Exercise 5, § 1

This solution was written by laguna-s-2.1-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

To show that $\emptyset = \tau_X(\tau_x(x \in X) \notin X)$, we work under contradiction. Let $Y = \tau_X(\tau_x(x \in X) \notin X)$. By definition of $\tau$, $Y$ is a term satisfying $(\tau_x(x \in Y)) \notin Y$. Suppose $\overline{\vdash} Y = \emptyset$, so there exists $z \in Y$. Then $\tau_x(x \in Y)$ is an element of $Y$ (since $Y$ is non-empty, by the properties of $\tau$ for collectivizing relations). But the defining condition of $Y$ requires that $\tau_x(x \in Y) \notin Y$, contradicting $z \in Y$. Hence no such $z$ exists, so $(\forall x)(x \notin Y)$. By Theorem 1 (no. 7), this implies $Y = \emptyset$.

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