**Proof.**
To prove part (1) choose a surjective étale morphism $V \to Y$ where $V$ is a scheme. For each $v \in V$ choose an algebraic closure $\kappa (v) \subset k_ v$. Consider the morphisms $\overline{x} : \mathop{\mathrm{Spec}}(k_ v) \to V \to Y$. By construction of $|Y|$ these cover $|Y|$.

To prove part (2) we will use the following two facts whose proofs we omit: (i) If $K$ is a field and $\overline{K}$ is algebraic closure then $|\overline{K}| \leq \max \{ \aleph _0, |K|\} $. (ii) For any algebraically closed field $k$ and any cardinal $\aleph $, $\aleph \geq |k|$ there exists an extension of algebraically closed fields $k'/k$ with $|k'| = \aleph $. Now we set $\aleph = \max \{ \lambda (X), \lambda (Y)\} ^+$. Here $\lambda ^+ > \lambda $ indicates the next bigger cardinal, see Sets, Section 3.6. Now (i) implies that the fields $k_ u$ constructed in the first paragraph of the proof all have cardinality bounded by $\lambda (X)$. Hence by (ii) we can find extensions $k_ u \subset k'_ u$ such that $|k'_ u| = \aleph $. The morphisms $\overline{x}' : \mathop{\mathrm{Spec}}(k'_ u) \to X$ cover $|X|$ as desired. To really finish the proof of (2) we need to show that the schemes $\mathop{\mathrm{Spec}}(k'_ u)$ are (isomorphic to) objects of $\mathit{Sch}_{fppf}$ because our conventions are that all schemes are objects of $\mathit{Sch}_{fppf}$; the rest of this paragraph should be skipped by anyone who is not interested in set theoretical considerations. By construction there exists an object $T$ of $\mathit{Sch}_{fppf}$ such that $\lambda (X)$ and $\lambda (Y)$ are bounded by $\text{size}(T)$. By our construction of the category $\mathit{Sch}_{fppf}$ in Topologies, Definitions 34.7.6 as the category $\mathit{Sch}_\alpha $ constructed in Sets, Lemma 3.9.2 we see that any scheme whose size is $\leq \text{size}(T)^+$ is isomorphic to an object of $\mathit{Sch}_{fppf}$. See the expression for the function $Bound$ in Sets, Equation (3.9.1.1). Since $\aleph \leq \text{size}(T)^+$ we conclude.

The notation $X_ s$ in part (3) means the fibre product $\mathop{\mathrm{Spec}}(\kappa (s)) \times _ S X$, where $s \in S$ is the point corresponding to $\overline{s}$. Hence part (2) follows from (4) with $Y = \mathop{\mathrm{Spec}}(\kappa (s))$.

Let us prove (4). Let $X \to Y$ be a morphism of algebraic spaces over $S$. Let $k$ be an algebraically closed field over $S$ of cardinality $> \lambda (X)$. Let $\overline{y} \in Y(k)$ and $x \in |X|$ which map to the same element $y$ of $|Y|$. We have to find $\overline{x} \in X(k)$ mapping to $x$ and $\overline{y}$. Choose a commutative diagram

\[ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } \]

with $U$, $V$ schemes over $S$ and vertical arrows surjective and étale, see Spaces, Lemma 65.11.6. Choose a $u \in |U|$ which maps to $x$, and denote $v \in |V|$ the image. We will think of $u = \mathop{\mathrm{Spec}}(\kappa (u))$ and $v = \mathop{\mathrm{Spec}}(\kappa (v))$ as schemes. Note that $V \times _ Y \mathop{\mathrm{Spec}}(k)$ is a scheme étale over $k$. Hence it is a disjoint union of spectra of finite separable extensions of $k$, see Morphisms, Lemma 29.36.7. As $v$ maps to $y$ we see that $v \times _ Y \mathop{\mathrm{Spec}}(k)$ is a nonempty scheme. As $v \to V$ is a monomorphism, we see that $v \times _ Y \mathop{\mathrm{Spec}}(k) \to V \times _ Y \mathop{\mathrm{Spec}}(k)$ is a monomorphism. Hence $v \times _ Y \mathop{\mathrm{Spec}}(k)$ is a disjoint union of spectra of finite separable extensions of $k$, by Schemes, Lemma 26.23.11. We conclude that the morphism $v \times _ Y \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ has a section, i.e., we can find a morphism $\overline{v} : \mathop{\mathrm{Spec}}(k) \to V$ lying over $v$ and over $\overline{y}$. Finally we consider the scheme

\[ u \times _{V, \overline{v}} \mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(\kappa (u) \otimes _{\kappa (v)} k) \]

where $\kappa (v) \to k$ is the field map defining the morphism $\overline{v}$. Since the cardinality of $k$ is larger than the cardinality of $\kappa (u)$ by assumption we may apply Algebra, Lemma 10.35.12 to see that any maximal ideal $\mathfrak m \subset \kappa (u) \otimes _{\kappa (v)} k$ has a residue field which is algebraic over $k$ and hence equal to $k$. Such a maximal ideal will hence produce a morphism $\overline{u} : \mathop{\mathrm{Spec}}(k) \to U$ lying over $u$ and mapping to $\overline{v}$. The composition $\mathop{\mathrm{Spec}}(k) \to U \to X$ will be the desired geometric point $\overline{x} \in X(k)$. This concludes the proof of part (4).

Part (5) is a formal consequence of parts (2) and (4) and Properties of Spaces, Lemma 66.4.4.
$\square$

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