Cont: Deeper than primes - Continuation 2

[doronshadmi]

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

You can't use Cantor's Theorem because it can't be used in order to prove that there is no bijection between S and Z exactly 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), called Z, with S.

You still do not understand that the mapping between S member and P(S) member that is involved with contradiction (according to Cantor's theorem), this mapping does not exist in the first please since the involved P(S) member is not a member of Z in the first place.

The same principle holds also in case of Cantor's diagonal argument, since the, so called, "missing R member in the list" is not a member of any possible proper subset of R (symbolized by T) in the first place (where T and R are bijective by Dedekind-infinite) so it can't be used in order to conclude that there is no bijection between N and T.

 

[jsfisher]

You can't use Cantor's Theorem because it can't be used in order to prove that there is no bijection between S and Z exactly 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), called Z, with S.

I used Cantor's Theorem strictly for its result that |S| < |P(S)|, nothing more. Would you deny that |S| < |P(S)|? If so, it falls to you to disprove Cantor.

As a direct consequence of |S| < |P(S)| we can conclude |S| < |Z|, but Cantor isn't needed any more for that part.

 

[jsfisher]

You still do not understand that the mapping between S member and P(S) member that is involved with contradiction (according to Cantor's theorem)....

Cantor's Theorem is that for any set S, |S| < |P(S)|. There is no mapping. There is no Z, either, for that matter.

 

[doronshadmi]

There is no mapping.

Wrong, you can't prove anything by Cantor's theorem if mapping is not involved, or by your own words

The proof shows that there is no injection from P(S) to S

where injection is a form of mapping.

 

[doronshadmi]

(1) Whether all infinite sets are also Dedekind-infinite depends on the underlying set theory.

Please explicitly demonstrate how a given infinite set is not Dedekind-infinite, by using some underlying set theory.

 

[doronshadmi]

I used Cantor's Theorem strictly for its result that |S| < |P(S)|, nothing more.

No, it is based on contradiction that does not exist in the first place, exactly because it can't handle with proper subsets of P(S) that have bijection with P(S).

The conclusion of Cantor's theorem, which according to it there are P(S) members that are beyond the range of all S members, is based on P(S) members that are not mapped with any S members in the first place (exactly as seen by any given proper subset of P(S) that does not include these "beyond the range" P(S) members in the first place and yet the considered P(S) proper subsets has bijection with P(S)).

 

[jsfisher]

There is no mapping.

Wrong, you can't prove anything by Cantor's theorem if mapping is not involved

You would be much more convincing if you didn't take things out of context and respond to something completely different from what the statement, in context, meant.

Cantor's Theorem provides no mapping.

or by your own words

The proof shows that there is no injection from P(S) to S

where injection is a form of mapping.

Yes, but Cantor's Theorem doesn't provide one. And for the case of an injection from P(S) to S, Cantor's Theorem guarantees there isn't one.

 

[jsfisher]

Please explicitly demonstrate how a given infinite set is not Dedekind-infinite, by using some the underlying set theory.

Now that you have discovered the term (Dedekind-infinite), maybe you should actually read up on it before you make any more mistakes using it.

 

[doronshadmi]

Now that you have discovered the term (Dedekind-infinite), maybe you should actually read up on it before you make any more mistakes using it.

Once again, and this time please do not avoid it. Please explicitly demonstrate how a given infinite set is not Dedekind-infinite, by using some underlying set theory.

 

[jsfisher]

Once again, and this time please do not avoid it. Please explicitly demonstrate how a given infinite set is not Dedekind-infinite, by using some underlying set theory.

Once again, and this time please do not avoid it, do a little reading on Dedekind-infinite (and Dedekind-finite) sets.

 

[doronshadmi]

Cantor's Theorem guarantees there isn't one.

It does not guarantee anything because it can't handle with Dedekind-infinite, which is an essential property of any infinite set.

If you disagree with me then please explicitly demonstrate how a given infinite set is not Dedekind-infinite, by using any underlying set theory that you wish.

 

[jsfisher]

In order to show that "there isn't one" one has no choice but to use the concept that is claimed not to be satisfied, where in this case the used concept is "injection".

No, we don't have to do anything extra to show "there isn't one." We have Cantor's Theorem at our disposal to tell us directly that |S| < |P(S)| for any set S. No additional work is required.

 

[doronshadmi]

Once again, and this time please do not avoid it, do a little reading on Dedekind-infinite (and Dedekind-finite) sets.

So you can't explicitly demonstrate how any possible given infinite set is not Dedekind-infinite, by using any underlying set theory that you wish.

 

[doronshadmi]

No, we don't have to do anything extra to show "there isn't one." We have Cantor's Theorem at our disposal to tell us directly that |S| < |P(S)| for any set S. No additional work is required.

No, Cantor's Theorem can't handle with Dedekind-infinite directly or indirectly.

If you disagree with me then please explicitly demonstrate how any possible given infinite set is not Dedekind-infinite, by using any underlying set theory that you wish.

 

[jsfisher]

No, Cantor's Theorem can't handle with Dedekind-infinite directly or indirectly.

Cantor's Theorem is well-established. Since you assert otherwise, it is up to you to prove your assertion.

If you disagree with me then please explicitly demonstrate how any possible given infinite set is not Dedekind-infinite, by using any underlying set theory that you wish.

Why? Whether there is an infinite set that is Dedekind-finite has nothing to do with Cantor's Theorem. Moreover, the obligation to discredit Cantor's Theorem is yours, not mine.

 

[doronshadmi]

Cantor's Theorem is well-established. Since you assert otherwise, it is up to you to prove your assertion.



Why? Whether there is an infinite set that is Dedekind-finite has nothing to do with Cantor's Theorem. Moreover, the obligation to discredit Cantor's Theorem is yours, not mine.

Cantor's Theorem can't handle with Dedekind-infinite directly or indirectly.

As long as you do not explicitly demonstrate how any possible given infinite set is not Dedekind-infinite, by using any underlying set theory that you wish, all you have is no more than hands weaving that can't handle with:

http://www.internationalskeptics.com/forums/showpost.php?p=11563267&postcount=2172

http://www.internationalskeptics.com/forums/showpost.php?p=11564766&postcount=2181

http://www.internationalskeptics.com/forums/showpost.php?p=11565143&postcount=2186

For the last time, please explicitly demonstrate that Dedekind-infinite is not an essential property of any infinite set, by using any underlying set theory that you wish.

If this time you can't explicitly demonstrate it, it will be clear that your posts on the issue at hand do not hold water, and in this case I am going to ignore any post of yours that does not handle with Dedekind-infinite, which I claim that it is an essential property of any infinite set, and you, jsfisher, did not explicitly demonstrated otherwise (you did not provide a counter example, which according to it Dedekind-infinite is NOT an essential property of any infinite set).

 

[jsfisher]

Cantor's Theorem can't handle with Dedekind-infinite directly or indirectly.

So you have said. However, you have done nothing to support this bare assertion.

Nothing in Cantor's Theorem nor its proof has any dependence upon Dedekind-finiteness (or, equivalently, Dedekind-infiniteness), yet you proclaim to the contrary. Perhaps you could point to where such a dependence occurs.

 

[doronshadmi]

So you have said. However, you have done nothing to support this bare assertion.

Nothing in Cantor's Theorem nor its proof has any dependence upon Dedekind-finiteness (or, equivalently, Dedekind-infiniteness), yet you proclaim to the contrary. Perhaps you could point to where such a dependence occurs.

jsfisher, all is needed to support my claim is already explicitly given in:

http://www.internationalskeptics.com/forums/showpost.php?p=11563267&postcount=2172

http://www.internationalskeptics.com/forums/showpost.php?p=11564766&postcount=2181

http://www.internationalskeptics.com/forums/showpost.php?p=11565143&postcount=2186

My claim is very simple:

There is no proof or theorem about infinite sets that does not depend on Dedekind-infiniteness because it is an essential property of any infinite set, no matter what underlying set theory is used.

Until this very moment you have done exactly nothing to provide a counter example which explicitly demonstrates a given infinite set that does NOT have this essential property.

 

[jsfisher]

There is no proof or theorem about infinite sets that does not depend on Dedekind-infiniteness because it is an essential property of any infinite set, no matter what underlying set theory is used.

As is your modus operandi, you assert much while proving nothing.

(1) You would need to prove your assertion that all infinite sets are also Dedekind-infinite. (Hint: It cannot be done without a choice axiom. The Axiom of Countable Choice would do.)
(2) You would need to prove your assertion that Cantor's Theorem is invalid. (Hint: Either a counter-example to Cantor's Theorem or a flaw in its proof would be sufficient evidence.)

The two are independent, by the way, since whether a set satisfies the criterion for being a Dedekind-finite (or Dedekind-infinite) set never enters the theorem's proof. The criterion in no way impacts the proof.

 

[doronshadmi]

As is your modus operandi, you assert much while proving nothing.

(1) You would need to prove your assertion that all infinite sets are also Dedekind-infinite. (Hint: It cannot be done without a choice axiom. The Axiom of Countable Choice would do.)
(2) You would need to prove your assertion that Cantor's Theorem is invalid. (Hint: Either a counter-example to Cantor's Theorem or a flaw in its proof would be sufficient evidence.)

The two are independent, by the way, since whether a set satisfies the criterion for being a Dedekind-finite (or Dedekind-infinite) set never enters the theorem's proof. The criterion in no way impacts the proof.

Again let's do it step by step.

Step 1: Please prove that one must use The Axiom Of Choice in order to determine that any given infinite set is Dedekind-infinite.