In the transitive closure of $\in$, is the empty set the least element?

Question

Consider a model $(M,\in)$ of ZFC set theory. Consider the transitive closure of $\in$, which I will denote by $R$. Now, it can be proven that $R$ is a strict partial order. In that strict partial order, is $\emptyset$ the least element?

Answer 1

Yes, and you can show that by induction on the rank. We have to show that $\emptyset R x$ for any $x \neq \emptyset$. We do this by induction on $\operatorname{rank}(x)$. Suppose $\operatorname{rank}(x) = \lambda$ and $\emptyset R y$ holds for any $y$ such that $\operatorname{rank}(y) < \lambda$.

Now, by assumption $x$ is non-empty, so let $y \in x$ be any element. This implies $y R x$. As $\operatorname{rank}(y) < \lambda$, we either have $y = \emptyset$ or $\emptyset R y$. By transitivity it follows that $\emptyset R x$.