Kory Heath skrev:
> On Jun 4, 2009, at 8:27 AM, Torgny Tholerus wrote:
>
>> How do you handle the Russell paradox with the set of all sets that
>> does
>> not contain itself? Does that set contain itself or not?
>>
>> My answer is that that set does not contain itself, because no set can
>> contain itself. So the set of all sets that does not contain
>> itself, is
>> the same as the set of all sets. And that set does not contain
>> itself.
>> This set is a set, but it does not contain itself. It is exactly the
>> same with the natural numbers, BIGGEST+1 is a natural number, but it
>> does not belong to the set of all natural numbers. The set of all
>> sets
>> is a set, but it does not belong to the set of all sets.
>>
>
> So you're saying that the set of all sets doesn't contain all sets.
> How is that any less paradoxical than the Russell paradox you're
> trying to avoid?
>
The secret is the little word "all". To be able to use that word, you
have to define it. You can define it by saying: "By 'all sets' I mean
that set and that set and that set and ...". When you have made that
definition, you are then able to create a new set, the set of all sets.
But you must be carefull with what you do with that set. That set does
not contain itself, because it was not included in your definition of
"all sets".
If you call the set of all sets for A, then you have:
For all x such that x is a set, then x belongs to A.
A is a set.
But it is illegal to substitute A for x, so you can not deduce:
A is a set, then A belongs to A.
This deductuion is illegal, because A is not included in the definition
of "all x".
--
Torgny Tholerus
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Received on Fri Jun 05 2009 - 09:14:35 PDT