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I was reading about the axiom of regularity on Wikipedia.

It is stated that:

In mathematics, the axiom of regularity (also known as the axiom of foundation) is an axiom of Zermelo–Fraenkel set theory that states that every non-empty set $A$ contains an element that is disjoint from $A$.

$$\forall x\left(x\neq \emptyset\implies\exists y\in x\left(y\cap x=\emptyset\right)\right)$$

How can this be correct?

If $A$ contains an element $x$, then $x$ can not be disjoint from $A$, because $x$ belongs both to $A$ and to the set containing only $x$.

What am I misunderstanding?

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bsky
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    Your question makes no sense. Syntactically. – Asaf Karagila Jun 01 '17 at 10:33
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    What does it mean that $x$ is "from" $A$, and what do you mean by "belongs"? – Asaf Karagila Jun 01 '17 at 10:48
  • Are you concerned about an element being part of two or more sets at the same time? Look at this: $A={{1,2},{2,3},2}$. The $2$ is element of both sets inside $A$ and of $A$ itself. Are you feeling uncomfortable with this? – M. Winter Jun 01 '17 at 10:50
  • I updated my question because I phrased it poorly. – bsky Jun 01 '17 at 10:52
  • The sets $A:={\varnothing}$ and $x:=\varnothing\in A$ are disjoint. So here $x$ is an element of $A$ disjoint from $A$ (i.e. the sets have no common elements). – drhab Jun 01 '17 at 10:53
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    An element $x$ is not the same as the set ${x}$ containing only this element. Is this the confusing point? If $x\in A$, then ${x}$ must not be in $A$, so $x$ can be disjoint from $A$. – M. Winter Jun 01 '17 at 10:54
  • The fact that $X$ belongs both to $A$ and to the set containing only $x$ (commonly denoted by ${x}$) tells us ${x}$ can not be disjoint from $A.$ It doesn't follow that $x$ itself is disjoint from $A.$ – bof Jun 01 '17 at 10:54
  • It is better to avoid mix-and-match with variables... Regularity: $∀A
    (A≠∅ \to ∃x∈A(x∩A=∅))$. If $A$ is not empty (this is the part $A≠∅$) it contains an element: call it $x$ (i.e. it is this is the part $∃x∈A$). What does it mean: "$x$ can not be from $A$" ? We have assumed that $A$ is not empty and we have called $x$ (one of) the element(s) of $A$.     – Mauro ALLEGRANZA Jun 01 '17 at 10:55
  • If for instance we have A = {1, 2, 3}, can we say that 1 is disjoint from A? – bsky Jun 01 '17 at 10:55
  • Depends on how $1$ is (and the other numbers are) defined as a set, but usually it is set $1:={\varnothing}$, so yes, they are disjoint. – M. Winter Jun 01 '17 at 10:56
  • @octavian That depends on how $1$ is defined. – drhab Jun 01 '17 at 10:56
  • @M.Winter yes that's what I have trouble with, the difference between x being disjoint from A and {x} being disjoint from A. – bsky Jun 01 '17 at 10:56
  • You are misunderstanding the fact that $x \in x$. This is not... The empty set is... empty, i.e. it has no elements. Thus, $x \notin \emptyset$, for every $x$, and also $\emptyset \notin \emptyset$. – Mauro ALLEGRANZA Jun 01 '17 at 10:57
  • @MauroALLEGRANZA but the axiom is about non-empty sets, right? – bsky Jun 01 '17 at 10:59
  • @octavian At the moment it is not completely clear to me what confuses you about the distinction between $x$ and ${x}$. Maybe you can edit your question to make this clear. – M. Winter Jun 01 '17 at 11:00
  • Thus, in general, it is not true that $x \in x$... The axiom asserts that this is so for every $x$. Apply the axiom to $A= { x }$. We have that $x \in { x }$ and thus ${ x }$ is not empty. Thus $∃y∈ { x } (y∩ { x } =∅))$. But ${ x }$ has only one element: $x$; thus $x ∩ { x } =∅$. Conclusion: $x \notin x$ (otherwise, form: $x \in x$ and $x \in { x }$ we get $x ∩ { x } = x$). – Mauro ALLEGRANZA Jun 01 '17 at 11:02

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One consequence of this is that $x \cap \{x\} = \emptyset$. This often confuses people, because it seems counter-intuitive.

But if the sets $x$ and $\{x\}$ have a common element, it must be $x$ itself, since $x$ is the only element of $\{x\}$. And this would lead us to conclude (reluctantly) that $x \in x$. Such a set would have "no foundation", since we'd have to "keep opening up" the set $x$ only to find still another one inside it: $x\in x \in x \in x\dots$

To avoid this, we devised a formula that says (in effect) "the buck stops somewhere". At first, it was believed we might need the set-equivalent of "atomic elements", or ur-elements, primitive objects that belonged to sets, but were not sets themselves. But mathematicians being what they are, found that set theory "made sense" without using ur-elements, and so they by and large abandoned them (why pack more luggage than you need?).

David Wheeler
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