Cont: Deeper than primes - Continuation 2

[jsfisher]

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)|.

That is my assumption as well. It is likely a simple corollary to Cantor's Theorem, but I haven't thought about it enough to prove it (nor am I likely to).

 

[doronshadmi]

Edit:

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.

You have missed (2), it is about 3 disjoint sets as folows:

1) Set S (which is infinite).

2) The set of |S| P(S) members that is in bijection with S.

3) Set K of |K| P(S) members that are not in the range of all S members.

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)|.

Here is a corrected version of my argument:

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|

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.

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 GCH is true, then |K| is not less than |P(S)| (there is no transfinite catdinality between |S| and |P(S)|).

9) If GCH is false, then |S| < |K| < |P(S)| (there is transfinite catdinality between |S| and |P(S)|).

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

 

[jsfisher]

Here is a corrected version of my argument:

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.

This is into word-salad territory. For any mapping there is a bijection?? I think you may have meant: For any mapping m from S to P(S), let K = P(S) \ range(m).

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|.

No, we don't. S < P(S) is established. Perhaps you can use that to prove things about your number 6.

8) If GCH is true, then |K| is not less than |P(S)| (there is not transfinite catdinality between |S| and |P(S)|).

9) If GCH is false, then |S| < |K| < |P(S)| (there is transfinite catdinality between |S| and |P(S)|).

Still false.

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

Still false.

 

[phiwum]

Edit:


You have missed (2), it is about 3 disjoint sets as folows:

1) Set S (which is infinite).

2) The set of |S| P(S) members that is in bijection with S.

3) Set K of |K| P(S) members that are not in the range of all S members.

It's just a matter of applying the theorem stated -- except that I said in the proof of the corollary that we let X = S, when I should have said X = ran(f).

Here is a corrected version of my argument:

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 GCH is true, then |K| is not less than |P(S)| (there is no transfinite catdinality between |S| and |P(S)|).

9) If GCH is false, then |S| < |K| < |P(S)| (there is transfinite catdinality between |S| and |P(S)|).

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

Statement (9) is simply not proved.

Statements (7) is nonsense, since one needn't show that. It follows from the above.

Statements (1) - (6) are just the corollary I stated. Let f:S -> P(S). It is a triviality that the following hold:

P(S) = ran(f) u P(S) \ ran(f)
Hence |P(S)| = |ran(f)| + |P(S) \ ran(f)|
Hence if S is infinite, |ran(f)| < |P(S) \ ran(f)|

 

[doronshadmi]

This is into word-salad territory. For any mapping there is a bijection?? I think you may have meant: For any mapping m from S to P(S), let K = P(S) \ range(m).

(2) simply says that for any mapping from all |S| members of set S to |S| members of set P(S) there is 1-to-1 and onto, simply because both mapped sets have |S| members.

(2) also says that in addition to these two disjoint sets there is another disjoint set, called K, which its |K| members are P(S) members that are not mapped with any S member in the considered mapping, for example:

S={a,b,c,...}

In case of the mapping

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

a,b,c,... are members of S,
{a},{b},{c},... are members of P(S) that are in bijection with S members,
and, for example, {{},{a,b},{a,c},...} are members of P(S) that are not mapped with any S member, in that particular example.

Generally, for any mapping between S and P(S), such 3 disjoint sets are found (with different members in case of the two disjoint sets that each one of them includes only part of the members of set P(S)).

 

[doronshadmi]

It's just a matter of applying the theorem stated -- except that I said in the proof of the corollary that we let X = S, when I should have said X = ran(f).



Statement (9) is simply not proved.

Statements (7) is nonsense, since one needn't show that. It follows from the above.

Statements (1) - (6) are just the corollary I stated. Let f:S -> P(S). It is a triviality that the following hold:

P(S) = ran(f) u P(S) \ ran(f)
Hence |P(S)| = |ran(f)| + |P(S) \ ran(f)|
Hence if S is infinite, |ran(f)| < |P(S) \ ran(f)|

Dear phiwum, if you wish to communicate with me, please express what you wish to say in plain English, as I do for example in http://www.internationalskeptics.com/forums/showpost.php?p=10950016&postcount=785, about the discussed subject (as given in plain English in http://www.internationalskeptics.com/forums/showpost.php?p=10949927&postcount=782).

Thank you.

 

[phiwum]

Dear phiwum, if you wish to communicate with me, please express what you wish to say in plain English, as I do for example in http://www.internationalskeptics.com/forums/showpost.php?p=10950016&postcount=785, about the discussed subject (as given in plain English in http://www.internationalskeptics.com/forums/showpost.php?p=10949927&postcount=782).

Thank you.

It is easiest to use simple symbolic notation to discuss mathematics. Perhaps you're not familiar with the notation I use. Let f: X -> Y

ran(f) = { y in Y | there is an x in X such that f(x) = y }
X \ Y = { x in X | x is not in Y }
X u Y = X union Y = { z | z in X or z in Y }

 

[jsfisher]

(2) simply says that for any mapping from all |S| members of set S to |S| members of set P(S) there is 1-to-1 and onto, simply because both mapped sets have |S| members.

False. The set size condition can be met without the mapping being either 1-to-1 or onto.

Let S = {..., -3, -2, -1, 0, 1, 2, 3,...}
Let Q = {..., {-3}, {-2}, {-1}, {0}, {1}, {2}, {3},...}
Clearly |S| = |Q| and Q is a subset of P(S).

Let m be a mapping from S to Q such that m: x -> {x²}

The mapping, m, is neither 1-to-1 nor onto.

 

[doronshadmi]

It is easiest to use simple symbolic notation to discuss mathematics. Perhaps you're not familiar with the notation I use. Let f: X -> Y

ran(f) = { y in Y | there is an x in X such that f(x) = y }
X \ Y = { x in X | x is not in Y }
X u Y = X union Y = { z | z in X or z in Y }

This is very nice but my argument is very simple:

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.

S is an infinite set.

In order to conclude that |S| < |P(S)|, |K| (as described in http://www.internationalskeptics.com/forums/showpost.php?p=10949927&postcount=782) must be explicitly determined as > |S| (which means that the term at least one is insufficient).

So please show that |K| > |S| by using the standard proof of Cantor's theorem.

 

[doronshadmi]

False. The set size condition can be met without the mapping being either 1-to-1 or onto.

Let S = {..., -3, -2, -1, 0, 1, 2, 3,...}
Let Q = {..., {-3}, {-2}, {-1}, {0}, {1}, {2}, {3},...}
Clearly |S| = |Q| and Q is a subset of P(S).

Please read all of http://www.internationalskeptics.com/forums/showpost.php?p=10950016&postcount=785 in order to understand my argument, as given in http://www.internationalskeptics.com/forums/showpost.php?p=10950582&postcount=789.

 

[jsfisher]

Please read all of http://www.internationalskeptics.com/forums/showpost.php?p=10950016&postcount=785 in order to understand my argument, as given in http://www.internationalskeptics.com/forums/showpost.php?p=10950582&postcount=789.

Please read and understand all of my post to see that you made a false statement.

That said, I have read and I do understand your "argument". It is based on false statements, bald assumptions, and bogus logic.

 

[jsfisher]

In order to conclude that |S| < |P(S)|, |K| (as described in http://www.internationalskeptics.com/forums/showpost.php?p=10949927&postcount=782) must be explicitly determined as > |S| (which means that the term at least one is insufficient).

You have yet to prove this as a requirement.

 

[phiwum]

This is very nice but my argument is very simple:

S is an infinite set.

In order to conclude that |S| < |P(S)|, |K| (as described in http://www.internationalskeptics.com/forums/showpost.php?p=10949927&postcount=782) must be explicitly determined as > |S| (which means that the term at least one is insufficient).

So please show that |K| > |S| by using the standard proof of Cantor's theorem.

Why must I use the standard proof of Cantor's theorem, when the corollary is obvious?

Cantor's theorem entails that |S| < |P(S)|. It is also a theorem that, for infinite sets X and Y, if |Y| <= |X|, |X| + |Y| <= |X|.

Consequently, if |P(S)\ran(f)| <= |ran(f)|, it follows that |ran(f)| + |P(S)\ran(f)| <= |ran(f)| < |P(S)|. Hence, |P(S)\ran(f)| > |ran(f)|

Done.

There is no reason at all that I must use the same reasoning as the proof of Cantor's theorem when the claim at issue is an obvious corollary.

(Minor point: I did use some facts about < on cardinals that we haven't discussed here, namely that it is a linear order.)

 

[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.

S is an infinite set.

H is a set that its members are not mapped with all |S| members of set S, and is has exactly one member for any mapping between S and P(S).

Here is an example, without loss of generality:

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

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

is {}.

In that case H={{}}, where a member of P(S) like {a,b} is not a member of H because a --> {a} or b --> {b}, so what you wrote above is irrelevant to my "weaker" argument.

The "weaker" version of my argument holds, is as follows:

For any mapping from set S to set P(S) there is a bijection from all |S| members of set S to |S| members of set P(S), and also there is set H that has exactly one member of set P(S) that not in the range of all the members of set S.

In other words, |S|+|H| (where |H| is exactly 1) holds for any mapping from set S to set P(S).

|S|+|H| does not mean that H member is added to set S.

For any mapping from set S to set P(S) the best that can be shown is |S|+|1|, but by the transfinite number system |S| = |S|+1, so for any mapping from set S to set P(S) it is not shown that |S|<|P(S)|, so the standard proof of Cantor's theorem does not hold if S is an infinite set.

EDIT:

The only set defined in the proof is a diagonal set

which has exactly one member for any mapping from set S to the diagonal set.

So the same result holds in case of Cantor's diagonal argument (https://en.wikipedia.org/wiki/Cantor's_diagonal_argument) because for any mapping from set S to the diagonal set the best that can be shown is |S|+|1| (H has exactly 1 member of the diagonal set that is not in the range of all S members, but by the transfinite number system |S| = |S|+1, so for any mapping from set S to the diagonal set it is not shown that |S|<|the diagonal set|.

 

[phiwum]

The "weaker" version of my argument holds, is as follows:

For any mapping from set S to set P(S) there is a bijection from all |S| members of set S to |S| members of set P(S), and also there is set H that has exactly one member of set P(S) that not in the range of all the members of set S.

In other words, |S|+|H| (where |H| is exactly 1) holds for any mapping from set S to set P(S).

The above is simply nonsense. Let f:S -> P(S)

What the argument shows is that P(S) \ ran(f) is non-empty. It does this by exhibiting an element of P(S) \ ran(f). There is no sense in which you can conclude P(S) \ ran(f) consists only of the element exhibited.

We may, of course, iterate the proof. In the case S = N, to find a second element, do the following: Let T be the element of P(N) that Cantor's proof showed was an element of P(N) \ ran(f).

f_1(0) = T
f_1(n + 1) = f

This is a new function from N -> P(N), and we may apply Cantor's argument again. We get a new element of P(N) not in the range of f_1. This new element, then, is a second element not in the range of f.

Iterate as much as you desire, though this method only gives countably many elements of P(N) \ ran(f). That there are uncountably many follows from what you've already shown is true: |P(N) \ ran(f)| > |ran(f)|.

 

[doronshadmi]

This new element, then, is a second element not in the range of f.

For any mapping from S to P(S) one and only one element is found.

Adding those elements is based only on intuition, and only intuition is insufficient in order to establish "real mathematics".

If you disagree with me, then rigorously (which meas formally\logically) show how these elements are added to each other.

EDIT:

Please pay attention to the fact that you can't formally\logically define the set of all members of P(S) that are not mapped with all S members, simply because any particular member of H (which is some member of P(S)) can be mapped with some member of S in another mapping from S to P(S), so the members of H can't be added to each other.

 

[phiwum]

For any mapping from S to P(S) one and only one element is found.

Adding those elements is based only on intuition, and only intuition is insufficient in order to establish "real mathematics".

If you disagree with me, then rigorously (which meas formally and logically) show how these elements are added to each other.

Sorry, but I've no idea what you're asking for.

All that us required is to show that the axioms of set theory entail the consequence. I've done that.

 

[doronshadmi]

Sorry, but I've no idea what you're asking for.

All that us required is to show that the axioms of set theory entail the consequence. I've done that.

Please pay attention to the fact that you can't formally\logically define the set of all members of P(S) that are not mapped with all S members, simply because any particular member of H (which is some member of P(S)) can be mapped with some member of S in another mapping from S to P(S), so the members of H can't be added to each other.

In other words, H formally\logically can have one and only one member, and therefore |S|=|S|+1 holds for all mappings from S to P(S), by the standard proof of Cantor's theorem (and so is the case about Cantor's diagonal set).

So, formally\logically what is called Cantor's theorem is no more than a conjecture (in case that S is an infinite set) if the standard proof is used.

 

[doronshadmi]

Also I think that we show that formally\logically there are cases where one can't define a set, which is the union of infinitely many disjoint sets.

 

[phiwum]

Please pay attention to the fact that you can't formally\logically define the set of all members of P(S) that are not mapped with all S members, simply because any particular member of H (which is some member of P(S)) can be mapped with some member of S in another mapping from S to P(S), so the members of H can't be added to each other.

In other words, H formally\logically can have one and only one member, and therefore |S|=|S|+1 holds for all mappings from S to P(S), by the standard proof of Cantor's theorem (and so is the case about Cantor's diagonal set).

So, formally\logically what is called Cantor's theorem is no more than a conjecture (in case that S is an infinite set) if the standard proof is used.

Sorry, but this is nonsense. I've already sketched the obvious proof that P(N) has at least countably many elements not in the image of any function f:N -> P(N).

I should never have bothered entering into this silly conversation. I will bow out now, since I see no real possibility of settling this fairly trivial issue.