Proof.
Assume (1). In this case the morphism of algebraic spaces \mathcal{X}_ k \to \mathop{\mathrm{Spec}}(k) is locally quasi-finite as a base change of f. Hence |\mathcal{X}_ k| is discrete by Morphisms of Spaces, Lemma 67.27.5. Conversely, assume (2). Pick a surjective smooth morphism V \to \mathcal{Y} where V is a scheme. It suffices to show that the morphism of algebraic spaces V \times _\mathcal {Y} \mathcal{X} \to V is locally quasi-finite, see Properties of Stacks, Lemma 100.3.3. The morphism V \times _\mathcal {Y} \mathcal{X} \to V is locally of finite type by assumption. For any morphism \mathop{\mathrm{Spec}}(k) \to V where k is a field
\mathop{\mathrm{Spec}}(k) \times _ V (V \times _\mathcal {Y} \mathcal{X}) = \mathop{\mathrm{Spec}}(k) \times _\mathcal {Y} \mathcal{X}
has a discrete space of points by assumption. Hence we conclude that V \times _\mathcal {Y} \mathcal{X} \to V is locally quasi-finite by Morphisms of Spaces, Lemma 67.27.5.
\square
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