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I am currently studying the Zermelo-Fraenekl Set Theory and I have some problems with understanding the Axiom of Regularity:

Firstly, I found this version of the axiom on Wiki enter image description here

I understand that x and y are all referring to sets. Then there is a problem: Consider the set {1,2,3}. This set is non-empty but there not does exist a "y" in x such that "y intersects x" is the empty set because none of the members of x are sets (In this case, it does not make sense to talk about "y intersect x" since intersection is a binary connective between two sets).

Secondly, here is another version of this axiom:

enter image description here

I have some troubles with understanding the second part of this experssion (after the implication arrow). Why is this version equivalent to the version above? Thanks so much if someone could give a hand. I really appreciate!

Asaf Karagila
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  • What are $1$, $2$, and $3$ in set theory? See https://en.wikipedia.org/wiki/Ordinal_number#Von_Neumann_definition_of_ordinals In set theory everything is a set. – user4894 Feb 26 '17 at 11:24
  • The first part of the second expression says : "$x$ is not empty" : $\exists a ( a \in x)$. – Mauro ALLEGRANZA Feb 26 '17 at 11:30
  • The second part says : "there is an element of $x$ (call it $y$) that has no elements "in common" with $x$" : does not exists $z$ which is both in $y$ and in $x$. This in turn means that the intersection of $y$ and $x$ (i.e. their "common part") is empty, i.e. $y \cap x = \emptyset$. – Mauro ALLEGRANZA Feb 26 '17 at 11:32
  • Thanks for your answers. For my first question, what if the members of a set are some other mathematical objects other than natural numbers? What do we do to them in order to turn them into sets? Also, what if the members of a set are non-mathematical objects? –  Feb 26 '17 at 11:47
  • Non-mathematical objects are not part of the mathematical universe. Since sets are objects of the mathematical universe, set theory is concerned with them, and with them alone. – Asaf Karagila Feb 26 '17 at 11:55
  • Note: can be interesting to note that the second version : $∃a(a∈x) \to ∃y(y∈x \land \forall z (z \in y \to z \notin x))$ applies also to versions of set theory with urelements (or individual), i.e. objects that are not sets. Consider your example $x= { 1,2,3 }$ and consider the numbers as individuals (that are not sets). Clearly, $x$ is not empty, and thus the antecedent of the axiom is satisfied. What about the consequent : $∃y(y∈x \land \forall z (z \in y \to z \notin x))$ ? 1/2 – Mauro ALLEGRANZA Feb 26 '17 at 13:49
  • The elements of $x$ are the "individuals" : $1,2,3$; but thet are not sets, and thus $z \in y$ is false for every $y \in x$. Thus, the conditional $∀z(z∈y→z∉x)$ is vacuously true and (due to the fact that $y \in x$ is true, because $x$ is not empty) the complete consequent $∃y(y∈x∧∀z(z∈y→z∉x))$ is true. Conclusion: also in your case the (suitable instance) of AR is satisfied. 2/2 – Mauro ALLEGRANZA Feb 26 '17 at 13:53
  • The "trick" is that (in versions of set th with "individuals") the two versions of the axiom are not equivalent; in case of individuals, $y \cap x$ is not defined if $y$ is not a set (and this is your example with $x= { 1,2,3 }$). – Mauro ALLEGRANZA Feb 26 '17 at 13:55

2 Answers2

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In set theory, everything is a set. Specifically, the most common interpretation of $1$ is $\{\{\}\}$, and $2$ becomes $\{\{\}, \{\{\}\}\}$. In fact, your set $\{0,1, 2, 3\}$ (with a small addition) is actually what we would denote as $4$.

As for why they are equivalent, one says "if $x$ is non-empty, there is an element in $x$ that doesn't intersect $x$", while the other says "if there is an element in $x$ (i.e., if $x$ is non-empty), then there is an element $y$ such that $y$ is an element of $x$ and at the same time, there is no set that is an element of both $x$ and $y$ at the same time (i.e. $x$ and $y$ do not intersect)".

Arthur
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  • Thanks for your answer. For my first question, what if the members of a set are some other mathematical objects other than natural numbers? What do we do to them in order to turn them into sets? Also, what if the members of a set are non-mathematical objects? –  Feb 26 '17 at 11:33
  • You can have a set theory that includes things like apples and bridges, but that wouldn't be ZF. – Arthur Feb 26 '17 at 11:47
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So in ZF set theory every object is a set, in particular $1,2,3$ denote sets ($1= \{\emptyset\}$ usually, for instance), so it does make sense to talk about their intersection.

The second version is equivalent because of the following : the right part of the implication states that $x$ is not empty and so corresponds with the $x\neq \emptyset$ part of the first version. The second part precisely states that their exists a $y$ such as in the first version : indeed $y\cap x=\emptyset$ is equivalent to $\neg \exists z, z\in y\cap x$, which is equivalent to $\neg \exists z, z\in x \land z\in y$.

Maxime Ramzi
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  • Thanks for your answer. For my first question, what if the members of a set are some other mathematical objects other than natural numbers? What do we do to them in order to turn them into sets? Also, what if the members of a set are non-mathematical objects? –  Feb 26 '17 at 11:39
  • This question doesn't really make sense : if you have "non mathematical objects", or even "things that are not sets", then it's not ZF set theory. It may be philosophy, or it may be some other kind of mathematical theory (in Neuman-Bernays-Gödel theory, there are sets and proper classes for instance, or in some variations of ZF, there are so-called "urelements"); but it will not be set theory in the usual sense – Maxime Ramzi Feb 26 '17 at 14:29