**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 66.27.5. 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 99.3.3. 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 66.27.5.
$\square$

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