Re: The definition of ordinal.

On May 3, 3:24 pm, zuhair <zaljo...@xxxxxxxxx> wrote:
Hi all,

In a previous post I've presented the following definition of ordinal.

x is transitive <-> Ayz((zey & yex) -> zex).

x is ordinal <-> ( x is transitive & Ay( yex -> y is transitive) ).

However this definition was proved to be equivalent to the standard
definition if and only if Regularity holds.

I think this definition can be modified to be independent of
Regularity. So I propose the following modification:

x is ordinal <-> ( x is transitive & Ay( yex -> y is transitive) &
~Ey(yex & yey) ).

Is that simple modification enough to make this definition
independent of regularity?

Zuhair

Though I don't know the answer to this question of mine, but yet I
think there can be a better definition.

x is ordinal <-> ( x is transitive & Ay( yex -> y is transitive) &
Ay( (y subset_of x & ~y=0) -> Ez(zey & z disjoint y) ) ).

I think this definition is equivalent to the standard definition of x
is ordinal with or without regularity.

Here is Dave Seaman's proof of equivalence between
the definition of x is ordinal <-> ( x is transitive & Ay(yex -> y is
transitive))
and the standard definition of x is ordinal. It demands regularity.


-----------------------------------
Call a set A a _pseudoordinal_ if A is transitive and each member of A
is
transitive.


It is clear that evey ordinal is also a pseudoordinal. The converse
may
fail in the absence of regularity: if a = {a}, then { 0, a } is a
counterexample. We therefore assume regularity in what follows.


Lemma. Suppose A is a pseudoordinal and L is an ordinal that is a
subset
of A. If A\L is nonempty, then L is a member of A.


Proof. Let E = A\L. By regularity, we may choose x in E such that x
is
disjoint from E, implying by transitivity that x must be a subset of
L.
But the only subset of L that can be a member of E is L itself,
Q.E.D.


Proposition. Let A be a pseudoordinal and let L be the least ordinal
that is not a member of A. Then A = L.


Proof. We are given that L is a subset of A. However, L is not a
member
of A, implying that A\L is empty.


Corollary. Every pseudoordinal is an ordinal.

----------------------------------------

So I think this new definition that I have stated , can be prooved in
the same manner, since it is clear that every ordinal according to
this new definition is well founded.

This definition is actually simpler than the standard one.

Zuhair

.

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