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Replies: 4   Last Post: May 24, 2012 4:55 AM

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Torben Mogensen

Posts: 60
Registered: 12/6/04
Posted: May 23, 2012 5:58 AM
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Graham Cooper <> writes:

> On May 22, 7:53 pm, (Torben gidius Mogensen) wrote:
>> Graham Cooper <> 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?


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