Re: Why Regularity?


Rupert wrote:
zuhair wrote:
Rupert wrote:
Jesse F. Hughes wrote:
"Rupert" <rupertmccallum@xxxxxxxxx> writes:

That implies that the set {m|mex or m=x'} exists, and since every
set is a member of itself it will be either x or x'.

Sorry, I don't follow.

Suppose that mex. That does not imply that m=x, does it? Every set
is self-membered, but not every set satisfies z={z}.

How does our resident philosopher define x'? I might have missed
that.


It was zuhair who confused me. First he said the successor of U was
{U,U'}, which as I proved can't exist, then elsewhere he said the
successor of x was {m|mex or m=x'}.

Suppose {m|mex or m=x'} exists. It's a member of itself, and it can't
be x. So it must be in x. So in any model of zuhair's axioms there are
infinite descending membership chains.

right, but all of these chains would make any difference. the axiom of
extentiality man,
Let me tell you what I mean.

for example 0={0}={0,{0}} = {0,{0},{0,{0}} } , etc... all of those are
nothing but zer0.

Now 1={0,0',1} were 0' is the complementary set of 0.

Now you can say that {0,0',1} = {0,0',1,{0,0',1}} =
{0,0',1{0,0',1},{0,0',1,{0,0',1}} } =.....

All of these cyclations should not bother you since by extensiality we
can reduce all of them
to {0,0',1}.

2= {0,0',1',2}

3= { 0,0',1',2',3}


You seem to keep changing your mind about how you define successor. I
thought the successor of x was {m|mex or m=x'}. That means 1=0 union
{0'}. It won't work if you start with 0 as the unique singleton.

So how is successor defined?

y=S(x) <-> y:(Amex -> mey) /\ x'ey

By axiom of Pairing if x is a set, y is a set, so is {x,y}
so {x,y} is the paired set of x and y.

Let me use this definition.

Axiom of infinity:

EN:QeN /\ ( Ax:xeN -> (y:(Amex -> mey) /\ x'ey) eN )

so the essence of this axiom is:

EN:QeN/\(Ax:xeN -> S(x) e N ).

Q which is the unique singlton set in this theory, which I used to
symbolize it as U.

For simplicity I will symbolize it as 0.

Now the ordinals are as fellow:

0 = {0}
1= {0,0',1}
2= {0,0',1,1',2}
3= {0,0',1,1',2,2',3}
..
..
..
..
i={0,0',1,1',2,2',3,3',........,(i-1),(i-1)',i}
..
..
..
w={0,0',1,1',2,2',3,3',........,w}
w+1={0,0',1,1',2,2',3,3',........,w,w',w+1}
..
..
..
..
N = the set of all ordinals.

The real problem with this method is ordinal 1 which is defined as
{0,0',1}.

Your objection was that number 1 defined in such a manner would be the
set of all sets.

Now according to this 1' would be 0( according to my new definition of
complementary set ).

and we will never move greater than 1, if your argument is true, then
all the ordinals

above 1 are nothing but different ways of expresing zero and 1.

My answer to this is that 1 is not the union of 0 and 0'.

{0,0',1} =/= Union {0,0',1}

There is a difference between axiom of pairing and axiom of union.

You may say that in this set theory that is not the case.

You can say that {0,0',1} is {0} U {0'} = {0,0',1} is the same as 0 U
0' = the set of all sets.

This is not correct because 0' is not singlton.

0' ={0,x,d,r,.......,0' } (by the way according to my new definition of
complementary set 0' is the set of all sets, since it contains 0 in it
).

Now 0' though in itself, yet it is not the only member in itself.
That's why {0} U {0' } is not the same as 0 U 0', because x U y = { m|
mex v mey }, and so 0 U 0'=0'.
But {0,0',1} = {m|mex v m=0' }=/=0', in fact {0,0',1} e0' , see the
difference.

Therefore your previous objection is not correct, it was build on the
confusion between the paired set, and the set union of a set.

Remined you of the recent axiomatization that I have already presented
elsewhere in this thread, it uses unrestricted comprehension wihtout
leading to Russell's paradox.

Zuhair

.

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