Proof.
Let u \in U and U \to \mathcal{X} be as in (2). Then \mathop{\mathrm{Spec}}(\kappa (u)) \to U is of finite type, and U \to \mathcal{X} is representable and locally of finite type (by Morphisms of Spaces, Lemmas 67.39.8 and 67.28.5). Hence we see (1) holds by Lemma 101.17.2.
Conversely, assume \mathop{\mathrm{Spec}}(k) \to \mathcal{X} is locally of finite type and represents x. Let U \to \mathcal{X} be a surjective smooth morphism where U is a scheme. By assumption U \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \to U is a morphism of algebraic spaces which is locally of finite type. Pick a finite type point v of U \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) (there exists at least one, see Morphisms of Spaces, Lemma 67.25.3). By Morphisms of Spaces, Lemma 67.25.4 the image u \in U of v is a finite type point of U. Hence by Morphisms, Lemma 29.16.4 after shrinking U we may assume that u is a closed point of U, i.e., (2) holds.
\square
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