|
|
Russell Easterly wrote:
> I have been reading up on the Axiom of Regularity,
> also called the Axiom of Foundation.
> I admit it, I do read this stuff sometimes.
>
> There are several ways the Axiom of Regularity
> can be stated, all of which are supposedly "equivalent".
> http://planetmath.org/encyclopedia/AxiomOfRegularity.html
>
> I want to look at two of the ways PlanetMath defines foundation:
>
> R1: For any non-empty set, X, there is some Y, element of X,
> such that Y intersect X is the empty set.
>
> In plain English, R1 says every set has a smallest element.
>
> R2: For any set X , there is no function F from omega to the
> transitive closure of X such that F(n+1) is an element of F(n).
>
I think you need the axiom of depending choice to prove these two are
equivalent.
> R2 basically says there are no infinite descending sets.
>
> It is obvious that R1 is not equivalent to R2.
> For example, I can state R1 in ZF without the Axiom of Infinity.
> I can't even state R2 without AoI.
>
Yes, you can, you can define the notion of a finite ordinal without
assuming the axiom of infinity. It's an ordinal that does not dominate
or equal any limit ordinal other than 0.
> At best, one might say R1 and R2 are equivalent in the
> presence of AoI. But, even this is false.
>
> I want to work in ZF without AoI using R1 for the
> Axiom of Regularity.
>
> Let set B be equal to any ordinal.
> A_0 = Union(B)
> A_i = A_(i-1) after removing the smallest element of A_(i-1) for i>0.
> Let A be the set of all A_i.
>
> R1 lets me prove A has a smallest element.
> Since A is partially ordered by inclusion, the intersection of A
> equals the smallest element of A.
You're confusing the membership relation with the inclusion relation.
> We know the intersection of A is the empty set.
> This means the empty set is a member of A.
> If A_z = {} there exists a member of A with exactly one element,
> A_y = {y} , and y must be the largest element in Union(B).
>
> Since I assumed B could be any ordinal, this proves
> every ordinal has a largest element.
>
> In ZF with AoI using R2 as the Axiom of Regularity,
> I can "prove" the set A doesn't exist.
>
> How can "set A has a smallest element" be equivalent to
> "set A doesn't exist"?
>
> R2 should be called the Axiom of Consistency.
> Essentially R2 says since ZF is consistent,
> especially AoI, there can't be any infinitely
> descending sets because such sets would
> prove ZF is inconsistent.
>
> How can the set A not exist?
> Obviously, I can define set A.
> What does it mean to say there is no function
> that defines set A? I prove A is finite.
> Is there an axiom that forbids finite descending chains?
>
> How can R2 be independent of the other axioms of ZF
> when AoI is required to even define R2?
>
> I looked at the "proof" R1 is equivalent to R2 given at PlanetMath.
> It says there must exist F(n+1) that is not a member of F(n).
> A_(y+1) = {} is not a member of A_y = {y}, so I don't see how
> this proves R1 and R2 are equivalent.
>
> This looks like a test of faith to me.
> If you can convince yourself R1 is equivalent to R2,
> you have what it takes to be a set theorist.
>
You need to learn a lot more about set theory. You're very confused.
R1 and R2 are equivalent in the presence of the axiom of depending
choice.
>
> Russell
> - 2 many 2 count
|
|