Let $L$ be a field and $K$ be a subfield of $L.$ Let $V$ be an affine algebraic $K$-variety in $\Bbb A^n (L).$ Then the coordinate ring $K[V]$ is defined by
$$K[V] = \frac {K[X_1,X_2, \dots ,X_n]} {\mathcal I(V)}$$
where $\mathcal I(V)$ is the vanishing ideal of $V.$ Now for any $(x_1,x_2,\dots,x_n) \in V$ set $p_x = \mathcal I_V \left ( \{(x_1,x_2, \dots,x_n) \} \right)$ the set of all functions $\psi \in K[V]$ such that $\psi (x_1,x_2, \dots, x_n) = 0.$ Then $p_x$ is a prime ideal $\ne K[V]$.
We thus have a mapping $\varphi : V \longrightarrow \mathrm {Spec} (K[V])$ defined by $x \mapsto p_x.$ In my book it has been claimed that $\varphi$ is continuous. In order to see it I take a closed subset $A$ of $\mathrm {Spec} (K[V]).$ Then $\exists$ an ideal $I$ of $K[V]$ such that $A=V(I)=\{p \in \mathrm {Spec} (K[V]) : p \supseteq I \},$ the zero set of the ideal $I$ in $\mathrm {Spec} (K[V]).$ Then
$$\begin{align} {\varphi}^{-1} (A) & = \{x \in V : \varphi (x) \in A \}.\\ & = \{x \in V : p_x \in A \}.\\ & = \{x \in V:p_x \supseteq I\}. \end{align}$$
But I don't understand why is the set $\{x \in V:p_x \supseteq I\}$ closed in $V.$ Please help me in this regard.
Thank you very much.
$\{x\in V : p_x \supset I \} = \mathscr V(I)$ since $I\subset p_x \iff x\in \mathscr V(I)$ where if $J$ is an ideal of $K[V]$ then define $\mathscr V(J)= \{p \in V : f(p)= 0 \forall f \in K[V] \}$
Note: This is well defined since elements of $K[V]$ give distinct functions on $V$.