Re: Small Set Theory,Updated.
- From: "zuhair" <zaljohar@xxxxxxxxx>
- Date: 16 Jan 2007 13:40:13 -0800
hagman wrote:
zuhair schrieb:
Hi All,
Small Set Theory:
Primitives:
e
=
Definitions:
Def.1) x is P_defined<->Ay(yex<->(P[y] & ~y=x)).
Def.2) x is P_embedded<->(x is P_defined & P[x]).
Def.3) x is P_not embedded<->(x is P_defined & ~P[x]).
were P is a Predicate in one variable, that doesn't use x,
nor it is equivalent to any Predicate Q in one variable that use x.
Assume that P is a Predicate in one variable, that doesn't use x,
let
Q:<-> (P & Ax(x=x))
I don't understand this what ':' is doing after Q. do you mean the
following
Q<-> (P&Ax(x=x)).
If you mean that, then you are greatly confused.
what do you mean by x here, it is certainly not the same as the
x in x is P_defined.
in your formula x is any set, more specifically is variable that ranges
over the domain of all sets, while x in x is P_defined is a specific
set defined by P.
Perhpas you mean the following:
Q<-> (P&Az(z=x)) , were x here mean the same x present
in x is P_defined.
if so then with my due respect to you, here Q is not equivalent to P.
equivalent means biconditioning. and it is clear that there is no
biconditioning between Q and P. unless P[y]<->y=x then of course
here Q is equivalent to P. and it cannot be used to defined x, and so
P cannot be used to define x.
Then Q is equivalent to P and uses x.
Read above. you are greatly confused.
Thus all definitions are void. --- ???
No, you confusion is void. my definitions are clear.
Axioms:
Ax.0: Extensionality:AxAy(x=y<->(Az(zex<->zey)))
Ax.1: Regularity: Ax(~x={} ->Ey(yex & ~Ez(zey & zex)))
With {} (or 0) as defined below by Ax.4 as a (and by Ax.0 the)
(~y=y)_defined set.
Ax.2: Concordance:
AxAz(((x is P_not embedded & z is Q_defined & (P<->Q)) -> z is Q_not
embedded)
&((x is P_embedded & z is Q_defined & (P<->Q)) ->z is Q_embedded))
Good goodness, this awfully long sentence better be equivalent to
AxAz((x is P_defined & z is P_defined) -> (P[x]<->P[z]))
Ok,
Yes of course, that is my definition of equivalence of predicates. I
Ax.3: Exclusion:AxAy(yex & y is P_defined -> ~P[x])
Ax.4: Comprehension:Ex(x is P_defined)
Ax.5: Infinity:EN:0eN&(xeN->S(x)eN).
With 0 as above and S(x) whatever your taste for a successor is.
The precise statement would however be interesting as it might
or might not explicitly postulate the existence of a set 1 with
Az(ze1<->z=0).
Ax.6: Axiom of choice: as in ZFC.
Theorums:
Theorum 1: Uniqueness: x is P_defined & y is P_defined -> x=y.
Proof: since P<->P, then any member of x(other than x or y) is a member
of y, and vice versa.
what remains is to prove that neither xey nor yex.
Since P<->P then, either x and y are P_embedded
or x and y are P_not embedded .(Ax.2).
If ~P[x],~P[y] then neither xey nor yex.( definition of P_defined)
Now if P[x],P[y] then also neither xey nor yex. (Ax.3).
Theorum proved.
Sounds OK.
Theorum2: Equivalence: x is P_defined & x is O_defined -> (P<->Q).
Proof: for a prove by negation let P<-/->Q.
we have ~xex, so we are left only with members other than x.
if P<-/->Q, then these members shouldn't be all the same in x as
defined
by P and in x as defined by Q.
becaue not every y that is true for P is true for Q.
IOW, You consider P and Q equivalent iff Ay(P(y)<->Q(y)).
said
it again and again and again, if they are biconditioned. But you didn't
seem
to read what I am saying.
But still I don't see your argument.
The sets other than x are the /trivial/ part, as for ~y=x we have
P(y) <-> yex
and
O(y) <-> yex
What's missing is P(x)<->O(x).
And that doesn't follow from your axioms.
Then ~x=x, which violate extensionality.
Therefore P<->Q.
Theorum proved.
Nope.
Theorum 3: Empty set: ExAy (~yex).
Let P(y)<->yey. By Comprehension x={}.
Theorum proved.
Yep.
Theorum 4: Universal set: ExAy (~y=x <-> yex)
In words: there exist a set x such that what is other than it is in
it, and what is in it is other than it.
Def.6) V=x.
Proof: Let P[y]<-> y=y, By comprehension x=V.
V is unique since if Y is a set that contains V as a member
then ~Y=Y.Ax.3. which violates Ax.0.
Theorum proved.
Oh-kaaay.
But we cannot yet exclude the possibility that V=0.
Theorum 5: Pairing: AaAbExAy ( x is (y=a v y=b)_defined &
y is P_defined -> ~P[x]).
?? For which P??
How about P(y):<-> aey ? Does this imply ~ae{a,b} ?
Proof. from the theorum1 of uniqueness: x is unique. and the rest
follows from Ax.3.
Theorum proved.
Don't believe
Theorums of Union,Separation,Replacement and Power. can be modefied in
a way similar to th.5.
Similar to theorem 5 => Don't believe.
All can be proved from theorum1. and Ax.3.
Theorum 10: Complementary set: Ax Ex' Ay ( yex<->~yex' ).
Proof.
if x is P_defined, then x' is (~P)_defined. since ~P is a predicate
then x' exist.Ax.3.
What's missing is ~xex' and ~x'ex (as we have ~xex and ~x'ex').
Theom proved.
Theorum 11: Ax (x is P_not embedded -> x' is (~P)_embedded).
Proof: since x is P_not embedded then ~P[x], accordingly xex'.
If x' is ~P_not embedded then P[x'], i.e x'ex. which violate Ax.3.
Therefore x' is P_embedded.
Theorum proved.
Theorum 12: Ax x is well founded.
This follows from Ax.1
Theorum 13: AxAyAz(yex&zey -> x is not a subset of z).
Proof: since y cannot be in z. Ax.3., and yex. then x cannot be a
subset of z.
Theorm proved.
Theorum 14. Ax ( x 1-1 y(yeN) -> x is P_not embedded).
in words. Every finite set is P_not embedded.
not proved yet.
elements of N are not necessarily finite
Theorum 15. Ax ( x is P_embedded -> x ~ 1-1 y (yeN) )
in words. Every x such that x is P_embedded then x is infinite.
not proved yet.
Zuhair
.
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