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March 12th, 2018, 09:11 AM   #1
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Axiom of Regularity is a Theorem

Zylo's Axiom (ZA):
A = B → A ∉ B
can be an axiom,
or Theorem::
Assume A = B and A ∈ B,
B={A} = {B} = {{A}} = {{B}} = {{{A}}} = ...... indefinitely → A ∉ B

Theorem (Extended Axiom of Regularity}
Every member of A which doesn't contain another member of A is disjoint from A.
Proof:
Lst B be such a member. Then B ∩ A = ∅ because no member of A belongs to B (ZA).

This is best illustrated by an example.
Let A = {B, {B}}
Then B ∩ A = 0 but {B} ∩ A = B.

Comment: I personally believe ZA is more fundamental, intuitive, and easy to remember than AR.

Last edited by skipjack; March 14th, 2018 at 10:56 AM.
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March 12th, 2018, 09:52 AM   #2
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Doesn't work. You haven't precluded $A \in B \in C \in A$. That's why the axiom of regularity is so clever. It covers all those types of cases.
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March 14th, 2018, 08:38 AM   #3
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Quote:
Originally Posted by Maschke View Post
Doesn't work. You haven't precluded $A \in B \in C \in A$.
You are right: OP is wrong, The problem, among other things, is Axiom of Regularity is wrong to begin with.

Axiom of Regularity* (AR): Every non-empty set A contains an element that is disjoint from A:
∀x(x≠∅→∃y∈x(y∩x=∅))
.
1) A={B}, by AR B∩A= ∅ → B ∉ B because B ∈ A, all B.

2) A={A,B}, B∩A = ∅ because B ∈ A but B ∉ B. So B disjoint from A, which satisfies AR, but not OK.

3) A ∉ B, B ∉ C, C ∉ A, B={A}, C={B}, A={C} →
A={{{A}}}, {{A}}∩A=∅, because {{A}} ∈ A but {{A}} ∉ {{A}}, which satisfies AR, but not OK.

4) I originally got stuck trying to prove, if A ∈ B, then B ∉ A, by assuming A ∈ B and B ∈ A. Couldn't do it using same method as 3):
A={{A}}, {A}∩A=∅, because {A} ∈ A but {A} ∉ {A}

In summary, all AR says is A ∉ A, which is axiomatic, but allows any other circular definition.

The only way I can see to clean it up is as follows:

Axiom (revised AR)
A cannot contain A: If A={....}, A cannot appear anywhere, directly or indirectly, inside the brackets.
or Theorem (revised AR Theorem) : A contains A leads to a circular definition: meaningless infinite string of memberships:

For example, If A={B, {A}}, then
A = {B, {{B, {A}}}} = {B, {{B, {{B, {A}}}}}}

Prove: if A ∈ B, then B ∉ A
Assume A ∈ B and B ∈ A. Then A={{A}} which violates revised AR. Therefore B ∉ A.

* https://en.wikipedia.org/wiki/Axiom_of_regularity
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March 14th, 2018, 10:11 AM   #4
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Quote:
Originally Posted by zylo View Post
You are right: OP is wrong,
Yes, I know.

Quote:
Originally Posted by zylo View Post
The problem, among other things, is Axiom of Regularity is wrong to begin with.
LOL.
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March 14th, 2018, 12:25 PM   #5
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ps -- Zylo, which part of Brian Scott's proof on Stackexchange didn't you understand? It seems short and clear to me.
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March 14th, 2018, 01:26 PM   #6
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An axiom can’t generally be “wrong.” It is a statement that is taken to be true as a preference, but not a requirement. The standard set of axioms for set theory tends to be ZFC, but there are other sets of axioms you can choose to adhere to if you like. For example, a finitist wouldn’t use a set of axioms that include the axiom of infinity.

You should note that the other axioms of ZF are independent from the axiom pf regularity. A model for ZF less the axiom of regularity may not also be a model of ZF, but it may be.

If you don’t like a model that adheres to the axiom of regularity you can choose to work with one that doesn’t. Your model may suffer from inconsistencies like Russell’s paradox, but maybe not if you have other axioms sufficient to avoid them (just as ZF avoids Russell’s paradox even without the axiom of regularity).

https://en.m.wikipedia.org/wiki/Russell%27s_paradox

What I suspect at this point is that you lack a fundamental understanding of the relationship between a set of axioms and models that either adhere to them or not. Before you go trying to assert that one of the most common axioms is flawed and trying to replace it with your own, I suggest you develop this understanding.
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March 14th, 2018, 03:05 PM   #7
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Quote:
Originally Posted by AplanisTophet View Post
If you don’t like a model that adheres to the axiom of regularity you can choose to work with one that doesn’t.
Non well-founded sets are a thing. Wish I knew more about them.

https://en.wikipedia.org/wiki/Non-we...ded_set_theory
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March 19th, 2018, 08:14 AM   #8
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Quote:
Originally Posted by zylo View Post
You are right:

Axiom of Regularity* (AR): Every non-empty set A contains an element that is disjoint from A:
∀x(x≠∅→∃y∈x(y∩x=∅))


1) A={B}, by AR B∩A= ∅ → B ∉ B because B ∈ A, all B.

2) A={A,B}, B∩A = ∅ because B ∈ A but B ∉ B. So B disjoint from A, which satisfies AR, but not OK.

............................

In summary, all AR says is A ∉ A, which is axiomatic, but allows any other circular definition.

The only way I can see to clean it up is as follows:

Axiom (revised AR)
A cannot contain A: If A={....}, A cannot appear anywhere, directly or indirectly, inside the brackets.
or Theorem (revised AR Theorem) : A contains A leads to a circular definition: meaningless infinite string of memberships.
Can anyone show why 2), obviously unacceptable, is ruled out by AR?
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