Cont: Deeper than primes - Continuation 2

[ctamblyn]

By using the term "at least" do you mean (in case that S is countably infinite) that there can be uncountable number of uncountable proper subsets of P(S) that are in bijection with P(S)?

I only mean that there is at least one.

 

[doronshadmi]

I only mean that there is at least one.

Do you agree that "at least one" also means that there are "more than one"?

 

[jsfisher]

Thanks for the clarification, jsfisher.

This relates to Doronshadmi's starting point for the current thread arc. It seems he's stumbled across the concept of Dedekind-infinite, and without understanding it he has attempted to discredit Cantor's Theorem.

At any rate, in set theories with an appropriate choice axiom (not necessarily as strong as the Axiom of Choice), then infinite and Dedekind-infinite are equivalent. Without it, there can be infinite sets that are Dedekind-finite.

 

[doronshadmi]

Without it, there can be infinite sets that are Dedekind-finite.

Can you please show an infinite set S of natural numbers that is not Dadekind-infinite (if it is not Dadekind-infinite it means that there is no bijection between S and all of its infinite proper subsets)?

 

[ctamblyn]

Do you agree that "at least one" also means that there are "more than one"?

No, but it doesn't rule it out.

 

[ctamblyn]

This relates to Doronshadmi's starting point for the current thread arc. It seems he's stumbled across the concept of Dedekind-infinite, and without understanding it he has attempted to discredit Cantor's Theorem.

At any rate, in set theories with an appropriate choice axiom (not necessarily as strong as the Axiom of Choice), then infinite and Dedekind-infinite are equivalent. Without it, there can be infinite sets that are Dedekind-finite.

Right, gotcha. Fields Medal Committee on standby...

 

[jsfisher]

For the particular set under consideration - essentially, the power set of the integers - I believe it is true even in ZF (or am I mistaken again?).

Whether it is the set of integers or just some generic infinite set S, Doron has been unclear. But, you are correct, if S is the set of integers (or any well-ordered set), then S infinite and S Dedekind-infinite are equivalent.

 

[doronshadmi]

No, but it doesn't rule it out.

So is it logically possible that there are more than one?

 

[ctamblyn]

Saying "at least one" means that there cannot be zero. There might be exactly one. There might be more than one. By saying "at least one", I'm not making any commitment one way or the other.

 

[ctamblyn]

In the specific case of the integers Z, the power set P(Z) has the cardinality of the continuum, c. And there are indeed infinitely many subsets of P(Z) that also have cardinality c. The same applies if Z is replaced with any countably infinite set.

 

[ctamblyn]

It look like there are several slightly different questions in flight at the same time here.

 

[doronshadmi]

It look like there are several slightly different questions in flight at the same time here.

Let's be focused on this question.

Do you agree that given any infinite set Z, there are |Z| proper subsets of Z that are in bijection with Z?

 

[ctamblyn]

Let's be focused on this question.

Do you agree that given any infinite set Z, there are |Z| proper subsets of Z that are in bijection with Z?

No, that is false. For example, see Asaf's answer here:

http://math.stackexchange.com/quest...-mathbbn-have-the-same-cardinality-as-mathbbn

 

[doronshadmi]

No, that is false. For example, see Asaf's answer here:

http://math.stackexchange.com/quest...-mathbbn-have-the-same-cardinality-as-mathbbn

Dear ctamblyn, let's change the question a little in order to clarify it, by using your at least.

Do you agree that given any infinite set Z, there are at least |Z| proper subsets of Z that are in bijection with Z?

 

[jsfisher]

Dear ctamblyn, let's change the question a little in order to clarify it, by using your at least.

Do you agree that given any infinite set Z, there are at least |Z| proper subsets of Z that are in bijection with Z?

Under what set theory?

 

[jsfisher]

Under what set theory?

How about we move things along and avoid the whole Dedekind-finite vs infinite issue by stipulating ZFC? If so, Doronshadmi gets the number of equi-numerous subsets he seems to be seeking, and then maybe we can move to the core matter he has been circling.

 

[doronshadmi]

Please look at this part, taken from http://math.stackexchange.com/quest...et-and-all-of-its-uncountable-proper-subsets# :

Edited by Locknar: 

SNIPed, breach of rule 4.

Here is my question again, by using also the term at least:

There are at least |P(S)| uncountable proper subsets of P(S) that are in bijection with P(S).

Can we use this mathematical fact in order to conclude (in case of Cantor's theorem about |S| < |P(S)|) that the P(S) members that are used in his proof in order to provide the needed contradiction, are actually not members in case of each one of the (at least) |P(S)| uncountable proper subsets of P(S) that are in bijection with P(S), in the first place?

 

[Little 10 Toes]

So you still have a problem with Cantor and his theorem?

 

[doronshadmi]

Here is the further dialog done between Noah and me in http://math.stackexchange.com/quest...set-and-all-of-its-uncountable-proper-subsets :

Edited by Locknar: 

SNIPed, breach of rule 4.

As clearly be seen, Noah currently avoids any further dialog without any concrete reason.

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

Let's clarify my argument by providing a concrete example, which is without loss of generality.

In case of, for example:

1 --> {1}
2 --> {2}
3 --> {3}
4 --> {4}
5 --> {5}
...

the given P(S) member that is not in f range is {}.

Since there are at least |P(S)| proper subsets of P(S) that are in bijection with P(S), it is clear that one of those proper subsets that are in bijection with P(S), does not have {} as its member in the first place, so the contradiction is not shown in the first place , and it is not shown over at least |P(S)| proper subsets of P(S) that are in bijection with P(S).

In other words, Cantor's theorem is insufficient in order to prove that |S| < |P(S)| among infinite sets.

 

[jsfisher]

Let's clarify my argument by providing a concrete example, which is without loss of generality.

In case of, for example:

1 --> {1}
2 --> {2}
3 --> {3}
4 --> {4}
5 --> {5}
...

the given P(S) member that is not in f range is {}.

Are you trying to say that f: S -> P(S) where f(a) = {a}? You could save us all a lot of trouble trying to decipher your posts if you just used, you know, math to express your math.

Ok, what of it? It is an injection, but not a surjection, from S to P(S). It provides an example of something that isn't a bijection between S and P(S).

How does this triviality relate to Cantor's Theorem? It certainly has nothing to do with the proof of the theorem since the proof does not consider any mapping from S to P(S) that isn't surjective.