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Re: ALL PERMUTATIONS OF INFINITY
Posted:
May 22, 2012 5:29 AM
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Graham Cooper <grahamcooper7@gmail.com> writes:
> On May 22, 2:46 am, torb...@diku.dk (Torben Ægidius Mogensen) wrote:
>> For the same reason, the set of all sets of naturals (P(N)
>> or, equivalently, 2^N) has the same size as the set of real numbers (R):
>> There is a bijection between sets of naturals and reals.
>>
>> There is a bijection between permutations of N and subsets of N, so the
>> set of permutations has the same size as the set of subsets.
>>
>> Torben
>
> Thanks for the NON-SEQUITUR of the Century.
>
> So there's a BIJECTION between O(2^n) and O(n!) sized sets now?
Only for infinite n.
> SETSIZE 1 {a} = 1 Permutation
> SETSIZE 2 {a,b} = 2! Permutations
> SETSIZE 3 {a,b,c} = 3! Permutations
> ...
True so far.
> SETSIZE |N| {1,2,3..} = 2^|N| Permutations
But this is a non sequitur. A property that holds for all finite values
does not necessarily hold for infinite values. Or, more geneally, a
property that holds for all elements of a sequence of values does not
necessarily hold for the limit of the sequence -- whether the limit is
finite or not.
> How convenient!
Not at all. It would be so much more convenient if a property that
holds for elements of a sequence would also hold in the limit. But,
alas, we are not so lucky.
Torben
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