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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.

Asaf Karagila
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MatteoBianchetti
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    Well, $A = {A,\emptyset}$ is not true. – Tobias Kildetoft Jun 25 '15 at 22:21
  • $A\neq \lbrace A\rbrace$ – sti9111 Jun 25 '15 at 22:22
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    If $A\in A$, then ${A}$ violates (R), since it does not have a $\in$-minimal element, – abc Jun 25 '15 at 22:22
  • @Tobias@sti9111: why is $A={A,\emptyset}$ false. It may be, but this looks like a legitimate definition of A. Consider the set defined at p. 262 by Hrbacek-Jech: $\pi={{{\pi}},{{\emptyset}}}$. They say that $\pi={{{\pi}},{{\emptyset}}}$ is a legitimate definition of a set if regularity does not hold. So why is my $A={A\emptyset}$ outright false. You may be right, but I am still not understanding. In the same page they also define $A={A}$ and they say that this can be true without regularity. – MatteoBianchetti Jun 25 '15 at 22:27
  • @Sebastian: why is $\emptyset$ not a $\in$-minimal element? It seems that this make the formula (R) true. – MatteoBianchetti Jun 25 '15 at 22:31
  • You ask to consider the 'set' $A={A,\varnothing}$. This isn't a set, this an equality. If you want to replace $x$ by ${A,\varnothing}$ in the formula, that's fine, but saying $A={A, \varnothing}$ is completely meaningless to me. Refer to 7 here. – Git Gud Jun 25 '15 at 22:33
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    $\emptyset$ 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 – abc Jun 25 '15 at 22:33
  • note that $\emptyset \notin {A}$ – abc Jun 25 '15 at 22:35
  • @ Git Gud: I ask you to consider the set $A$ and I define $A$ to be ${A,\emptyset}$. This kind of definition is normal practice in set-theory, if I am not mistaken. – MatteoBianchetti Jun 25 '15 at 22:37
  • @orient: You are mistaken. That kind of definition is not normal practice in set theory, and in most models of set theory (even without regularity) there is no set that satisfies the equation $A={A,\varnothing}$. – hmakholm left over Monica Jun 25 '15 at 22:39
  • @Sebastian: I agree that you can derive the denial of (R) from the existence of $A$. And I agree that one way to derive this denial is to proceed as you do in your comment. However, I am confused because $A$ has an element that is disjoint from itself and this should make (R) true, not false. – MatteoBianchetti Jun 25 '15 at 22:39
  • @orient: The axiom starts with an $\forall x$ -- so it can only be true if it holds for every value of $x$. Simply finding one value of $x$ that has the property is not enough to make the axiom true. – hmakholm left over Monica Jun 25 '15 at 22:41
  • Well no, (R) has a universal quantifier for x, so for (R) to be true every(!) set must have a $\in$-minimal element. – abc Jun 25 '15 at 22:41
  • @Henning Makholm: see the definitions from Hrbacek-Jech. If by 'common practice' you mean that people usually accept (R), then I agree that you do not find definitions of set that violate (R). But this does not make the definition illegitimate. I am asking why $A$ violates (R) if it contains a element disjoint from itself. – MatteoBianchetti Jun 25 '15 at 22:41
  • $A = {A,\emptyset}$ is not a definition (why would there be any set satusfying it, and why would there be a unique one?) – Tobias Kildetoft Jun 25 '15 at 22:43
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    @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. – hmakholm left over Monica Jun 25 '15 at 22:44
  • @Sebastian and Henning Makholm: I see the point now about (R). Thanks very much. To Henning and Tobias about definitions: definitions do not have to be consistent. You can define N to be the greatest natural numbers. It refers to nothing, but it is still a definition. Anyway this is another problem. – MatteoBianchetti Jun 25 '15 at 22:44
  • Anyone who says that "$\pi={{{\pi}},{{\emptyset}}}$" is a legitimate definition in standard set theory is simply wrong. It works as a definition in certain variant set theories, especially ones that accept Aczel's Anti-Foundation Axiom. Unless Hrbacek-Jech is speaking specifically about such a theory, I think you must be misunderstanding what they say. – hmakholm left over Monica Jun 25 '15 at 22:47
  • @orient: No, you can't define $N$ to be the greatest natural number, because there is no such thing. That combination fails to be a definition because there is nothing it defines. – hmakholm left over Monica Jun 25 '15 at 22:48
  • I did not say that $\pi$-etc. is used in standard ST. Anyway, from my point of view, the problem of what counts as a legitimate definition does not have to do with consistency. – MatteoBianchetti Jun 25 '15 at 22:49
  • To discover that N does not exist, you need to define it first. – MatteoBianchetti Jun 25 '15 at 22:50
  • @orient: You don't "discover that $N$ does not exist". You discover that there is no greatest natural number. The phrase "the greatest natural number" does not refer to anything and therefore cannot be used to define anything. And you certainly do not need to introduce a nonsensical definition in order to speak about the fact that there is no greatest natural number. – hmakholm left over Monica Jun 25 '15 at 22:52
  • A definition is the act of giving a name to something that you already know exists. If you want to give a name to a given phrase, such as "the greatest natural number", your wish only becomes a definition once you can supply proof that this phrase determines a unique thing for you to bestow the name on. And the onus is on YOU who want to make the definition to supply such proof. Unless you can deliver it, your attempt to connect a name to the phrase does not constitute a definition. – hmakholm left over Monica Jun 25 '15 at 22:55
  • If one wants to insist that a definition must be consistent, the discussion may become only about words or a too complicated ontological discussion. I would at least say that, if one wants to insist that definition must be consistent, then she should say that we are not sure that any math definition exists, since we do not know whether math is consistent. This may be right, but not obviously so. I will think about that. Thanks. – MatteoBianchetti Jun 25 '15 at 22:58

4 Answers4

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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.

Cameron Buie
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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 $A$.

Henry
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  • I disagree. Such a set $A$ is perfectly consistent with Regularity on its own. We need other axioms for it to cause problems. – Cameron Buie Jun 25 '15 at 23:00
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    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. – Henry Jun 25 '15 at 23:03
  • 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. – Cameron Buie Jun 25 '15 at 23:10
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    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 $A \in A$). – chharvey Jun 26 '15 at 01:36
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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.

MatteoBianchetti
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  • @CameronBuie: what is (b)? I was trying to disambiguate (a). It seems to me that people can use that expression in two senses. – MatteoBianchetti Jun 26 '15 at 02:27
  • 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). – Cameron Buie Jun 26 '15 at 04:04
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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.

chharvey
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