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Let $X$ be a locally finite type $k$-scheme. Let $p$ be a closed point. We define the dimension of $X$ at $p$ to be the largest dimension of an irreducible component of $X$ that contains $p$, denoted $\dim_p X$. Then the function $\phi:p \mapsto \dim_p X$ is upper semicontinuous. To see this it is enough to show that for $n$ any positive integer $\phi^{-1}[0, n)$ is an open set. Indeed, $\phi^{-1}[0, n)$ is just the set of closed points of $X$ minus those closed points that lie on irreducible components of dimension $\ge n$. If there are finitely many such irreducible components their union is a closed set and the inverse image is open. But what if there are infinitely many of them? (Reference: Definition 11.2.I, Vakil November 2017)

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  • $\begingroup$ Doesn't locally finite type imply only finitely many irreducible components through any point? $\endgroup$ – Mohan Oct 2 '19 at 13:11
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Lemma: For any point $x$ in any locally noetherian scheme $X$, there exists a neighborhood of $x$ which intersects only finitely many irreducible components of $X$.

Proof: Take a noetherian affine open neighborhood $U$ of $x$. As $\mathcal{O}_X(U)$ is a noetherian ring, it has finitely many minimal primes, so $U$ has finitely many irreducible components. As every irreducible component of $X$ intersects $U$ in an irreducible component or misses $U$ entirely, we have the claim.

To apply this to the situation at hand, $X$ is locally noetherian as any scheme locally of finite type over a locally noetherian scheme is again locally noetherian. So there's a neighborhood of $p\in X$ which only has finitely many irreducible components, so only finitely many of them can meet $p$ and your problem is solved.

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