Cont: Deeper than primes - Continuation 2

[doronshadmi]

For any given (infinite) set S, there are many such sets, Z, that can satisfy both criteria.

Do you also agree that for any given (infinite) set P(S), there are many such sets, Z, that can satisfy both criteria?

 

[jsfisher]

For any given (infinite) set S, there are many such sets, Z, that can satisfy both criteria.

Do you also agree that for any given (infinite) set P(S), there are many such sets, Z, that can satisfy both criteria?

Since P(S) is determined by S, do you not agree that my statement includes yours?

 

[doronshadmi]

Since P(S) is determined by S, do you not agree that my statement includes yours?

Let's put it this way: Dedekind-infinite holds for (infinite) S or P(S) sets.

Do you agree about what I wrote above?

 

[jsfisher]

Let's put it this way: Dedekind-infinite holds for (infinite) S or P(S) sets.

Do you agree about what I wrote above?

Let's stay on the initial path, shall we?

Given any infinite set S, its power set, P(S), exists. Given such a powerset, then, there are many, many sets, Z, that satisfy the following conditions:

(z) Z is a subset of P(S).
(a) There are infinitely many members of P(S) members that are not members of Z. (I.e., the set difference P(S)\Z is an infinite set.)
(b) There is bijection between Z and P(S).​

...and before you launch into a quibble about proper subsets, note that (z) and (a) taken together assure Z is a proper subset of P(S).

 

[doronshadmi]

note that (z) and (a) taken together assure Z is a proper subset of P(S).

(b) There is bijection between Z and P(S).

EDIT:

In that case given any P(S) member that is not mapped with some S member, one concludes that this P(S) member is simply not a member of Z, so one can't use it in order to prove that there is no bijection between S and Z (as you have said, Z is used here as a symbol for any passible proper subset of P(S) such that there is bijection between Z and P(S)).

 

[jsfisher]

In that case given any P(S) member that is not mapped with some S member....

What mapping would that be?

 

[doronshadmi]

What mapping would that be?

Exactly the same "mapping" that can't be defined between a given P(S) member and a given S member, whether it is given by Cantor's theorem (https://en.wikipedia.org/wiki/Cantor's_theorem) or Cantor's diagonal argument (https://en.wikipedia.org/wiki/Cantor's_diagonal_argument).

EDIT:

In both cases the considered members that are used to prove that there is no bijection between the considered sets, are simply members that are not included in proper subsets, which have bijection with a given set that Cantor climes that its cardianlity is strictly bigger than the cardinality of a given other set (where examples of the other sets are the infinite sets S or N, etc.).

 

[jsfisher]

Exactly the same "mapping" that can't be defined between a given P(S) member and a given S member

So, you are using a mapping that cannot be defined. Interesting.

If the mapping doesn't exist, how do you propose using it?

 

[doronshadmi]

Some example:

N={1, 2, 3, 4, 5, ...}
j={10 , 200, 3000, 40000, 500000, ...}

There is bijection between j and N.

K={100 , 2000, 30000, 400000, 5000000, ...}

The N members that are not j members, can't be used in order to prove that there is no bijection between j and K.

 

[doronshadmi]

So, you are using a mapping that cannot be defined. Interesting.

If the mapping doesn't exist, how do you propose using it?

You can't, yet Cantor used such impossibility in order to conclude that there are (infinitely many) more real numbers than natural numbers or there are more (infinitely many) P(S) members than S members.

 

[jsfisher]

You can't, yet Cantor used such impossibility in order to conclude that there are (infinitely many) more real numbers than natural numbers or there are (infinitely many) P(S) members than S members.

You still do not understand Cantor's standard proof.

The proof shows that there is no injection from P(S) to S. It is a proof by contradiction in which the hypothetical existence of such an injection leads to a contradiction. Nothing more.

 

[doronshadmi]

You still do not understand Cantor's standard proof.

The proof shows that there is no injection from P(S) to S. It is a proof by contradiction in which the hypothetical existence of such an injection leads to a contradiction. Nothing more.

You steel do not understand that there is no contradiction, because the considered hypothetical existence of such an injection does not exist in the first place in case of any possible proper subset of P(S) that has bijection with P(S), and we explicitly consider only the mapping of such proper subset of P(S), with S.

In other words, "Cantor's standard proof by contradiction" is insufficient in order to prove that there is no bijection between S and Z (where Z is a symbol for any possible proper subset of P(S) such that there is bijection between Z and P(S)).

Moreover, the same insufficiency holds also in the case of Cantor's diagonal argument.

So jsfisher, you still did not answer to http://www.internationalskeptics.com/forums/showpost.php?p=11557530&postcount=2155 , and it can be done only if you show that Cantor's methods hold also in case of the fundamental property of any infinite set, known as Dedekind-infinite (https://en.wikipedia.org/wiki/Dedekind-infinite_set).

 

[jsfisher]

You steel do not understand that there is no contradiction

You are back to denying Cantor's Theorem, then. You are again invited to demonstrate the flaw in the theorem. Either show a specific counter example or show a specific mistake in the standard proof.

We've been down this path before; you've failed each and every time; but have at it again, if you like.

 

[jsfisher]

"Cantor's standard proof by contradiction" is insufficient in order to prove that there is no bijection between S and Z (where Z is a symbol for any possible proper subset of P(S) such that there is bijection between Z and P(S)).

Cantor's Theorem is sufficient. You don't have to resort to the theorem's proof method to use the theorem.

 

[doronshadmi]

Cantor's Theorem is sufficient.

It is not sufficient since it doesn't handle with Dedekind-infinite, which is a basic property of any infinite set.

 

[doronshadmi]

jfisher, you wrote two posts.

No one of them actually address in details my arguments in http://www.internationalskeptics.com/forums/showpost.php?p=11563267&postcount=2172.

So I hope that your third reply will be right to the point by actually using detailed information about the issue at hand, and not just pointless hands weaving as done in your last two posts.

-----

In order to be clear, I will not reply to any of your hands weaving posts about the issue at hand.

 

[jsfisher]

It is not sufficient since it doesn't handle with Dedekind-infinite, which is a basic property of any infinite set.

Cantor's Theorem is simply that |S| < |P(S)|, so which aspect of |S| < |P(S)| doesn't work for you? Did you find a counter-example? Or perhaps you found a mistake in the proof of Cantor's Theorem?

 

[doronshadmi]

Cantor's Theorem is simply that |S| < |P(S)|, so which aspect of |S| < |P(S)| doesn't work for you? Did you find a counter-example? Or perhaps you found a mistake in the proof of Cantor's Theorem?

It can't handle with Dedekind-infinite, which is a basic property of any infinite set.

The details are already given in http://www.internationalskeptics.com/forums/showpost.php?p=11563267&postcount=2172 and once again you simply ignore them.

 

[jsfisher]

It can't handle with Dedekind-infinite, which is a basic property of any infinite set.

It is always curious when you come across a new term then you try to wedge it in just about everywhere, whether it belongs or not.

(1) Whether all infinite sets are also Dedekind-infinite depends on the underlying set theory. If you accept the Axiom of Choice (even in a lesser form), then infinite and Dedekind-infinite are equivalent. Without it, no.

(2) Cantor's Theorem has no weakness regarding infinite sets nor Dedekind-infinite sets. The theorem is true for all sets, whether any flavor of infinite or not. You are welcome to show otherwise, either by specific counter-example or by error in the proof itself.

 

[jsfisher]

In other words, "Cantor's standard proof by contradiction" is insufficient in order to prove that there is no bijection between S and Z (where Z is a symbol for any possible proper subset of P(S) such that there is bijection between Z and P(S)).

Given any set S and any set Z such that Z is a proper subset of P(S) and there exists a one-to-one and onto function f: Z -> P(S) (i.e., there is a bijection between Z and P(S)), prove there is no one-to-one and onto function g: S -> Z.

Proof:
(1) Assume g exists.
(2) The composite function, f of g exists.
(3) Since f and g are each one-to-one, the composite function is one-to-one. Since f and g are each onto, the composite function is onto.
(4) Since g: S -> Z and f: Z -> P(S), then f of g: S -> P(S).
(5) Since f of g is one-to-one and onto, there is a bijection between S and P(S).
(6) By Cantor's Theorem, there is no bijection between S and P(S).
(7) The function g does not exist.

QED