Re: A simple question?




Randy Poe wrote:
zuhair wrote:
All of what I am saying is that if a set subsets its power set, then
its power set also subsets its power.

What does "subsets" mean as a verb? Normally, subset is a noun.

Does "A subsets B" mean "A is a subset of B"?


In symboles:

x subsets P(x) -> P(x) subsets P(P(x) ). do anybody have an objection
to this?

If "x subsets P(x)" means "x is a subset of P(x)", then it is not
in general true. For instance, if x is the set {1}, then the subsets
of P(x) are {}, {{1}}, neither of which is x.

Accordingly we say that x= {1} is not a set which subsets its power
set.

I call every set that subsets its power set a " subpower".

accordingly x={1} is not a subpower, because as you demonstrated above,
it doesn't subset its power set.

My definition of cardinality that is: The equivalence class of
subpowers , under equivalence relation "bijection". is therefore not
Frege's definition of cardinality. which is what it is intended to be,
ie not the same as Frege's definition of cardinality.

if every x subsets its power set, then there is no usefullness in me
stating the above definition as a new one, since by then it will only
be another way of stating Frege's definition of cardianlity.

To make U understand the matters more , lets take a more complex
example , that is my definition of ordinals . which is a set that
subsets its power set and every member of it also subsets its power
set.

x is an ordinal <-> x subsets P(x) /\ Am:mex -> m subsets P(m).

This also can be writtin in the following manner.

x is an ordinal <-> x: Amex,Anem,Azen -> nex /\ zem

In ZFRC , this definition is equivalent to the standard definition of
ordinals.

This was proved by Dave Seaman.

But without the axiom of regularity , it is not equivalent to the
standard definition.

Have a nice time excersizing what I have mentioned.

Zuhair

- Randy

what do you mean general.

x subsets P(x) <-> x:Amex,Anem -> nex.

Of course not every x fulfills this definition. That what this
definition is made for.

Zuhair

.



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