Cont: Deeper than primes - Continuation 2

[ctamblyn]

I disagree. In my argument S is at least an infinite set, which means that it can be in bijection with at least |S| of its proper subsets.

Choose whatever set S you like, I don't mind.

Now choose any function you like which maps elements of your set S to elements of P(S), and call it f.

Now construct the set T_f = { all x in S such that x is not contained in f(x) }.

Now answer the question, yes/no: do you agree that there is no y in S such that T_f = f(y)?

 

[jsfisher]

I disagree. In my argument ... S is at least an infinite set

If the more general statement is true, there is no reason to consider the special case.

 

[jsfisher]

Wrong jsfisher, {} is just an example of a member of set P(S) that is claimed by Cantor's theorem not to be in the range of all S members

Ok, then show me where I am wrong. Where exactly in Cantor's Theorem (and you really mean its proof) does this thing you claim happen? Feel free to use the version of the proof ctamblyn presented, or the one I prefer that begins "Assume there exists a function, g, that is a bijection from S to P(S)", or any other complete proof you like; just make an explicit connection from your claim to the proof.

 

[doronshadmi]

Now answer the question, yes/no: do you agree that there is no y in S such that T_f = f(y)?

Now answer the question, yes/no: do you agree that there is no y in S such that T_f = f(y), simply because T_f is not a member of any proper subset of set P(S) that is in bijection with set P(S)? (the needed details are already given in http://www.internationalskeptics.com/forums/showpost.php?p=11584177&postcount=2251).

 

[doronshadmi]

Ok, then show me where I am wrong. Where exactly in Cantor's Theorem (and you really mean its proof) does this thing you claim happen? Feel free to use the version of the proof ctamblyn presented, or the one I prefer that begins "Assume there exists a function, g, that is a bijection from S to P(S)", or any other complete proof you like; just make an explicit connection from your claim to the proof.

All is needed is already given in http://www.internationalskeptics.com/forums/showpost.php?p=11584469&postcount=2254.

 

[jsfisher]

Now answer the question, yes/no: do you agree that there is no y in S such that T_f = f(y), simply because T_f is not a member of any proper subset of set P(S) that is in bijection with set P(S)?

Of course there is no bijection between T_f and P(S), but that is an irrelevant fact as to why there does not exist a y such that T_f = f.

 

[Little 10 Toes]

{} is the set that is defined by Cantor's theorem, simply because the defined subset of P(S) (that Cantor's theorem claims that it is not in the range of all S members) includes exactly all the S members that do not have their image in the P(S) members that they are mapped with.

Please provide proof of your claim.

 

[doronshadmi]

Of course there is no bijection between T_f and P(S), but that is an irrelevant fact as to why there does not exist a y such that T_f = f.

T_f is constructed as follows:

It includes exactly all the S members that do not have their image in the P(S) members that they are mapped with.

Since my argument is only about infinite sets, and some T_f can be also finite sets, then any considered mapping between T_f and P(S) is irrelevant to my argument.

 

[jsfisher]

T_f is constructed as follows:

It includes exactly all the S members that do not have their image in the P(S) members that they are mapped with.

Their image? No. T_f is the set of all x in S that are not members of f(x).

Since my argument is only about infinite sets, and some T_f can be also finite sets, then any considered mapping between T_f and P(S) is irrelevant to my argument.

You are the only one bringing up mappings between T_f and P(S). I have no idea why.

The two characteristics of T_f that are important are (1) T_f is a member of P(S), and (2) for no x in S does f(x) = T_f.

 

[doronshadmi]

The two characteristics of T_f that are important are (1) T_f is a member of P(S), and (2) for no x in S does f(x) = T_f.

There is one important thing:

For at least |P(S)| proper subsets of P(S) that are in bijection with P(S) (because of Dedekind-infinite property among infinite sets), T_f is not their member.

Therefore Cantor's theorem is insufficient in order to prove that |S| < |P(S)|

 

[Little 10 Toes]

Doron, are you complaing about the empty set? Why havent you answered my previous post?

Edit : Would you that the following definitions are definitions that you use?

[Wp]power set[/wp]: In mathematics, the power set (or powerset) of any set S is the set of all subsets of S, including the empty set and S itself.

Proper subset: A proper subset of a set A is a subset of A that is not equal to A. In other words, if B is a proper subset of A, then all elements of B are in A but A contains at least one element that is not in B. (http://mathinsight.org/definition/proper_subset)

 

[jsfisher]

There is one important thing:

For at least |P(S)| proper subsets of P(S) that are in bijection with P(S) (because of Dedekind-infinite property among infinite sets), T_f is not their member.

Therefore Cantor's theorem is insufficient in order to prove that |S| < |P(S)|

You have yet to link your conclusion to your bare assertion.

All you need to is show which step in the proof of Cantor's Theorem falls apart because Dedekind gave a name to a certain type of set.

 

[doronshadmi]

You have yet to link your conclusion to your bare assertion.

All you need to is show which step in the proof of Cantor's Theorem falls apart because Dedekind gave a name to a certain type of set.

It falls apart even before it takes any step exactly because in case that S is an infinite set, P(S) has at least |P(S)| proper subsets that are bijective with P(S) (exactly because of Dedekind-infinite property) where no one of them includes T_f as its member, in the first place.

Because of this simple mathematical fact Cantor's theorem is insufficient in order to prove that |S| < |P(S)|, in case that S is an infinite set.

 

[jsfisher]

It falls apart even before it takes any step exactly because in case that S is an infinite set, P(S) has at least |P(S)| proper subsets that are bijective with P(S) (exactly because of Dedekind-infinite property) where no one of them includes T_f as its member, in the first place.

You still trot out that bear assertion. Be that as it may, since all these proper subsets of P(S) play no role in Cantor's Theorem, how does your little factoid, above, relate to Cantor's Theorem?

For any f: S -> P(S), T_f is a member of P(S), and that is sufficient to show that there is no bijection from S to P(S). The subsets are not relevant to the proof.

 

[Little 10 Toes]

So doron has problems with the empty set.

 

[doronshadmi]

For any f: S -> P(S), T_f is a member of P(S), and that is sufficient to show that there is no bijection from S to P(S). The subsets are not relevant to the proof.

For any f: S -> P(S), T_f is not a member of at least |P(S)| proper subsets that are bijective with P(S), and that mathematical fact is sufficient to show that T_f can't be used in order to conclude that there is no bijection from S to P(S).

 

[jsfisher]

For any f: S -> P(S), T_f is not a member of at least |P(S)| proper subsets that are bijective with P(S)

Prove it.

and that mathematical fact is sufficient to show that T_f can't be used in order to conclude that there is no bijection from S to P(S).

You have provided no linkage from these subsets that concern you so and Cantor. Until you do, it is just an irrelevant (and unproven) factiod.

 

[doronshadmi]

Prove it.

http://math.stackexchange.com/quest...f-set-s-for-which-there-is-a-bijection-with-s

You have provided no linkage from these subsets that concern you so and Cantor. Until you do, it is just an irrelevant (and unproven) factiod.

Since Cantor's theorem is involved also with infinite sets, it must deal with Dedekind-infinite property, which according to it at least |P(S)| proper subsets of P(S) (in case that S is an infinite set) that are bijective with P(S), do not include T_f as their member.

This fundamental mathematical fact can't be avoided by any attempt to prove something about infinite sets, and Cantor's theorem certainty avoids this fundamental mathematical fact, and therefore it is insufficient in order to prove that |S| < |P(S)|, in case that S is an infinite set.

 

[Little 10 Toes]

Obvious rebuttal: nowhere in that post is power set (or powerset) mentioned.

Edit: i am refering to the stack exchange post. Please try again.

 

[doronshadmi]

Obvious rebuttal: nowhere in that post is power set (or powerset) mentioned.

Edit: i am refering to the stack exchange post. Please try again.

Currently any mathematician "does its best" (as clearly seen in http://math.stackexchange.com/quest...f-set-s-for-which-there-is-a-bijection-with-s or in the last comments with my dialog with Noah Schweber in http://math.stackexchange.com/quest...set-and-all-of-its-uncountable-proper-subsets) in order to avoid the mathematical fact that given any infinite set, it has infinitely many proper subsets that are bijective with it, where the number of these proper subsets is at least the cardinality of the considered infinite set.

As for P(S) (where S is an infinite set), the proof is seen in http://www.internationalskeptics.com/forums/showpost.php?p=11588290&postcount=2281.