**Proof.**
Assume $f$ is quasi-finite at $x$. By assumption there exist opens $U \subset X$, $V \subset S$ such that $f(U) \subset V$, $x \in U$ and $x$ an isolated point of $U_ s$. Hence $\{ x\} \subset U_ s$ is an open subset. Since $U_ s = U \cap X_ s \subset X_ s$ is also open we conclude that $\{ x\} \subset X_ s$ is an open subset also. Thus we conclude that $x$ is an isolated point of $X_ s$.

Note that $X_ s$ is a Jacobson scheme by Lemma 29.16.10 (and Lemma 29.15.4). If $x$ is isolated in $X_ s$, i.e., $\{ x\} \subset X_ s$ is open, then $\{ x\} $ contains a closed point (by the Jacobson property), hence $x$ is closed in $X_ s$. It is clear that there is no point $x' \in X_ s$, distinct from $x$, specializing to $x$.

Assume that $x$ is closed in $X_ s$ and that there is no point $x' \in X_ s$, distinct from $x$, specializing to $x$. Consider a pair of affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$, $\mathop{\mathrm{Spec}}(R) = V \subset S$ with $f(U) \subset V$ and $x \in U$. Let $\mathfrak q \subset A$ correspond to $x$ and $\mathfrak p \subset R$ correspond to $s$. By Lemma 29.15.2 the ring map $R \to A$ is of finite type. Consider the fibre ring $\overline{A} = A \otimes _ R \kappa (\mathfrak p)$. Let $\overline{\mathfrak q}$ be the prime of $\overline{A}$ corresponding to $\mathfrak q$. Since $\mathop{\mathrm{Spec}}(\overline{A})$ is an open subscheme of the fibre $X_ s$ we see that $\overline{q}$ is a maximal ideal of $\overline{A}$ and that there is no point of $\mathop{\mathrm{Spec}}(\overline{A})$ specializing to $\overline{\mathfrak q}$. This implies that $\dim (\overline{A}_{\overline{q}}) = 0$. Hence by Algebra, Definition 10.121.3 we see that $R \to A$ is quasi-finite at $\mathfrak q$, i.e., $X \to S$ is quasi-finite at $x$ by definition.

At this point we have shown conditions (1) – (3) are all equivalent. It is clear that (4) implies (1). And it is also clear that (2) implies (4) since if $x$ is an isolated point of $X_ s$ then it is also an isolated point of $U_ s$ for any open $U$ which contains it.
$\square$

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