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From: quasi on 1 Aug 2010 17:36
On Sun, 1 Aug 2010 13:09:11 -0700 (PDT), David R Tribble
<david(a)tribble.com> wrote:

>David R Tribble writes:
>>> That was my feeling, but is there a specific line of argument
>>> about how we know that? Is there a particular property (or lack
>>> of one) of Cs that renders it a class?
>>
>
>Aatu Koskensilta wrote:
>>> For any x, there's a definable (with x as a parameter) bijection between
>>> Cs(x) and the totality of all sets. By replacement, it follows that
>>> Cs(x) can't be a set. David Hartley already outlined an explicit
>>> definition of a suitable bijection: just take the set y to {x,y}.
>>
>
>Butch Malahide wrote:
>> Maybe you need replacement to prove that Cs(x) is not a set for any
>> possible choice of x. However, unless x is something very exotic, you
>> can get by with just separation, pairing, and union. Namely, if Cs(x)
>> is a set, then consider the set
>> R = {x} union {A in CS(x)| A not in A}.
>> Since the assumption "R not in R" leads to a contradiction, we must
>> have R in R, and in fact R = x; i.e.,
>> x = {x} union {A| x in A and A not in A}.
>
>I expected Russel's paradox to be at play somewhere in
>there. I didn't know where to start, though, not being an
>expert.

As a simple (essentially equivalent to the above, I think) way of
seeing that your class can't be a set, just consider its cardinality,
assuming that it was a set. There are injections of any given set into
your class, so the cardinality of your class would have to be greater
or equal to any given cardinal (in particular, greater or equal to the
cardinality of its own power set).

quasi
From: Butch Malahide on 1 Aug 2010 17:30
On Jul 31, 7:25 pm, Butch Malahide <fred.gal...(a)gmail.com> wrote:
>
> Maybe you need replacement to prove that Cs(x) is not a set for any
> possible choice of x. However, unless x is something very exotic, you
> can get by with just separation, pairing, and union. Namely, if Cs(x)
> is a set, then consider the set
> R = {x} union {A in Cs(x) | A not in A}.
> Since the assumption "R not in R" leads to a contradiction, we must
> have R in R, and in fact R = x; i.e.,
> x = {x} union {A | x in A and A not in A}.

That's silly. You're assuming separation, pairing, and union? Why
don't you just start by observing that the union of Cs(x) is the whole
universe, and then derive the Russell paradox?
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