**Proof.**
Let $X = \bigcup _{i \in I} Z_ i$ be the decomposition of $X$ into its irreducible components. There are finitely many of them (see Lemmas 10.31.3 and 10.31.5). Let $I' = \{ i \mid x \in Z_ i\} $, and let $T = \bigcup _{i \not\in I'} Z_ i$. Then $U = X \setminus T$ is an open subset of $X$ containing the point $x$. The number (2) is $\max _{i \in I'} \dim (Z_ i)$. For any open $W \subset U$ with $x \in W$ the irreducible components of $W$ are the irreducible sets $W_ i = Z_ i \cap W$ for $i \in I'$ and $x$ is contained in each of these. Note that each $W_ i$, $i \in I'$ contains a closed point because $X$ is Jacobson, see Section 10.35. Since $W_ i \subset Z_ i$ we have $\dim (W_ i) \leq \dim (Z_ i)$. The existence of a closed point implies, via Lemma 10.114.4, that there is a chain of irreducible closed subsets of length equal to $\dim (Z_ i)$ in the open $W_ i$. Thus $\dim (W_ i) = \dim (Z_ i)$ for any $i \in I'$. Hence $\dim (W)$ is equal to the number (2). This proves that (1) $ = $ (2).

Let $\mathfrak m \supset \mathfrak p$ be any maximal ideal containing $\mathfrak p$. Let $x_0 \in X$ be the corresponding point. First of all, $x_0$ is contained in all the irreducible components $Z_ i$, $i \in I'$. Let $\mathfrak q_ i$ denote the minimal primes of $S$ corresponding to the irreducible components $Z_ i$. For each $i$ such that $x_0 \in Z_ i$ (which is equivalent to $\mathfrak m \supset \mathfrak q_ i$) we have a surjection

\[ S_{\mathfrak m} \longrightarrow S_\mathfrak m/\mathfrak q_ i S_\mathfrak m =(S/\mathfrak q_ i)_{\mathfrak m} \]

Moreover, the primes $\mathfrak q_ i S_\mathfrak m$ so obtained exhaust the minimal primes of the Noetherian local ring $S_{\mathfrak m}$, see Lemma 10.26.3. We conclude, using Lemma 10.114.4, that the dimension of $S_{\mathfrak m}$ is the maximum of the dimensions of the $Z_ i$ passing through $x_0$. To finish the proof of the lemma it suffices to show that we can choose $x_0$ such that $x_0 \in Z_ i \Rightarrow i \in I'$. Because $S$ is Jacobson (as we saw above) it is enough to show that $V(\mathfrak p) \setminus T$ (with $T$ as above) is nonempty. And this is clear since it contains the point $x$ (i.e. $\mathfrak p$).
$\square$

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