Cont: Deeper than primes - Continuation 2

[jsfisher]

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.

Step 0: Stop trying to reverse the burden of proof.

by the way, I did not say the Axiom of Choice was a requirement. Please read what was written, not what you'd like it to say.

 

[doronshadmi]

Step 0: Stop trying to reverse the burden of proof.

by the way, I did not say the Axiom of Choice was a requirement. Please read what was written, not what you'd like it to say.

You wrote:

Hint: It cannot be done without a choice axiom.

Step 1: Please prove that a choice axiom must be used in order to show that a given infinite set is Dedekind-infinite.

 

[jsfisher]

You made the initial claim....your burden.

 

[doronshadmi]

You made the initial claim....your burden.

You made the initial claim that a choice axiom must be used in order to show that a given infinite set is Dedekind-infinite, so:

Step 1: Your initial claim....your burden.

 

[jsfisher]

You made the initial claim that a choice axiom must be used in order to show that a given infinite set is Dedekind-infinite, so:

Step 1: Your initial claim....your burden.

Nice try, but the initial claims regarding Dedekind-finiteness were yours.

 

[doronshadmi]

Nice try, but the initial claims regarding Dedekind-finiteness were yours.

Please show some of my posts where the words "Dedekind-finiteness" or "Dedekind-finite" are used as parts of my arguments, which are explicitly only about infinite sets (so no finitisim of any kind is found in my arguments).

 

[jsfisher]

Please show some of my posts where the words "Dedekind-finiteness" or "Dedekind-finite" are used as parts of my arguments, which are explicitly only about infinite sets (so no finitisim of any kind is found in my arguments).

You do realize that Dedekind-finite and Dedekind-infinite are complements, right? Or do you just want to quibble your way out of supporting your claims?

 

[doronshadmi]

You do realize that Dedekind-finite and Dedekind-infinite are complements, right? ...

It is irrelevant because of the simple fact that only infinite sets are in bijection with their proper subsets, and my argument is explicitly only about infinite sets.

Step 1: Please prove that a choice axiom must be used in order to show that a given infinite set is Dedekind-infinite.

If once again you are going to avoid an explicit answer about infinite sets, it will be clear that you can't support your arguments about the independence of Cantor's theorem of Dedekind-infiniteness.

 

[doronshadmi]

Let's ask about the current issue and hand as follows:

Is there an underlying set theory that enables to prove that there is bijection between an uncountable set and any of its proper subsets?

If the answer is yes, then please write such proof.

If the answer is no, then please explain why it can't be done.

Thank you.

 

[Little 10 Toes]

Why do you need an answer?

 

[doronshadmi]

Let's do it clearer:

Is there an underlying set theory that enables to provide a general proof that there is bijection between an uncountable set and all of its uncountable proper subsets?

The question is about a given uncountable set, where all of its uncountable proper subsets are taken separately (they are not taken as one collection).

If the answer is yes, then please write such proof.

If the answer is no, then please explain why it can't be done.

If the answer is unknown or undefined, then please explain why it is unknown or undefined?

Thank you.

 

[Little 10 Toes]

Again, why do you need an answer.

To be clearer, what are you trying to prove claim?

 

[doronshadmi]

My question is still unclear so let's improve it:

Let S be a countably infinite set and let P(S) be its uncountable power set.

Is there an underlying set theory which proves that all the uncountable proper subsets of P(S) are in bijection with P(S)?

If the answer is yes, then please write such proof.

If the answer is no, then please explain what it can't be done?

Thank you.

 

[ctamblyn]

Just a quick fly-by post: doronshadmi, your question seems equivalent to "is the continuum hypothesis true", which cannot be proved or disproved in ZF (with or without the axiom of choice). So the answer is "yes", "no", or "don't know" depending on what set theory you use.

*recloaks*

 

[phiwum]

Delete.

 

[doronshadmi]

Just a quick fly-by post: doronshadmi, your question seems equivalent to "is the continuum hypothesis true", which cannot be proved or disproved in ZF (with or without the axiom of choice). So the answer is "yes", "no", or "don't know" depending on what set theory you use.

*recloaks*

Thank you ctamblyn, your answer is helpful.

Given P(S) such that S is countably infinite set, can we prove (by some set theory) that there are uncountable proper subsets of P(S) that are in bijection with P(S)?

 

[ctamblyn]

Yes, every infinite set (of any cardinality) can be put into 1-1 correspondence with at least one of its proper subsets.

 

[doronshadmi]

Yes, every infinite set (of any cardinality) can be put into 1-1 correspondence with at least one of its proper subsets.

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)?

 

[jsfisher]

Yes, every infinite set (of any cardinality) can be put into 1-1 correspondence with at least one of its proper subsets.

While true in ZFC, this is not true in Z and ZF.

 

[ctamblyn]

Thanks for the clarification, 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?).