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Cantor’s construction method and P(S) members
 Posted: 09 April 2011 08:09 AM [ Ignore ]
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http://www.mathacademy.com/pr/prime/articles/cantor_theorem/index.asp is a clear example of Cantor’s theorem as a proof by contradiction, which leads to contradiction if one tries to define mapping between an explicit P(S) member and S member, because of the construction rules of the explicit P(S) member (the member of S must be AND can’t be a member of the explicit P(S) member, according to the construction rules of the explicit P(S) member, under Cantor’s theorem).

Also since |P(S)| is not less than |S| (because it is trivial to show that all S members (for example {a,b,c,d,...}) are at least mapped with {{a},{b},{c},{d},...} P(S) members), then by using this fact and the contradiction shown above, one must conclude that P(S) is a larger set than S.

———————————————————————

But this is not the only way to look at this case, for example, we are using Cantor’s construction method to systematically and explicitly define P(S) members, for example:

By using the trivial mapping between {a,b,c,d,...} S members and {{a},{b},{c},{d},...} P(S) members, we explicitly define P(S) member {}.

Also by using the mapping between {a,b,c,d,...} S members and {{},{a},{b},{c},...} P(S) members, we explicitly define P(S) member {a,b,c,d,...}.

Actually by using Cantor’s construction method independently of Cantor’s theorem, we are able to explicitly define the all P(S) members between {} and {a,b,c,d,...}.

As a result, there is a bijection between S and P(S) members, as follows:

a ↔ {}
b ↔ {a,b,c,d,...}
c ↔ some explicit P(S) member, which is different than the previous mapped P(S) members

...

Please be aware of the fact that this construction method has nothing to do with Cantor’s theorem exactly because the construction is used independently of Cantor’s theorem and therefore it is not restricted to the logical terms of Cantor’s theorem.

In other words, the distinction between countable and uncountable infinite sets has no basis, according to this non-standard reasoning.

Actually Dedekind’s definition for infinite sets (which according to it there is a bijection between any given infinite set and some proper subset of it) is true also in the case of the mapping between the members of P(S) and the members of S, where S is indeed a proper subset of P(S).

The bijection between the members of P(S) and the members of S is equivalent to the bijection between the members of [B]N[/B] and the members of even numbers, or the members of [B]Q[/B] and the members of [B]N[/B], etc ...

[ Edited: 09 April 2011 11:35 AM by DoronShadmi ]
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 Posted: 09 April 2011 08:52 AM [ Ignore ]   [ # 1 ]
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Seems you just add the missing element to your set and repeat that endlessly. Or don’t I understand your mapping? What is the difference between your idea and a full Hilbert Hotel, with another countable infinite set of guests, which can be placed there without any problem?

PS I am afraid kwann will soon be here…

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“The light is on, but there is nobody at home”

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 Posted: 09 April 2011 10:45 AM [ Ignore ]   [ # 2 ]
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GdB - 09 April 2011 08:52 AM

Seems you just add the missing element to your set and repeat that endlessly. Or don’t I understand your mapping? What is the difference between your idea and a full Hilbert Hotel, with another countable infinite set of guests, which can be placed there without any problem?

PS I am afraid kwann will soon be here…

I show an explicit construction method of all P(S) members, without missing any one of them, which enables to define a function of one-to-one and onto between all P(S) members and all S members (where S is a proper subset of P(S)) exactly as shown, for example, between N members and the members of the set of even numbers (which is a proper subset of N), or between the members of set Q and set N (where N is a proper subset of set Q).

In general, I show the consistency of Dedekind’s definition of infinite sets on all infinite sets, without exceptional.

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 Posted: 09 April 2011 11:01 AM [ Ignore ]   [ # 3 ]
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Sounds great

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 Posted: 09 April 2011 12:30 PM [ Ignore ]   [ # 4 ]
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GdB - 09 April 2011 11:01 AM

Sounds great

Thank you.

Like any new view of some already agreed reasoning, it must be carefully re-examined.

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 Posted: 09 April 2011 03:20 PM [ Ignore ]   [ # 5 ]
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Posters here might wish to review this epic train wreck (now approaching 15,000 posts with no end in sight).

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 Posted: 10 April 2011 12:46 AM [ Ignore ]   [ # 6 ]
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the PC apeman - 09 April 2011 03:20 PM

Posters here might wish to review this epic train wreck (now approaching 15,000 posts with no end in sight).

Lot of work to do in the free will threads…. Wow 375 pages!