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The axiom of regularity says:

(R) $\forall x[x\not=\emptyset\to\exists y(y\in x\land x\cap y=\emptyset)]$.

From (R) it follows that there is no infinite membership chain (imc).

Consider this set: $A=\{A,\emptyset\}$.

I am confused because it seems to me that A violates and does not violate (R).

It seems to me that A does not violate (R) because $\emptyset\in A$ and $\emptyset\cap A=\emptyset$.

It seem to me that A violates (R) because we can define the imc $A\in A\in A\in ...$

I checked some sources, but I am still confused. Can anyone help me? Thanks.

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    $\begingroup$ Well, $A = \{A,\emptyset\}$ is not true. $\endgroup$ – Tobias Kildetoft Jun 25 '15 at 22:21
  • $\begingroup$ $A\neq \lbrace A\rbrace$ $\endgroup$ – sti9111 Jun 25 '15 at 22:22
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    $\begingroup$ If $A\in A$, then $\{A\}$ violates (R), since it does not have a $\in$-minimal element, $\endgroup$ – Achilles Jun 25 '15 at 22:22
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    $\begingroup$ $\emptyset$ is a $\in$-minimal element of A, but by the pairing axiom you can form the singelton set $\{A\}$, which does not have a $\in$-minimal element, because the only possible choice would be A, but $A\cap\{A\}$ contains A $\endgroup$ – Achilles Jun 25 '15 at 22:33
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    $\begingroup$ @orient: What makes the "definition" illegetimate is that it is not a definition. It is an equation, that is, a property that any given set may or may not have -- such a thing only becomes a definition if you have an argument that there is exactly one thing that satisfies the equation. Since you have no such argument (and indeed such an argument is not possible from the usual axioms) your equation is not a definition. $\endgroup$ – Henning Makholm Jun 25 '15 at 22:44
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Suppose there were a set $A$ with the property $A=\{A,\emptyset\}$. Clearly $A \not = \emptyset$ since $A$ would have at least one element and possibly two.

Now let $x=\{A\}$. Clearly $x \not = \emptyset$ since $x$ would have exactly one element. So the axiom of regularity would imply $\exists y\,(y\in x\land x\cap y=\emptyset)$.

Since $x=\{A\}$ would only have one element, this would mean $y\in x$ would imply $y=A$ and then $x\cap y = \{A\} \cap \{A,\emptyset\} = \{A\} = x \not = \emptyset$.

So the property $A=\{A,\emptyset\}$ is inconsistent with the axiom of regularity for any set $A$.

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  • $\begingroup$ I disagree. Such a set $A$ is perfectly consistent with Regularity on its own. We need other axioms for it to cause problems. $\endgroup$ – Cameron Buie Jun 25 '15 at 23:00
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    $\begingroup$ You are saying that we need more axioms, for example when I said "let $x=\{A\}$". That may be true - but I do not think it was really the point being asked. $\endgroup$ – Henry Jun 25 '15 at 23:03
  • $\begingroup$ It seems that was precisely the point of confusion: why doesn't $A$ conflict with Regularity (as expressed), even though it yields an infinite membership chain? The problem is that different versions of Regularity are equivalent in the presence of other axioms, but not necessarily equivalent a priori. $\endgroup$ – Cameron Buie Jun 25 '15 at 23:10
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    $\begingroup$ The set $A$ itself does not conflict with Regularity, since indeed there exists an element inside whose intersection with $A$ is empty. However the set $\{A\}$ does conflict with Regularity: there does not exist any element inside $\{A\}$ whose intersection with $\{A\}$ is empty (namely because $A \in \{A\}$ and $A \in A$). $\endgroup$ – chharvey Jun 26 '15 at 1:36
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Let's suppose that there is a set $A$ such that $$A=\{A,\emptyset\}.$$ Now, consider the set $$B:=\{A,A\}.$$ This set exists by Pairing (if $A$ exists), and we can show by Extensionality that $$B=\{A\}.$$ Now we apply Regularity to get our contradiction.

You have correctly observed that such a set $A$ does not violate Regularity. However, Regularity is not our only axiom.

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The set $A=\{A, \emptyset\}$ itself does not violate Regularity, since indeed there exists an element inside $A$ whose intersection with $A$ is empty. $$\emptyset\cap A = \emptyset$$ (Incidentally this property holds true for any set $A$.)

However the set $\{A\}$ does violate Regularity: every element (there is only one) inside $\{A\}$ intersects $\{A\}$ in a nonempty way. $$A \cap \{A\} = A \ne \emptyset$$ because $A \in \{A\}$ and $A \in A$.

Therefore although it may seem that $A$ doesn’t violate regularity, it has a consequence that does.

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The other answers and the various comments clarified this important point. There are two senses of the expression

(a) 'A violates (R)'.

First, (a) may mean: if you let x=A in (R), then you get a false sentence. This is not true: A does not violate (R) in this sense.

Second, (a) may mean: if you assume that A exists (plus the axioms of ZFC minus (R)), then the denial of (R) holds. This is true (as people showed). A violates (R) in this sense. Many thanks.

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  • $\begingroup$ @CameronBuie: what is (b)? I was trying to disambiguate (a). It seems to me that people can use that expression in two senses. $\endgroup$ – orient Jun 26 '15 at 2:27
  • $\begingroup$ Now that you've edited your answer, I see that you've caught on to the key point. ^_^ Your original phrasing seemed to have both "senses" saying precisely the same thing (albeit in different words). $\endgroup$ – Cameron Buie Jun 26 '15 at 4:04

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