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Cont: Deeper than primes - Continuation 2

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jsfisher said:
Prove it.
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It is trivial, given any Tf (which is a member of P(S)), it is not a member of any proper subset of P(S) that is bijective with P(S).

Since there are at least |P(S)| such Tf members, there can't be but at least |P(S)| proper subsets of P(S) that are bijective with P(S).
 
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doronshadmi said:
http://math.stackexchange.com/quest...f-set-s-for-which-there-is-a-bijection-with-s
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As already observed, the stackexchange post doesn't address your claim. Moreover, your respondent over there has made an unjust assumption. Can you figure out what it was?

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 Tf as their member.
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The "Dedekind-infinite property", as you call it, is simply a term for sets that have a equi-numerous proper subset. If a set is Dedekind-infinite, it has such a subset; it may have more, but it is not guaranteed. Not all infinite sets are Dedekind-infinite.

Be that as it may, Cantor's Theorem is about |S| < |P(S)|. It is does not, nor does it need to interest itself in any of the proper subsets of P(S). They are irrelevant.

Cantor's Theorem "deals", as you like to say, with all finite and all infinite sets. Dedekind-infinite sets happen to be included in that, so, yeah, Cantor's Theorem "deals" with them, too.
 
doronshadmi said:
It is trivial, given any Tf (which is a member of P(S)), it is not a member of any proper subset of P(S) that is bijective with P(S).
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That is easy to disprove.

Let S be the set of natural numbers and f(x) = {x} for all x in S.
Then Tf = {}.

Let U = P(S) \ {0} .

Tf is a member of U and there is a bijection from U to P(S).
 
doronshadmi said:
Please read again the last comments of Hans Hüttel in http://math.stackexchange.com/quest...f-set-s-for-which-there-is-a-bijection-with-s
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I am a participant in this thread as a member, and so the following statements are made as a member, not as a moderator:

It is generally frowned upon to cross-post exchanges between two forums. Stackexchange is a forum; the International Skeptics Forum is a forum. It would be improper for you, Doronshadmi, to proxy a discussion between the two.

If Hans Hüttel were to join the ISF, folks here would be more than happy to discuss with him topics of mutual interest (including the validity of Cantor's Theorem).

Finally, since you directed us to a specific question at stackexchange, it is fair to ask: Are you and "doromshadmi" at stackexchange one in the same?
 
jsfisher said:
That is easy to disprove.

Let S be the set of natural numbers and f(x) = {x} for all x in S.
Then Tf = {}.

Let U = P(S) \ {0} .

Tf is a member of U and there is a bijection from U to P(S).
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Nothing is disproved simply because it is irrelevant to http://www.internationalskeptics.com/forums/showpost.php?p=11588290&postcount=2281, because all we care is about proper subsets of P(S) that are bijective with P(S), and since there are at least |P(S)| members where each one of them is not a member of at least one proper subset of P(S) that is bijective with P(S), there can't be but at least |P(S)| proper subsets of P(S) that are bijective with P(S) (where, again, my argument is only about S as any infinite set).
 
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jsfisher said:
Finally, since you directed us to a specific question at stackexchange, it is fair to ask: Are you and "doromshadmi" at stackexchange one in the same?
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Yes, here and in stackexchange it is me in spite of my typo mistake as I wrote it in stackexchange.

doromshadmi and doronshadmi is me.
 
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jsfisher said:
The "Dedekind-infinite property", as you call it, is simply a term for sets that have a equi-numerous proper subset. If a set is Dedekind-infinite, it has such a subset; it may have more, but it is not guaranteed.
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The mathematical fact (in case that S is any infinite set) that there are at least |P(S)| proper subsets of P(S) that are bijective with P(S), is enough in order to show that Cantor's theorem is insufficient in order to prove that |S| < |P(S)|.
 
doronshadmi said:
The mathematical fact (in case that S is any infinite set) that there are at least |P(S)| proper subsets of P(S) that are bijective with P(S), is enough in order to show that Cantor's theorem is insufficient in order to prove that |S| < |P(S)|.
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(1) It is not a mathematical fact. There are infinite sets that are Dedekind-finite.
(2) Even if it were a mathematical fact, nothing about Cantor's Theorem is impacted by it.
 
jsfisher said:
(1) It is not a mathematical fact. There are infinite sets that are Dedekind-finite.
(2) Even if it were a mathematical fact, nothing about Cantor's Theorem is impacted by it.
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The claim that Cantor's diagonal argument and Cantor's theorem prove anything (in case that S is any infinite set) is equivalent to the claim that the only way to cross a given yard from one side to the other side, must be done only through a given door that stands in the middle of this yard.

This claim actually avoids the fact that one actually can ignore the door in order to cross the yard from one side to the other side, where this ability is equivalent to the fact that given any P(S) member that is claimed not to be in the range of all S members, it simply can be ignored since it is not a member of at least |P(S)| proper subsets of P(S) that are bijective with P(S).

In other words jsfisher, Cantor's diagonal argument and Cantor's theorem are naïve about infinite sets.
 
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Until you can provide some connection between Cantor's Theorem and these proper subsets, there is nothing to consider.

Cantor remains unharmed by your latest attempt to discredit.
 
jsfisher said:
Until you can provide some connection between Cantor's Theorem and these proper subsets, there is nothing to consider.
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Already done in http://www.internationalskeptics.com/forums/showpost.php?p=11588215&postcount=2278 (including my detailed comments in the link there, about the issue at hand).

Moreover, it is very simply done also in http://www.internationalskeptics.com/forums/showpost.php?p=11590213&postcount=2291, but by your reply it is clear that http://www.internationalskeptics.com/forums/showpost.php?p=11590213&postcount=2291 is beyond the "glass ceiling" of your mathematical\logical reasoning.
 
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By the way, Doronshadmi, are you assuming there's a bijection between S and any of these proper subsets of P(S) you are constructing?

You have said repeatedly that for f: S -> P(S) where f(x) = {x}, the empty set, {}, is the missing member. Do you assume that f: S -> P(S) \ {} is a bijection?
 
jsfisher said:
By the way, Doronshadmi, are you assuming there's a bijection between S and any of these proper subsets of P(S) you are constructing?

You have said repeatedly that for f: S -> P(S) where f(x) = {x}, the empty set, {}, is the missing member. Do you assume that f: S -> P(S) \ {} is a bijection?
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What I show is that Cantor's theorem is insufficient in order to prove (in case that S is any infinite set) that |S| < |P(S)|, exactly because for each given P(S) member that is claimed to be out of the range of all S members, it is omitted from P(S) and we get two results:

1) The contradiction that is used in order to conclude that there is a given P(S) member that is claimed to be out of the range of all S members, is not satisfied.

2) The omitted P(S) member does not prevent the bijection between P(S) and a proper subset of P(S) that one of the P(S) members is not included in it.

Generally there are at least such |P(S)| proper subsets of P(S) that are bijective with P(S), and this mathematical fact prevents the conclusion that |S| < |P(S)| (in case that S is any infinite set).

In other words, Cantor's work on the issue at hand is no more than a conjecture.
 
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jsfisher said:
You didn't answer my question. Instead, you continue to speak of "the omitted P(S) member."
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"the omitted P(S) member" shows that Cantor's work on the issue at end in no more than a conjecture.

As you know very well assuming X is not the same as proving X so your question is mathematically meaningless.
 
Dear jsfisher, in http://www.internationalskeptics.com/forums/showpost.php?p=11588514&postcount=2285 you had no problem to distinguish between "doromshadmi" and "doronshadmi".

Why do you follow after Cantor's mathematical notions about the issue at hand, which clearly do not distinguish between infinite sets and finite sets (more precisely, Cantor's mathematical notions ignore the mathematical fact that given any non-empty set, if it is bijective with at least one of its proper subsets, then it is an infinite set, otherwise it is a finite set)?

More details about the issue at hand can be found in the dialog (more than 50 comments) between Hans Hüttel and me in http://math.stackexchange.com/quest...f-set-s-for-which-there-is-a-bijection-with-s .
 
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