If $x \in |X|$, then $\dim _ x(|X|) = \dim _ x(X)$, and
$\dim (|X|) = \dim (X)$.
Proof. Proof of (1). Choose a scheme $U$ with a point $u \in U$ and an étale morphism $h : U \to X$ mapping $u$ to $x$. By definition the dimension of $X$ at $x$ is $\dim _ u(|U|)$. Thus we may pick $U$ such that $\dim _ x(X) = \dim (|U|)$. Let $d$ be an integer. If $\dim (U) \geq d$, then there exists a sequence of nontrivial specializations $u_ d \leadsto \ldots \leadsto u_0$ in $U$. Taking the image we find a corresponding sequence $h(u_ d) \leadsto \ldots \leadsto h(u_0)$ each of which is nontrivial by Lemma 66.12.1. Hence we see that the image of $|U|$ in $|X|$ has dimension at least $d$. Conversely, suppose that $x_ d \leadsto \ldots \leadsto x_0$ is a sequence of specializations in $|X|$ with $x_0$ in the image of $|U| \to |X|$. Then we can lift this to a sequence of specializations in $U$ by Lemma 66.12.2.
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