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I was reading through the ZF axioms and got to the axiom of foundation which my textbook defined as: $$\forall x(x\neq\emptyset\to\exists y\in x(y\cap x=\emptyset)).$$ Which I found quite confusing. I think I understand what it's saying but the $\exists y$ there feels slightly weird, wouldn't we want it to be $\forall y\in x$?

Constructing a set $A=\{A,\emptyset\}$ isn't disallowed with the axiom (since there exists a $y\in x$ such that $y$ is disjoint from $x$) which undermines its purpose? Or am I understanding it wrong?

Reverie
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  • Sorry, by accident I pushed the wrong button. For completeness let me repeat: If $A={A,\varnothing}$ then $A\in A$. But then the set ${A}$ has no $\epsilon$-minimal element (which cannot happen under the axiom). – drhab Jun 03 '24 at 06:51
  • $[\forall x(x\ne \emptyset\to \forall y\in x(y\cap x=\emptyset))]$ is in contradiction with [empty set + axiom of pair], because $x={{\emptyset},\emptyset}$ is a counterexample. – Sassatelli Giulio Jun 03 '24 at 06:57
  • $[\exists A, A={A,\emptyset}]$ implies $[\exists A, A\in A]$, which is in contradiction with [foundation + pair] see here. – Sassatelli Giulio Jun 03 '24 at 09:28

2 Answers2

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No, because $\{\{\varnothing\},\varnothing\}$ is an example of a set which has an element that has a non empty intersection.

Asaf Karagila
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The idea behind the axiom of foundation is that if you had a set violating it, i.e., a non-empty set $A$ such that every member of $A$ also contains a member of $A$, then you can

  • pick $x_0 \in A$ (it is non-empty),
  • because $x_0 \in A$, pick $x_1 \in x_0$ such that $x_1 \in A$,
  • repeating, pick $x_2 \in x_1$ such that $x_2 \in A$
  • and so on

building an infinite sequence

$x_0 \ni x_1 \ni x_2 \ni \dots$

In your hypothetical example $A = \{A, \emptyset\}$, the infinite sequence that could be built is

$A \ni A \ni A \ni \dots$

The set of elements of this sequence is the singleton $\{A\}$. And if you apply the axiom of regularity to $\{A\}$, you get a contradiction, because the only member of $\{A\}$ is $A = \{A\}$, which intersects $\{A\}$ (they have $A$ as a common member).

So, the problem with your initial reasoning is that rather than $A$, you should apply the axiom to $\{A\}$ to get a contradiction.

Jean Abou Samra
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