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From: Aatu Koskensilta on 31 Jul 2010 07:02
David R Tribble <david(a)tribble.com> 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?

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}.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 31 Jul 2010 07:12
David R Tribble <david(a)tribble.com> writes:

> I meant a universe of sets of a finite size, e.g., the universe of
> sets having no more than three members. Perhaps this is a specific
> kind of "model"?

I take it you have in mind something like the least fixed point S_n of
the operation

P_n(X) = X union {A subset X | A is of cardinality <= n}

for a given n. This is just the familiar cumulative hierarchy with sets
of cardinality > n dropped -- we obtain S_n as the limit of the sequence

S_n(0) = the empty set
S_n(i+1) = P_n(S_n(i)) = S_n(i) union {A subset S_n(i) | card(A) <= n}

by setting S_n = union of S_n(i) for i in omega.

There are infinitely many sets in S_n for n > 0, since {} is in S_n and
{{}}, {{{}}}, and so on are all one element sets. In S_1 there is, for
any set A in S_1, only one set having A as an element, namely {A}, so
the set having this set as its element is the collection of all sets (in
S_1) having A as an element, and is in S_1. For n > 1 the collection of
all sets in S_n having a given set A in S_n as an element is infinite,
since there are infinitely many sets in S_n, and for any set X in S_n,
{A,X} is a set in S_n having A as an element. So the set of all sets
containing A as an element, not being of cardinality <= n, is not in S_n
when n > 1.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Butch Malahide on 31 Jul 2010 20:22
On Jul 31, 6:02 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote:
> David R Tribble <da...(a)tribble.com> 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?
>
> 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}.

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}.
From: Butch Malahide on 31 Jul 2010 20:25
On Jul 31, 7:22 pm, Butch Malahide <fred.gal...(a)gmail.com> wrote:
> On Jul 31, 6:02 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote:
>
> > David R Tribble <da...(a)tribble.com> 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?
>
> > 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}.

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}.

From: David R Tribble on 1 Aug 2010 16:09
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.
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