My Math Forum ZFC Axiom of regularity

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 March 7th, 2018, 09:21 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,431 Thanks: 105 ZFC Axiom of regularity "In mathematics, the axiom of regularity (also known as the axiom of foundation) is an axiom of Zermeloâ€“Fraenkel set theory that states that every non-empty set A contains an element that is disjoint from A. In first-order logic, the axiom reads: $\displaystyle {\displaystyle \forall x\,(x\neq \varnothing \rightarrow \exists y\in x\,(y\cap x=\varnothing ))}$ "* Which says every set x contains a set y (Set intersection is only defined for sets). x={1,2}. So natural numbers can't be members of a set? *https://en.wikipedia.org/wiki/Axiom_of_regularity Note: Decided to do this because I discovered I could copy it from google Latex coming through.
March 7th, 2018, 09:47 AM   #2
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Quote:
 Originally Posted by zylo x={1,2}. So natural numbers can't be members of a set?
Question makes no sense in this context.

Unless you're forgetting that $0 = \emptyset$, $1 = \{0\}$, and $2 = \{0, 1\}$ in the usual convention.

March 7th, 2018, 10:01 AM   #3
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 Originally Posted by Maschke Question makes no sense in this context. Unless you're forgetting that $0 = \emptyset$, $1 = \{0\}$, and $2 = \{0, 1\}$ in the usual convention.
Forgetting? I never knew it, but have seen VN ordinal numbers.

There is a difference between a natural number and its representation as a set.

So is it saying every set has to contain a set?

Last edited by skipjack; March 8th, 2018 at 07:08 AM.

March 7th, 2018, 10:11 AM   #4
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Quote:
 Originally Posted by zylo There is a difference between a natural number and it's representation as a set.
Most definitely. Agree.

And "it's" is a contraction of "it is." The form you wanted is "its." Please make a note of it.

Quote:
 Originally Posted by zylo So is it saying every set has to contain a set.
I must be missing your point. Are you saying the empty set violates regularity? But the definition you posted specifically says the set in question must be nonempty.

March 7th, 2018, 10:30 AM   #5
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 Originally Posted by zylo $\displaystyle {\displaystyle \forall x\,(x\neq \varnothing \rightarrow \exists y\in x\,(y\cap x=\varnothing ))}$ "*
Which says $\displaystyle y\cap x$ exists, so y has to be a set, and y has to belong to x. So a set has to contain a set?

EEDIT
"ZFC is intended to formalize a single primitive notion, that of a hereditary well-founded set, so that all entities in the universe of discourse are such sets. Thus the axioms of ZFC refer only to pure sets and prevent its models from containing urelements (elements of sets that are not themselves sets)"

https://en.wikipedia.org/wiki/Zermel...kel_set_theory

There may be a problem with that which I brought up many moons ago. Thanks for response.

Last edited by zylo; March 7th, 2018 at 10:38 AM.

March 7th, 2018, 10:35 AM   #6
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Quote:
 Originally Posted by zylo Which says $\displaystyle y\cap x$ exists, so y has to be a set, and y has to belong to x. So a set has to contain a set?
It says x intersect y is empty. You are making no sense at all.

And in set theory, all sets are "pure" sets. They are either empty or they contain other sets. There are no ur-elements (objects that are not sets) in ZFC. There are set theories with ur-elements.

March 7th, 2018, 09:54 PM   #7
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Quote:
 Originally Posted by zylo EEDIT "ZFC is intended to formalize a single primitive notion, that of a hereditary well-founded set, so that all entities in the universe of discourse are such sets. Thus the axioms of ZFC refer only to pure sets and prevent its models from containing urelements (elements of sets that are not themselves sets)" https://en.wikipedia.org/wiki/Zermel...kel_set_theory There may be a problem with that which I brought up many moons ago. Thanks for response.
Are you saying that your question was indeed about urelements? That if a set isn't empty, it contains something; and that something is necessarily another set. Yes that is indeed a good question. Some set theories have urelements and some don't. Standard set theory, ZFC, is a pure set theory.

I was inspired to read the Wiki page on urelements and I ended up learning something. Zermelo's original 1908 version of set theory had urelements. I hadn't known that.

You know when it comes to the creation of set theory, Cantor gets the credit but Zermelo did most of the heavy lifting as far as I can tell.

Last edited by Maschke; March 7th, 2018 at 10:00 PM.

March 10th, 2018, 08:02 AM   #8
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Thanks, What motivated the OP was that the OP link didn't specify what an element was.

It also says that you can prove from AR that a set can't belong to itself. How?

Found something in an old thread
Quote:
 Originally Posted by zylo AR, But I felt it had a flaw a$\displaystyle \notin$a. Proof 1) Let x={a} 2) x$\displaystyle \cap$a=0, AR requirement 3) members of x: a 4) members of a: b (a has to have a member) 5) x$\displaystyle \cap$a=0 implies b$\displaystyle \neq$a 6) $\displaystyle \therefore$ a can not be a member of itself. Problem with proof: Step 4) You are not excluding b=a. If you exclude b=a, the proof is circular. * https://en.wikipedia.org/wiki/Axiom_of_regularity https://en.wikipedia.org/wiki/Zermel...kel_set_theory
Unfortunately I can't find the source of the proof. I thought it was in one of the links but it's not there (anymore?).

Do you have a proof?

March 10th, 2018, 08:05 AM   #9
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Quote:
 Originally Posted by zylo Do you have a proof?
See Brian Scott's checked answer here ...

https://math.stackexchange.com/quest...ontaining-sets

March 10th, 2018, 09:22 AM   #10
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Quote:
 Originally Posted by Maschke See Brian Scott's checked answer here ... https://math.stackexchange.com/quest...ontaining-sets
A=A, so you can't have A$\displaystyle \epsilon$A. Regardless of AR. Your assumption has to be viable. You can't assume something that can't be true in order to prove it can't be true.

Last edited by zylo; March 10th, 2018 at 09:43 AM. Reason: remove "fundamental mistake in set theory"

 Tags axiom, regularity, zfc

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