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Re: ALL PERMUTATIONS OF INFINITY
Posted:
May 22, 2012 5:29 AM


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 NONSEQUITUR 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



