Re: Simplifying M theories.
- From: zuhair <zaljohar@xxxxxxxxx>
- Date: Fri, 19 Jun 2009 17:40:34 -0700 (PDT)
On Jun 19, 6:54 pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Jun 19, 3:25 pm, zuhair <zaljo...@xxxxxxxxx> wrote:
6: Infinity: Ax((Ay( y in x -> y is a universe)
and x is countable) -> H(Union x)).
Please define 'countable' and 'Union'.
countable is defined in the convensional manner.
x is countable iff ( x equinumerous to Omega or x subnumerous to
Omega ).
Omega is the class of all finite ordinals.
Union (x) is the set of all members of members of x.
Define: y=Ux <-> Az( z in y <-> Em(m in x and z in m)).
7: Inflation:
Ax Ay ( (x is a set and Az(z in y -> z is a universe) and y
equinumerous x ) -> y is a set).
Please define 'equinumerous'.
It is the same convensional definition.
This theory contain all axioms of ZF-Regulariy-choice.
The proof of that is trivial I guess.
What I mean by axioms of ZF-Regulariy-choice is when we apply them to
"sets" in this theory.
theorm of pairing of sets:
The unordered pair of sets is a set.
(x is a set & y is a set) -> {x,y} is a set.
Proof: by definition there should exist a universe of which x is a
member , and there exist a universe which contain y as a member. Now
from connectivity axiom either the universe which contain x is a
member of the universe which contain y or the converse, in both
situations the members of all of one of the universes will be in the
other ( since universes are transitive sets ), so the net result is
that there will exist a universe which contain both x and y as members
in it, so {x,y} would be a subset of this universe, and so it will be
a member of the universe that contain this universe as a member in it,
and thus {x,y} is a set.
Proof for every class x, there exist Ux.
Since members of any class x are sets, then the members of these sets
are also sets (because universes are transitive), thus there will
exist a class of all of them from comprehension, which is the union
class of x.
for any 'set' x, Ux is a set.
This is simple because Ux will be a subset of the universe which
contain x as a member, thus it will be a member in the next universe
(i.e. the universe which contain the universe which contain x as a
member) and thus a set.
For any 'set' x, the power class of x is a set.
this is simple also since every subset of x is also a set, then the
class of all subsets of x will be a subclass of the universe which
contain x as a member, and thus a member of the next universe, thus a
set.
Infinity is proved easily, since the class X={0,Power0,
powerpower0,.....} would be equinumerous to Omega, thus from axiom of
infinity Union X has the property H, and it is clear that X has the
two properties of transitiveness, and of having every subset of a
member of it as a member in it, thus UX is a Universe by definition,
and thus it has a power set of which Ux would be a member, of course
this power set of UX is a universe also thus UX is a set. it is clear
that Omega is a subclass of X, and thus a member of PowerUX, thus
Omega is a set, proving infinity.
the proof of separation on sets is clear.
The proof of replacement directly comes from axiom of inflation.
Zuhair
MoeBlee
.
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