Re: ZFC in 4 Axioms.


MoeBlee wrote:
zuhair wrote:
I am not understanding this, union and pairing are theorums in this
theory, not axioms.

Perhaps they are and perhaps you proved them earlier in this thread.
But I'm coming in at this point fresh with your revised axioms. If you
like, you may post your proofs of union and pairing from your revised
axioms.

Now my modified Ax.3 is
Ax.3)Comprehension: Ex Ay ( yex<->(P(y) /\ ~ y=x /\ ~(y is embeded) )

were y is embeded is defined as below:
y is embeded <-> y:(Ax:x in y<->P(x)) /\ P(y).

Now for sets a and b, it can be proved that x={a,b} exists
Let P(y)<-> y=a v y=b
then by Ax.3. x={a,b} exists.

so theorum of pairing in this theory is as follows:

Aa Ab Ex Ay ( yex<->(y=a v y=b) /\ ~ y=x /\ ~(y is embeded) )




Your P is not followed by (y), why?

Most authors include '(y)' as you have, but it is not strictly
necessary, so I prefer to omit it in order to have a cleaner
formulation. However, we should be very specific as to what YOU mean by
'(y)'. Do you mean that y MUST occur free in the formula P(y)? Usually,
such formulations do not require that y MUST occur free in the formula.
But if do require that y occurs free in the formula, then you must
mention that. Also, are there any restrictions as to the variable x
occurring free in the formula P(y)?

However I discovered the the exclusion by ~y=V is not enough for this
theory.
I need to exlude every set that fulfills its inclusive formula.

i.e X: (Ax: x in X<-> P(x) ) /\ P(X) <-> X is embeded.

and then I will write comprehension as follows:

Ax.3)Comprehension: Ex Ay ( yex<->(P(y) /\ ~ y=x /\ ~(y is embeded) )"

Then please FORMALLY define 'is embedded' or declare it to be
primitive. (Doing neither leaves it primitive by default.)

But I defined "X is embeded" as

X is embeded <-> X:(Ax:x in X<->P(x)) /\ P(X).

informally speaking a set is embeded if and only if it fulfills the
very same formula every member of it should fulfill in order to be in
it.

Knowing that now,do you think that I should define "is embeded" or list
it as a primitive?

Example: the set of all sets. here it is V: A x in V<-> (x = x)
Since V=V. then V fulfilled the same formula of inclusion of every x in
it. Therefore V is an embeded set. Since I applied restrictions on
comprehension such as to exclude such sets from being members of other
sets. Then no set has an embeded set as a member.
Also, since comprhension also excludes every set from being a member in
itself by the restriction ~y=x in Ax.3. as a consequence of this later
restriction an embeded set is not in itself. Without the restriction
~y=x in Ax.3 all embeded sets would have been members of themselfs.

Embeded sets are like proper classes, they cannot be members of sets.





Why have you still not learned that you have to define EVERY predicate
and/or operation symbol that is not primitive?

MoeBlee

.

Relevant Pages

  • Re: ZFC in 4 Axioms.
    ... But I'm coming in at this point fresh with your revised axioms. ... I need to exlude every set that fulfills its inclusive formula. ... Then no set has an embeded set as a member. ...
    (sci.math)
  • Re: ZFC in 4 Axioms.
    ... An embeded set unlike what its name might suggest, is not a set that is ... It is a set that fulfills the formula of inclusion of members in it. ... The set of all ordinals. ...
    (sci.math)