Proof.
It is clear that (1) implies (2) and (2) implies (3). Let us prove that (3) implies (1). Suppose that the base change X_ T \to T is not open for some morphism of schemes g : T \to S. Then we can find some affine opens V \subset S, U \subset X, W \subset T with f(U) \subset V and g(W) \subset V such that U \times _ V W \to W is not open. If we can show that this implies \mathbf{A}^ n \times U \to \mathbf{A}^ n \times V is not open, then \mathbf{A}^ n \times X \to \mathbf{A}^ n \times S is not open and the proof is complete. This reduces us to the result proved in the next paragraph.
Let A \to B be a ring map such that A' \to B' = A' \otimes _ A B does not induce an open map of spectra for some A-algebra A'. As the principal opens give a basis for the topology of \mathop{\mathrm{Spec}}(B') we conclude that the image of D(g) in \mathop{\mathrm{Spec}}(A') is not open for some g \in B'. Write g = \sum _{i = 1, \ldots , n} a'_ i \otimes b_ i for some n, a'_ i \in A', and b_ i \in B. Consider the element h = \sum _{i = 1, \ldots , n} x_ i b_ i in B[x_1, \ldots , x_ n]. Assume that D(h) maps to an open subset under the morphism
\mathop{\mathrm{Spec}}(B[x_1, \ldots , x_ n]) \longrightarrow \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n])
in order to get a contradiction. Then D(h) would map surjectively onto a quasi-compact open U \subset \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n]). Let A[x_1, \ldots , x_ n] \to A' be the A-algebra homomorphism sending x_ i to a'_ i. This also induces a B-algebra homomorphism B[x_1, \ldots , x_ n] \to B' sending h to g. Since
\xymatrix{ \mathop{\mathrm{Spec}}(B[x_1, \ldots , x_ n]) \ar[d] & \mathop{\mathrm{Spec}}(B') \ar[l] \ar[d] \\ \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n]) & \mathop{\mathrm{Spec}}(A') \ar[l] }
is cartesian the image of D(g) in \mathop{\mathrm{Spec}}(A') is equal to the inverse image of U in \mathop{\mathrm{Spec}}(A') and hence open which is the desired contradiction.
\square
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