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Replies: 1   Last Post: May 21, 2012 12:56 PM

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Torben Mogensen

Posts: 60
Registered: 12/6/04
Posted: May 21, 2012 12:56 PM
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Graham Cooper <> writes:

> On May 21, 10:45 am, "K_h" <> wrote:
>> "Graham Cooper"  wrote in message

>> > How many different ways can you order <1,2,3,4....> ?
>> 2^aleph_0 ways, assuming the axiom of choice and defining the factorial of
>> the cardinality of a set S to be equal to the cardinality of the set of all
>> bijections of S to itself.

> The presented algorithm can list all computable permutations of N.
> All 1X2X3X4X5X6X7... of them!
> The set of binary string 2^oo=2X2X2X2X2...
> is much smaller.

You can not assume that a property that holds for finite numbers also
holds for infinite numbers. For example, n+1 > n for all (finite)
naturals, but oo+1 = oo (which I'm sure you agree).

So you can not conclude that n! > 2^n for naturals n>2 implies that c! >
2^c for infinite cardinals c. In fact, for infinite cardinals c, c! =
2^c (if c! is defined as the number of permutations of c and 2^c is
defined as the number of subsets of c).


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