be replaced by |{a,b,c}| < |{{a},(b},{c}}|+|{{}}| in order to clarify that no member of set P(S) is added to set S, in order to show the inequality between |{a,b,c}| and |{{a},(b},{c}}|+|{{}}|.

Since no member of set P(S) is added to set S even if S is infinite, then, for example, |{a,b,c,...}| ≠ |{{a},(b},{c},...}|+|{{}}| holds (according to the logic that is used by Cantor in what is called Cantor's theorem).

 

[jsfisher]

For distinct |P(S)| mappings (where each mapping is between all |S| members of S and |S| members of P(S))

Only |P(S)| possible mappings? From where did you get that?

...there is exactly one distinct member of P that is not in the range of each mapping, or in other words

Exactly one? Wrong again.

 

[doronshadmi]

Only |P(S)| possible mappings? From where did you get that?

Edit:
You are right, in order to be clearer it has to be "at least distinct |P(S)| mappings".

Exactly one?

Yes, for each mapping (of at least distinct |P(S)| mappings) from all |S| members of set S to |S| members of set P(S), there is exactly one distinct P(S) member that is not in the range of all |S| members of set S, whether S is finite, or not.

If you disagree with this, then please explicitly show that Cantor's method provides more than one distinct P(S) member that is not in the range of all |S| members of set S, for each mapping (of at least distinct |P(S)| mappings).

Some example (out of at least distinct |P(S)| mappings) is as follows:

If S={a,b,c,...} then the distinct result of the distinct mapping

a --> {a}
b --> {b}
c --> {c}
...

is {} (which is a member of P(S) that logically can't be mapped with any member of S) where the existence of {} is guaranteed (if ZF is considered) by the axiom of replacement.

So by this logic we get |S| ≠ |S|+|{{}}| (where |S| is a transfinite number), and so is the case for the rest of the distinct |P(S)| mappings, except the fact that instead of the particular |S| ≠ |S|+|{{}}| case, the general form|S| ≠ |S|+|{#}| (where # is a placeholder for any possible different single member of set P(S)) is defined.

EDIT:

More details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10947308&postcount=748.

 

[doronshadmi]

I have a link mistake in post http://www.internationalskeptics.com/forums/showpost.php?p=10947308&postcount=748.

Instead of the wrong link in the following part:

"More details are given in http://www.internationalskeptics.com/forums/showthread.php?p=10947308#post10947308, where (|S|+1) = (|S|+|{some distinct member of P(S)}|)."

it has to be

"More details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10946712&postcount=743, where (|S|+1) = (|S|+|{some distinct member of P(S)}|)."

 

[jsfisher]

Edit:
You are right, in order to be clearer it has to be "at least distinct |P(S)| mappings".

Still wrong.

Yes, for each mapping (of at least distinct |P(S)| mappings) from all |S| members of set S to |S| members of set P(S), there is exactly one distinct P(S) member that is not in the range of all |S| members of set S, whether S is finite, or not.

Still wrong.

If you disagree with this, then please explicitly show that Cantor's method provides more than one distinct P(S) member that is not in the range of all |S| members of set S, for each mapping (of at least distinct |P(S)| mappings).

Cantor's method? You mean the standard proof? The standard proof shows that for any map from S to P(S) there is at least one member of the power set not in the range of the map. Exactly one and at least one are far from the same thing.

 

[doronshadmi]

The standard proof shows that for any map from S to P(S) there is at least one member of the power set not in the range of the map.

In that case you have to explicitly show that for any map from S to P(S) there is also more than one member of the power set not in the range of the map.

If you are unable to explicitly show it, than at least one member is no more than wishful thinking.

 

[jsfisher]

In that case you have to explicitly show that for any map from S to P(S) there is more than one member of the power set not in the range of the map.

If you are unable to explicitly show it, than at least one member is no more than wishful thinking.

It is not my "proof" that is contingent on it being exactly one, but yours, so the burden would be yours to show "exactly one".

Nonetheless, consider S = {A, B}. The power set of S, P(S) is {{}, {A}, {B}, {A,B}}.

There are 16 distinct mappings from S to P(S). Twelve of them are injective, and four are not. I'll leave it as an exercise to identify all 16 mappings.

Each of the twelve injective mappings leaves two members of P(S) outside the range of the mapping. Please note that 2 is different from exactly 1.

Each of the remaining four mappings leaves three members of P(S) outside the range of the mapping. Please note that 3 is different from exactly 1.

 

[doronshadmi]

It is not my "proof" that is contingent on it being exactly one, but yours, so the burden would be yours to show "exactly one".

Nonetheless, consider S = {A, B}. The power set of S, P(S) is {{}, {A}, {B}, {A,B}}.

There are 16 distinct mappings from S to P(S). Twelve of them are injective, and four are not. I'll leave it as an exercise to identify all 16 mappings.

Each of the twelve injective mappings leaves two members of P(S) outside the range of the mapping. Please note that 2 is different from exactly 1.

Each of the remaining four mappings leaves three members of P(S) outside the range of the mapping. Please note that 3 is different from exactly 1.

It can't be done if S is an infinite set, and this is exactly why what is called Cantor's proof is based a contradiction that provides exactly one distinct P(S) member that is not in the range of S member, for any distinct mapping.

In the case of infinite sets one can't know the number of P(S) members that are outside the range of all S members, and his\her best is to conclude (according the the logic (based on contradiction) that is used by Cantor) that |S| ≠ |S|+|{#}| (where # is a placeholder for any possible different single member of set P(S)).

Furthermore, if what is called Cantor's proof (which is based a contradiction) is used in case that S is finite, only one mapping between S and P(S) is needed (which provides exactly one distinct member of P(S) that is not in the range of S members) in order to conclude that |S| ≠ |S|+|{#}| (where # is a placeholder for any possible different single member of set P(S)).

Moreover, you are able to count the number of P(S) members that are outside the range of all S (where S is finite) for any given mapping, exactly because the only option in case of a finite set is |S| ≠ |S|+|{#} (where # is a placeholder for any possible different single member of set P(S)).

Edit:

Generally, you can't use only finite sets, in order to conclude valid things about infinite sets, so please try again.

It is not my "proof" that is contingent on it being exactly one, but yours, so the burden would be yours to show "exactly one".

Please use what is called Cantor's proof (which is based a contradiction) and explicitly show more than one distinct P(S) member that is outside the range of S, for any distinct mapping from |S| members of set S to |S| members of set P(S).

 

[jsfisher]

It can't be done if S is an infinite set, and this is exactly why what is called Cantor's proof is based a contradiction that provides exactly one distinct P(S) member that is not in the range of S member, for any distinct mapping.

The standard proof of the theorem shows the existence of one member of the power set outside the range of a mapping. The proof no where suggests that that one member is the only one.

In the case of infinite sets one can't know the number of P(S) members that are outside the range of all S members

You need it to be exactly one for all possible mappings for your argument to even begin to make sense. But it isn't, even if you don't "know" it.

Revisit your identity mapping for S = {a, b, c, ...}. The mapping is f(x) : x -> {x}. You correctly observed that {} is outside the range of this mapping. So is {a,b}, and so is {a,c}, and so is....

Exactly one is nowhere to be found.

 

[doronshadmi]

Revisit your identity mapping for S = {a, b, c, ...}. The mapping is f(x) : x -> {x}. You correctly observed that {} is outside the range of this mapping. So is {a,b}, and so is {a,c}, and so is....

Exactly one is nowhere to be found.

You are right jsfisher, there is indeed at least one member of P(S) outside the range of all S members, for each distinct mapping , where S is an infinite set.

Thank you for the discussion and for your patience all along it.

Since you agree that only one mapping is insufficient in order to conclude that |S| < |P(S)| (where S is infinite) then what is the minimal number of mappings that are needed in order to conclude that |S| < |P(S)| (in case that S is finite, one mapping is sufficient)?

I am aware of the fact that S can be the power set of another set, so the question is about the number of mappings between |S| and |P(S)|, and I think that it is at least |P(S)| mappings.

Is there a proof that it is at least |P(S)| mappings?

 

[doronshadmi]

Let's try a stronger version of my original idea.

By using Cantor's proof, do we also prove for any mapping between all |S| members of S and |S| members P(S) that the number of P(S) members that are not mapped with any S member, is not less than |P(S)|?

If the answer is no, then for each given mapping |S| there is |#| (where # is a placeholder for any transfinite number < |P(S)|) and we get the inequality |S| ≠ |S|+|#| and also the equation |S| = |S|+|#|, as observed as follows:

|S|+|#| does not mean that |#| members of set P(S) have to be added to set S in order to conclude that |S|+|#| ≠ |S|.

But according to the transfinite number system |S|+|#| = |S|, so two possible results that do not "agree" with each other are both parts of the transfinite number system.

So in order to avoid it Cantor's proof must show that |#| is not less than |P(S)|.

Edit:

The standard proof shows that for any map from S to P(S) there is at least one member of the power set not in the range of the map. Exactly one and at least one are far from the same thing.

at least one is insufficient by the stronger version, because in order to prove that |S| < |P(S)| |#| must be determined by Cantor's proof as not less than |P(S)|.

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

If |#| can't be determined by Cantor's proof, then we can't conclude for sure that |S| < |P(S)|.

 

[jsfisher]

Let's try a stronger version of my original idea.

By using Cantor's proof, do we also prove for any mapping between all |S| members of S and |S| members P(S) that the number of P(S) members that are not mapped with any S member, is not less than |P(S)|?

No. The proof says nothing about the size of the set other than it must be non-empty.

However, not establishing the cardinality of any unmapped subset of P(S) in no way discredits the proof. The proof shows that no bijective mapping exists. The magnitude by which all of those non-bijective mappings "miss" is not relevant.

The proof establishes that |S| < |P(S)|. There is no uncertainty in the conclusion.

 

[Roofgardener]

Ummm.... I'm probably stating the obvious here, but...

I assume you have assumed that |N| is both Real, and greater than zero ?

I mean.... if not, then your sequence wouldn't necessarily converge at infinity ?