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Re: ALL PERMUTATIONS OF INFINITY
Posted:
May 23, 2012 5:58 AM
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Graham Cooper <grahamcooper7@gmail.com> writes:
> On May 22, 7:53 pm, torb...@diku.dk (Torben gidius Mogensen) wrote:
>> Graham Cooper <grahamcoop...@gmail.com> writes:
>> > You must have ASSUMED the amount of subsets of N is 2X2X2... = |R|
>>
>> Not at all. This is, in fact, fairly easy to prove: We use the theorem
>> that there exists a bijection between two sets A and B if there exists
>> an injection from A to B and an injection from B to A. So all we need
>> to do to prove that |P(N)| = |R| is to find injections both ways.
>>
>> [...]
>>
>> Since we have injections both ways, the sets are equipotent (i.e.,
>> having the same size).
>
> Yes Yes I know all that, you threw a red herring that there is a
> bijection between permutations and powersets,
Whay is that a red herring?
> That's like saying there's a bijection between {a,b,c} and {a,b,c,d,e}
There isn't. Do you even know what a bijection is?
Torben
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