Le 20-nov.-07, à 12:14, Torgny Tholerus a écrit :
>
> Bruno Marchal skrev:
>>
>> To sum up; finite ordinal and finite cardinal coincide. Concerning
>> infinite "number" there are much ordinals than cardinals. In between
>> two different infinite cardinal, there will be an infinity of ordinal.
>> We have already seen that omega, omega+1, ... omega+omega,
>> omega+omega+1, ....3.omega, ... 4.omega .... ....omega.omega .....
>> omega.omega.omega, .....omega^omega ..... are all different ordinals,
>> but all have the same cardinality.
>>
> Was it not an error there? 2^omega is just the number of all subsets
> of
> omega, and the number of all subsets always have bigger cardinality
> than
> the set.
Yes, that is true.
> So omega^omega can not have the same cardinality as omega.
But addition, multiplication, and thus exponentiation are not the same
operation for ordinals and cardinals. I should have written
omega"^"omega, or something like that. That is why I have written
3.omega instead of 3*omega.
We can come back on ordinal later, but now I will focus the attention
on the cardinals, and prove indeed that 2^omega, or 2^N, or
equivalently the infinite cartesian product (of sets)
2X2X2X2X2X2X2X2X... , is NOT enumerable (and indeed vastly bigger that
the ordinal omega"^"omega.
You can look at the thread on the growing functions for a little more
on the ordinals. Actually my point was to remind people of the
difference between ordinal and cardinal, and, yes, they have different
addition, multiplication, etc.
Bruno
http://iridia.ulb.ac.be/~marchal/
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Received on Tue Nov 20 2007 - 09:41:51 PST