Lemma 33.25.5. Let $k$ be a field. Let $d \geq 0$. Let $W \subset \mathbf{A}^ d_ k$ be nonempty open. Then there exists a closed point $w \in W$ such that $k \subset \kappa (w)$ is finite separable.
Proof. After possible shrinking $W$ we may assume that $W = \mathbf{A}^ d_ k \setminus V(f)$ for some $f \in k[x_1, \ldots , x_ d]$. If the lemma is wrong then $f(a_1, \ldots , a_ d) = 0$ for all $(a_1, \ldots , a_ d) \in (k^{sep})^ d$. This is absurd as $k^{sep}$ is an infinite field. $\square$
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