**Proof.**
It is clear that (1) implies (2). Let us prove that (2) implies (1). Suppose that the base change $X_ T \to T$ is not closed for some scheme $T$ over $S$. By Schemes, Lemma 26.19.8 this means that there exists some specialization $t_1 \leadsto t$ in $T$ and a point $\xi \in X_ T$ mapping to $t_1$ such that $\xi $ does not specialize to a point in the fibre over $t$. Set $Z = \overline{\{ \xi \} } \subset X_ T$. Then $Z \cap X_ t = \emptyset $. Apply Lemma 32.14.1. We find an open neighbourhood $V \subset T$ of $t$, a commutative diagram

\[ \xymatrix{ V \ar[d] \ar[r]_ a & T' \ar[d]^ b \\ T \ar[r]^ g & S, } \]

and a closed subscheme $Z' \subset X_{T'}$ such that

the morphism $b : T' \to S$ is locally of finite presentation,

with $t' = a(t)$ we have $Z' \cap X_{t'} = \emptyset $, and

$Z \cap X_ V$ maps into $Z'$ via the morphism $X_ V \to X_{T'}$.

Clearly this means that $X_{T'} \to T'$ maps the closed subset $Z'$ to a subset of $T'$ which contains $a(t_1)$ but not $t' = a(t)$. Since $a(t_1) \leadsto a(t) = t'$ we conclude that $X_{T'} \to T'$ is not closed. Hence we have shown that $X \to S$ not universally closed implies that $X_{T'} \to T'$ is not closed for some $T' \to S$ which is locally of finite presentation. In order words (2) implies (1).

Assume that $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is closed for every integer $n$. We want to prove that $X_ T \to T$ is closed for every scheme $T$ which is locally of finite presentation over $S$. We may of course assume that $T$ is affine and maps into an affine open $V$ of $S$ (since $X_ T \to T$ being a closed is local on $T$). In this case there exists a closed immersion $T \to \mathbf{A}^ n \times V$ because $\mathcal{O}_ T(T)$ is a finitely presented $\mathcal{O}_ S(V)$-algebra, see Morphisms, Lemma 29.21.2. Then $T \to \mathbf{A}^ n \times S$ is a locally closed immersion. Hence we get a cartesian diagram

\[ \xymatrix{ X_ T \ar[d]_{f_ T} \ar[r] & \mathbf{A}^ n \times X \ar[d]^{f_ n} \\ T \ar[r] & \mathbf{A}^ n \times S } \]

of schemes where the horizontal arrows are locally closed immersions. Hence any closed subset $Z \subset X_ T$ can be written as $X_ T \cap Z'$ for some closed subset $Z' \subset \mathbf{A}^ n \times X$. Then $f_ T(Z) = T \cap f_ n(Z')$ and we see that if $f_ n$ is closed, then also $f_ T$ is closed.
$\square$

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