I see it stated as something very clear, but i can't figure it out. i found a proof in Jech(old version) which goes through the concept of restricted formulas, which i don't quite understand. (i'm not sure i understand the notation in his Lemma 10.1, the statement numbered 10.5) Is there a way to show it directly?
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$\begingroup$ ok, i edited, thx $\endgroup$– Ur Ya'arCommented Feb 24, 2013 at 11:22
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2 Answers
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Let $\mathbf{M}$ be transitive, and suppose $x , y \in \mathbf{M}$ are distinct. Then, by extensionality in $V$, there must be (in $V$) a $u$ which belongs to either $x \setminus y$ or $y \setminus x$. In either case the transitivity of $\mathbf{M}$ implies that $u \in \mathbf{M}$, and so $\mathbf{M} \models \neg ( u \in x \leftrightarrow u \in y )$, or, more meaningfully $\mathbf{M} \models \neg ( \forall u ) ( u \in x \leftrightarrow u \in y )$. Therefore $$\mathbf{M} \models ( \forall x ) ( \forall y ) ( x \neq y \rightarrow \neg ( \forall u ) ( u \in x \leftrightarrow u \in y ))$$ which (logically equivalent to) the Axiom of Extensionality in Jech's text.
answered Feb 24, 2013 at 10:45
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$\begingroup$ I'm confused, doesn't the statement " there must be a u which belongs to either x∖y or y∖x" rely on the Axiom of Extentionality? When you say that if x and y are distinct then there is an element which is in one but not in the second, you're assuming that they are distinguished by their elements, which is what the axiom say... $\endgroup$– Ur Ya'arCommented Feb 24, 2013 at 11:05
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$\begingroup$ @UrBen-Ari-Tishler: Yes and no. The axioms of set theory are meant to reflect our intuitive notions about what sets are and what we can do with them. Perhaps the most basic intuitive notion about sets is that a set is completely determined by its elements, which is a non-technical way of expressing Extensionality. So really the Axiom of Extensionality is a description of the Universe of Sets (non-Platonism be damned!). [cont...] $\endgroup$ Commented Feb 24, 2013 at 11:30
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1$\begingroup$ @UrBen-Ari-Tishler: Well, you have to start somewhere. But note that we can look at this all purely formally, and note that the class $\mathbf M$ is really hiding a formula $\varphi (x,p)$ of the language of set theory, and saying "$\mathbf M$ satisfies the Axiom of Extensionality" is just a more natural way of saying that "the relativization of the Axiom of Extensionality to $\mathbf M$" is a theorem of ZFC." $\endgroup$ Commented Feb 24, 2013 at 12:43
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1$\begingroup$ Second, @AnduinWilde, $x=y$ is not defined in that way - equivalence is a primitive notion, stating that $x,y$ simply denote the same element. In the context of set theory we want to have extensionality - that the = relation is characterized using the $\in$ relation. What needs to be shown is that the model $M$ also "knows" that this characterization holds. $\endgroup$– Ur Ya'arCommented Jan 2, 2022 at 10:30
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1$\begingroup$ @AnduinWilde Well I admit I don't know what were the convensions in the early history of ZFC. Currently, in the two main textbooks of set theory (Kunen and Jech) the axiom of extensionality is indeed stated with only one direction and not with iff. $\endgroup$– Ur Ya'arCommented Jan 3, 2022 at 20:25
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The $\in$-relation of transitive models is the true $\in$ of the universe, and we know that the universe satisfies extensionality.
Now suppose $M$ is a transitive model, and that $x,y\in M$. By transitivity we know that $x,y\subseteq M$ as well. Therefore $M$ knows all of their elements. If $x\neq y$ then without loss of generality there is some $z\in x\setminus y$, but $M$ knows about $z$ and since $M$ agrees with the universe about $\in$, $M$ also knows that $z\in x\setminus y$.
So whenever $M$ is transitive and $x,y\in M$ are distinct elements, $M\models x\neq y$.
answered Feb 24, 2013 at 15:31