Cont: Deeper than primes - Continuation 2

[jsfisher]

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 ?

You may be misinterpreting the notation. The N would be a set, and |N|, the cardinality of that set.

If N = {}, then |N| = 0.
If N = {1, 2, 3, 4, ...}, then |N| is Aleph₀ (the smallest "infinity").

So, as you can see, |N| need not be greater than zero nor need it be real. This is why I think you may have interpreted things different than they were intended.

What sequence were you referring to?

 

[Little 10 Toes]

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.

This is wrong. Set S = {a}. P(S) = { {}, {a} }. There is one member of P(S) that does not directly match the members of S.

Example #2: Set S = {a,b}. P(S) = { {}, {a}, {b}, {a,b} }. There are two members of P(S) that do not directly match the members of S.

 

[doronshadmi]

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.

Please take your time to understand the conclusion about the transfinite number system (shown in http://www.internationalskeptics.com/forums/showpost.php?p=10947860&postcount=758) as a result of "not establishing the cardinality of any unmapped subset of P(S)" (as done as a part of what is called Cantor's proof).

Without explicitly show that for any mapping between all |S| members of set S and |S| members of set P(S), there are not less than |P(S)| unmapped members of set P(S) (where S is an infinite set), the transfinite number system itself is not rigorously established, and so is the case about what is called Cantor's proof.

Intuitively I think that |#| is not less than |P(S)|, but using only intuition is insufficient in order to rigorously establish the transfinite number system or what is called Cantor's proof.

Currently your last reply to me is based only on intuition.

 

[Little 10 Toes]

Please define "Cantor's proof".

 

[phiwum]

Please take your time to understand the conclusion about the transfinite number system (shown in http://www.internationalskeptics.com/forums/showpost.php?p=10947860&postcount=758) as a result of "not establishing the cardinality of any unmapped subset of P(S)" (as done as a part of what is called Cantor's proof).

Without explicitly show that for any mapping between all |S| members of set S and |S| members of set P(S), there are not less than |P(S)| unmapped members of set P(S) (where S is an infinite set), the transfinite number system itself is not rigorously established, and so is the case about what is called Cantor's proof.

I'm not about to enter this discussion, because I did my time on the Usenet group sci.math for many long years. But it sure isn't surprising to see another long thread over the validity of Cantor's theorem.

Before anyone misunderstands me, the proof of Cantor's theorem is relatively simple, certainly well-studied, and confirmed in very many mathematical theorem checkers (as well as proved by some theorem provers).

Just dropping by to say that I've certainly seen this kind of, in the end, tedious discussion before. Good luck to those who have taken up the charge to prove the fairly obvious: that Cantor's theorem is an obvious theorem of ZF.

 

[jsfisher]

Without explicitly show that for any mapping between all |S| members of set S and |S| members of set P(S), there are not less than |P(S)| unmapped members of set P(S) (where S is an infinite set), the transfinite number system itself is not rigorously established, and so is the case about what is called Cantor's proof.

If you want to discredit Cantor's Theorem or the standard proof used to establish Cantor's Theorem, then work with the standard proof. (And please stop with the meaningless term, "Cantor's proof". Cantor proved many things and therefore has many proofs that can be attributed to him.)

The standard proof shows unequivocally that S and P(S) cannot be equi-numerous. The magnitude of the difference is not exposed in the proof nor is it relevant to the proof.

 

[doronshadmi]

If you want to discredit Cantor's Theorem or the standard proof used to establish Cantor's Theorem, then work with the standard proof. (And please stop with the meaningless term, "Cantor's proof". Cantor proved many things and therefore has many proofs that can be attributed to him.)

Come on jsfisher, you know very well that our last discussion is about the standard proof of Cantor's theorem.

The standard proof shows unequivocally that S and P(S) cannot be equi-numerous.

What is written above is true only if one rigorously shows that for any mapping between all |S| members of set S and |S| members of set P(S), there are not less than |P(S)| unmapped members of set P(S) (where S is an infinite set).

Without it the cardinalitiy of the set of unmapped members of P(S) is not rigorously determined, such that it can be |S| or |P(S)| (if CH it true).

In case that this cardinality is |S| we get the inequality |S| < |S|+|S| by the logic that is used in the standard proof of Cantor's theorem.

But according to the transfinite number system, it is shown that |S| = |S|+|S|, so there are two results that do not "agree" with each other in the transfinite number system.

So, in order to rigorously establish the transfinite number system and also to prove that |S| < |P(S)| by what is called the standard proof of Cantor's theorem, there is no choice but to rigorously show that for any mapping between all |S| members of set S and |S| members of set P(S), there are not less than |P(S)| unmapped members of set P(S) (where S is an infinite set).

The magnitude of the difference is not exposed in the proof nor is it relevant to the proof.

jsfisher, what is called the standard proof of Cantor's theorem 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, so what is called the standard proof of Cantor's theorem actually can't prove that |S|<|P(S)| and it also shows that the transfinite number system is not well established, as long as it can't rigoriusly show that there are not less than |P(S)| unmapped members of set P(S) (where S is an infinite set) for any map from S to P(S).

Again, I intuitively think that there are not less then unmapped P(S) members for any map from S to P(S) (where S is an infinite set), but only intuition is not enough in what is called "real mathematics", and you are definitely using only intuition by claiming that at least one unnamed member of P(S) is needed in order to rigorously determine (or in other words, to prove) that |S|<|P(S)|.

More details are given in http://www.internationalskeptics.com/forums/showpost.php?p=10947860&postcount=758.

 

[jsfisher]

Come on jsfisher, you know very well that our discussion is about the standard proof of Cantor's theorem.

Are you opposed to clear communication?

What is written above is true only if one rigorously shows that for any mapping between all |S| members of set S and |S| members of set P(S), there are not less than |P(S)| unmapped members of set P(S) (where S is an infinite set).

Why?
Why are you demanding that a proof of Cantor's Theorem provide conclusive information about sets it does not even consider? The only set defined in the proof is a diagonal set; it's sole use is to disprove the existence of a bijection.

For that matter, why do you assume there is no cardinal number between |S| and |P(S)|?

For that matter, why do you demand a proof Cantor's Theorem resolve all questions in the transfinites? Cantor's Theorem gives rise to the transfinites; it is not dependent upon them. If you can construct a contradiction, fine, that would doom the theorem. However, imagining there is something unknown is not that.

 

[doronshadmi]

Why?
Why are you demanding that a proof of Cantor's Theorem provide conclusive information about sets it does not even consider? The only set defined in the proof is a diagonal set; it's sole use is to disprove the existence of a bijection.

For that matter, why do you assume there is no cardinal number between |S| and |P(S)|?

Cantor's Theorem: The cardinality of any set is less than the cardinality of its power set.

My argument is not about Cantor's theorem (as defined above).

I am talking only about the case that S is an infinite set, so my argument is not about the cardinally of any set. In that case (if CH is true) there is no cardinal number between |S| and |P(S)|.

But even if CH is false, still for any given mapping from S to P(S), the number of P(S) members that are out of the range of S members, must rigorously by defined as not less than |P(S)| in order to conclude that |S| < |P(S)| (where my argument is only about the case that S as an infinite set).

For that matter, why do you demand a proof Cantor's Theorem resolve all questions in the transfinites? Cantor's Theorem gives rise to the transfinites; it is not dependent upon them. If you can construct a contradiction, fine, that would doom the theorem. However, imagining there is something unknown is not that.

You are missing again my argument which shows that in case that S is an infinite set, this part of Cantor's theorem and its implications on the transfinite number system, is based only on intuition, if for any given mapping from S to P(S), the number of P(S) members that are out of the range of S members, is not rigorously defined (at least one can't be considered as rigorously defined).

EDIT:

jsfisher, without replying in details to the content of http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767 (including its link), there is no real discussion between us on this subject, so please take your time to understand my argument (which is only about infinite sets) and then please reply in details (according to what is written in http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767 (including its link)) about it.

Thank you.

 

[jsfisher]

still for any given mapping from S to P(S), the number of P(S) members that are out of the range of S members, must rigorously by defined as not less than |P(S)| in order to conclude that |S| < |P(S)| (where my argument is only about the case that S as an infinite set).

Why? Why must it be rigorously defined? The lack of bijection is sufficient to prove they are not equi-numerous (and the existence of a trivial injection is sufficient to establish the inequality).

You are demanding a condition that is unnecessary. Can you show otherwise?

 

[doronshadmi]

Why? Why must it be rigorously defined? The lack of bijection is sufficient to prove they are not equi-numerous (and the existence of a trivial injection is sufficient to establish the inequality).

You are demanding a condition that is unnecessary. Can you show otherwise?

Once again jsfisher, without replying in details to the content of http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767 (including its link), there is no real discussion between us on this subject, so please take your time to understand my argument (which is only about infinite sets) and then please reply in details (according to what is written in http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767 (including its link)) about it.

Please be aware that what I call "a stronger version of my idea" (as shown in http://www.internationalskeptics.com/forums/showpost.php?p=10947860&postcount=758, which is the link in http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767) is only about infinite sets.

Thank you.

 

[jsfisher]

Once again jsfisher, without replying in details to the content of http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767 (including its link), there is no real discussion between us on this subject, so please take your time to understand my argument (which is only about infinite sets) and then please reply in details (according to what is written in http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767 (including its link)) about it.

Thank you.

As I have stated before, you may choose to respond to my posts, or not; that is your choice. The same options are open to me with respect to your posts.

If you will not justify the necessity of an unnecessary requirement, they you are correct, there is no real discussion, here.

 

[doronshadmi]

If you will not justify the necessity of an unnecessary requirement,...

It is justified in http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767 (including its link).

Please be aware that what I call "a stronger version of my idea" (as shown in http://www.internationalskeptics.com/forums/showpost.php?p=10947860&postcount=758, which is the link in http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767) is only about infinite sets.

If you refuse to read, and at least try to understand and reply in details to my justification, then there is no real discussion between us about this subject.

Currently this is exactly what you are doing, you refuse to read, and at least try to understand and reply in details to my justification.

 

[doronshadmi]

I wish to correct some mistake written in http://www.internationalskeptics.com/forums/showpost.php?p=10949200&postcount=769.

I wrote:

"But even if CH is false, still for any given mapping from S to P(S), the number of P(S) members that are out of the range of S members, must rigorously by defined as not less than |P(S)| [this is a mistake] in order to conclude that |S| < |P(S)| (where my argument is only about the case that S as an infinite set)."

It has to be:

"But even if CH is false, still for any given mapping from S to P(S), the number of P(S) members that are out of the range of S members, must rigorously by defined as at least not equal to |S| in order to conclude that |S| < |P(S)| (where my argument is only about the case that S as an infinite set)."

 

[doronshadmi]

EDIT:

I'll try to explain my argument as simple as a can:

1) S is an infinite set.

2) For any mapping from S to P(S) there is a bijection from all |S| members of S to |S| members of P(S), and also there is set K of all P(S) members that are not in the range of all S members.

3) In that case we get the expression |S| $ |S| + |K|, where $ is a placeholder for = or < .

4) If |K|=|S| then according to the transfinite number system $ is replaced by = as follows: |S| = |S| + |K|

5) If |K|=|S| and one claims that the term at least one is sufficient in order to conclude that |S| < |P(S)|, then $ is replaced also by < as follows: |S| < |S| + |K|, because |K| satisfies the term at least one.

6) In that case |S| = |S| + |K| and |S| < |S| + |K| are both hold in the transfinite number system.

7) In order to avoid (6) one must to explicitly show that |K| > |S|.

8) If CH is true, then |K| is not less than |P(S)|.

9) If CH is false, then |S| < |K| < |P(S)|.

10) In both (8) and (9) |K| must be explicitly defined as > |S|, in order to conclude that |S| < |P(S)|.

 

[phiwum]

EDIT:

I'll try to explain my argument as simple as a can:

1) S is an infinite set.

2) For any mapping from S to P(S) there is a bijection from all |S| members of S to |S| members of P(S), and also there is set K of all P(S) members that are not in the range of all S members.

3) In that case we get the expression |S| $ |S| + |K|, where $ is a placeholder for = or < .

4) If |K|=|S| then according to the transfinite number system $ is replaced by = as follows: |S| = |S| + |K|

5) If |K|=|S| and one claims that the term at least one is sufficient in order to conclude that |S| < |P(S)|, then $ is replaced also by < as follows: |S| < |S| + |K|, because |K| satisfies the term at least one.

6) In that case |S| = |S| + |K| and |S| < |S| + |K| are both hold in the transfinite number system.

7) In order to avoid (6) one must to explicitly show that |K| > |S|.

8) If CH is true, then |K| is not less than |P(S)|.

9) If CH is false, then |S| < |K| < |P(S)|.

10) In both (8) and (9) |K| must be explicitly defined as > |S|, in order to conclude that |S| < |P(S)|.

Your conclusion (10) is very oddly stated, since it uses the word "defined" for something which is a consequence.

You're right that, in general, if X and Y are disjoint infinite sets and |X u Y| = |X| + |Y| > |X|, then |X| < |Y|. This is the only fact you've used, really, and there is nothing here to do with power sets.

By the way, you keep referring to CH when you mean GCH (generalized continuum hypothesis). CH just says |P(N)| = Aleph_1, whereas GCH says that for all n, |P(Aleph_n)| = Aleph_{n+1}. GCH is independent of ZFC+CH.

ETA: (9) is false. In fact, I suspect that it is always the case that |K|=|P(S)|, but I'm not positive about that.

 

[jsfisher]

It is justified in http://www.internationalskeptics.com/forums/showpost.php?p=10949034&postcount=767 (including its link).

No, it isn't. You merely claim knowing some unknown is necessary, even though nothing in Cantor's Theorem depends on it.

Claiming is not showing.

 

[jsfisher]

7) In order to avoid (6) one must to explicitly show that |K| > |S|.

Given that Cantor proved unconditionally that |S| < |P(S)|, there is nothing to "avoid".

...
9) If CH is false, then |S| < |K| < |P(S)|.

False.

 

[doronshadmi]

Your conclusion (10) is very oddly stated, since it uses the word "defined" for something which is a consequence.

Ok, let's change "defined" by "determined".

You're right that, in general, if X and Y are disjoint infinite sets and
|X u Y| = |X| + |Y| > |X|, then |X| < |Y|. This is the only fact you've used, really, and there is nothing here to do with power sets.

To what sentence do you refer?

By the way, you keep referring to CH when you mean GCH (generalized continuum hypothesis). CH just says |P(N)| = Aleph_1, whereas GCH says that for all n, |P(Aleph_n)| = Aleph_{n+1}. GCH is independent of ZFC+CH.

Thank you. I'll correct it.

ETA: (9) is false.

Let's see:

GCH states that there is no transfinite cardinality which is greater than |S| and smaller than |P(S)|, so if this statement is false, there is transfinite cardinality which is greater than |S| and smaller than |P(S)|, and this is exactly what is written in (9) (also CH is corrected to GCH).

 

[phiwum]

Ok, let's change "defined" by "determined".


To what sentence do you refer?

Your entire argument.

Theorem: if X and Y are disjoint infinite sets and |X u Y| = |X| + |Y| > |X|, then |X| < |Y|.

Corollary: Let f:S -> P(S) and K = P(S) \ ran(f). Then |S| < |K|.

Proof of corollary: Let X = S and K = Y in the above theorem.

Thank you. I'll correct it.


Let's see:

GCH states that there is no transfinite cardinality which is greater than |S| and smaller than |P(S)|, so if this statement is false, there is transfinite cardinality which is greater than |S| and smaller than |P(S)|, and this is exactly what is written in (9) (also CH is corrected to GCH).

No, what (9) says is that |P(S) \ ran(f)| < |P(S)|. GCH does not imply that. It is possible (and I think true for every f, though I might be wrong) that |P(S) \ ran(f)| = |P(S)|.