**Proof.**
(1) implies (2) implies (3) implies (4). We will now show (4) implies (1). As in the proof of Lemma 32.15.2, we can reduce to the case $S$ is affine, since properness is local on the base, and if $X \to S$ satisfies (4), then $X_\alpha \to S_\alpha $ does as well for open $S_\alpha \subset S$ and $X_\alpha = f^{-1}(S_\alpha )$.

Now $S$ is a Noetherian scheme, and so $X$ is as well, since $X \to S$ is of finite type. Now we may use Chow's lemma (Cohomology of Schemes, Lemma 30.18.1) to get a surjective, proper, birational $X' \to X$ and an immersion $X' \to \mathbf{P}^ n_ S$. We wish to show $X \to S$ is universally closed. As in the proof of Lemma 32.14.3, it is enough to check that $X' \to \mathbf{P}^ n_ S$ is a closed immersion. For the sake of contradiction, assume that $X' \to \mathbf{P}^ n_ S$ is not a closed immersion. Then there is some $y \in \mathbf{P}^ n_ S$ that is in the closure of the image of $X'$, but is not in the image. So $y$ is in the closure of the image of an irreducible component $X_0'$ of $X'$, but not in the image. Let $\bar X_0' \subset \mathbf{P}^ n_ S$ be the closure of the image of $X_0'$. As $X' \to \mathbf{P}^ n_ S$ is an immersion of Noetherian schemes, the morphism $X'_0 \to \bar X_0'$ is open and dense. By Algebra, Lemma 10.118.13 or Properties, Lemma 28.5.10 we can find a discrete valuation ring $A$ dominating $\mathcal{O}_{\bar X_0', y}$ and with identical field of fractions $K$. It is clear that $K$ is the residue field at the generic point of $X_0'$. Thus the solid commutative diagram

32.15.3.1
\begin{equation} \label{limits-equation-solid} \xymatrix{ \mathop{\mathrm{Spec}}K \ar[r] \ar[d] & X' \ar[r] \ar[d] & \mathbf{P}^ n_ S \ar[d] \\ \mathop{\mathrm{Spec}}A \ar@{-->}[r] \ar@{-->}[ru] \ar[urr] & X \ar[r] & S\\ } \end{equation}

Note that the closed point of $A$ maps to $y \in \mathbf{P}^ n_ S$. By construction, there does not exist a set theoretic lift to $X'$. As $X' \to X$ is birational, the image of $X'_0$ in $X$ is an irreducible component $X_0$ of $X$ and $K$ is also identified with the function field of $X_0$. Hence, as $X \to S$ is assumed to satisfy (4), the dotted arrow $\mathop{\mathrm{Spec}}(A) \to X$ exists. Since $X' \to X$ is proper, the dotted arrow lifts to the dotted arrow $\mathop{\mathrm{Spec}}(A) \to X'$ (use Schemes, Proposition 26.20.6). We can compose this with the immersion $X' \to \mathbf{P}^ n_ S$ to obtain another morphism (not depicted in the diagram) from $\mathop{\mathrm{Spec}}(A) \to \mathbf{P}^ n_ S$. Since $\mathbf{P}^ n_ S$ is proper over $S$, it satisfies (2), and so these two morphisms agree. This is a contradiction, for we have constructed the forbidden lift of our original map $\mathop{\mathrm{Spec}}(A) \to \mathbf{P}^ n_ S$ to $X'$.
$\square$

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