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The lecture slides for my course define the Zermelo-Fraenkel axiom of extensionality as follows:

Slide 1:

Extensionality: x and y have the same elements

Slide 2:

∀x[x ∈ y ⇔ x ∈ z] iff x and y are members of the same sets

Slide 3:

∀x[x ∈ y IFF x ∈ z] iff ∀x[y ∈ x IFF z ∈ x]

I understand that "$∀x.[x ∈ y ⇔ x ∈ z]$" means $z$ and $y$ are the same set, but why does the formula in slide 3 implies that $x$ and $y$ are members of the same set?

I thought maybe it was a typo and they meant to say "z and y" instead of "x and y", but even then, I don't really understand why the right-hand side of the formula ($∀x[y ∈ x ⇔ z ∈ x]$) is needed. It does not appear in the definition given in the coursebook:

Extensionality. Two sets are equal if they have the same members. $$(∀z, z ∈ x \ \ IFF \ \ z ∈ y) ⇒ x = y.$$

user51462
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1 Answers1

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Extensionality is the last formula of your post: $∀z(z∈a \leftrightarrow z∈b) \to a=b$ and reads: "if sets $a$ and $b$ have the same elements, they are equal".

To assert that "$a$ and $b$ are members of the same sets" we need: $\forall z (a \in z \leftrightarrow b \in za)$. As you can easily verify, the formula is different from 2) above.

Using Substitution axiom for equality [ in the form: $a=b \to (\varphi[a/w] \to \varphi [b/w])$ with formula $(w \in z)$ as $\varphi$], we derive: $a=b \to (a \in z \to b \in z)$.

Using $a=b \to b=a$ and Generalization we conclude with:

$a=b \to ∀z(a \in z \leftrightarrow b \in z)$.

Using Substitution axiom again we derive: $a=b \to ∀z(z∈a \leftrightarrow z∈b)$ that together with Extensionality produces:

$∀z(z∈a \leftrightarrow z∈b) \leftrightarrow a=b$.

Thus, the correct conclusion is:

$∀z(z∈a \leftrightarrow z∈b) \to ∀z(a \in z \leftrightarrow b \in z)$.

The gist is, provided the usual logical axioms for equality, that Extensionality implies that if two sets have the same elements, they are members of the same sets.

IMO we need some additonal principle, like e.g. Pair axiom, to conclude from $∀z(a \in z \leftrightarrow b \in z)$, by contradiction, that: $∀z(z∈a \leftrightarrow z∈b)$.

See also Equality in set theory.

  • Thank you @Mauro, I see now that the axiom in the slides is actually the definition of equality. May I ask why you used $⇒$ instead of $⇔$ (the second "iff" in slide 3)? Also, I understand that $x$, $y$ and $z$ are all sets, but I don't quite understand what it means for $z$ to be an element of $x$ and $y$ ($∀z(z∈a↔z∈b)$) while simultaneously containing $x$ and $y$ ($∀z(a∈z↔b∈z)$). If possible, could you please provide an example that makes this clearer? – user51462 Jul 22 '22 at 13:21
  • @user51462 - "May I ask why you used ⇒ instead of ⇔ (the second "iff" in slide 3)?" Because I'm not able to derive the other part of the biconditional :-) Maybe you can check on your slides. – Mauro ALLEGRANZA Jul 22 '22 at 13:22
  • The two "x and y" do not necessarily are the same sets... Example: the set of Even numbers and the set of Multiple of 2 are the same set of naturals, because they contain the same numbers. This implies that every set that contains the set of Even will also contain the set of Mult of 2. – Mauro ALLEGRANZA Jul 22 '22 at 13:23