**Proof.**
Conditions (4) and (5) are equivalent for any algebraic space by definition, see Properties of Spaces, Definition 64.10.2. Observe that any $Y$ as in (2) and (3) is decent by Lemma 66.6.6. Thus it suffices to prove the equivalence of (1) and (4) as then the equivalence with (2) and (3) follows since the dimension of the local ring of $Y$ at $y$ is equal to the dimension of the local ring of $X$ at $x$. Let $f : U \to X$ be an étale morphism from an affine scheme and let $u \in U$ be a point mapping to $x$.

Assume (1). Let $u' \leadsto u$ be a specialization in $U$. Then $f(u') = f(u) = x$. By Lemma 66.12.1 we see that $u' = u$. Hence $u$ is a generic point of an irreducible component of $U$. Thus $\dim (\mathcal{O}_{U, u}) = 0$ and we see that (4) holds.

Assume (4). The point $x$ is contained in an irreducible component $T \subset |X|$. Since $|X|$ is sober (Proposition 66.12.4) we $T$ has a generic point $x'$. Of course $x' \leadsto x$. Then we can lift this specialization to $u' \leadsto u$ in $U$ (Lemma 66.12.2). This contradicts the assumption that $\dim (\mathcal{O}_{U, u}) = 0$ unless $u' = u$, i.e., $x' = x$.
$\square$

## Comments (0)