= Solution
The <spectrum of a ring> $A$ is the set of its proper <prime ideals>. For an <ideal> $I$, set $V(I)=\{\mathfrak p:I\subseteq\mathfrak p\}$. These are the closed sets of the <Zariski topology>. They satisfy
$$
V(0)=\operatorname{Spec}A,\qquad V(A)=\varnothing,\qquad
\bigcap_\lambda V(I_\lambda)=V\!\left(\sum_\lambda I_\lambda\right),\qquad
V(I)\cup V(J)=V(IJ).
$$
The union identity follows because a <prime ideal> containing $IJ$ must contain $I$ or $J$: otherwise choose $a\in I\setminus\mathfrak p$ and $b\in J\setminus\mathfrak p$, contradicting $ab\in\mathfrak p$. These identities verify the closed-set axioms. The <maximal spectrum> has the subspace topology, whose closed sets are $V(I)\cap\operatorname{MaxSpec}A$.
For a finitely generated $\mathbb C$-algebra, the needed form of the <Hilbert Nullstellensatz> is
$$
\boxed{\sqrt I=\bigcap_{\substack{\mathfrak m\supseteq I\\\mathfrak m\ \mathrm{maximal}}}\mathfrak m.}
$$
Here an empty intersection means $A$. We prove this formula rather than assume the desired correspondence.
First establish the residue-field fact. If a <field> $L$ is a <finitely generated algebra> over $\mathbb C$, choose a maximal algebraically independent subset $t_1,\ldots,t_r$ of a finite generating set, with remaining generators $u_1,\ldots,u_s$. Each $u_j$ is algebraic over $\mathbb C(t_1,\ldots,t_r)$. Clearing the finitely many denominators in their monic equations gives a nonzero polynomial $d$ such that each $u_j$ is integral over
$$
R=\mathbb C[t_1,\ldots,t_r,1/d].
$$
Since $L$ is a <field>, it contains $1/d$, and $L=R[u_1,\ldots,u_s]$ is an <integral extension> of $R$. If a <field> is integral over a domain, the domain is itself a <field>: for $0\ne a\in R$, an integral equation for $a^{-1}$, multiplied by $a^{n-1}$, expresses $a^{-1}$ in $R$.
But if $r>0$, choose $c\in\mathbb C$ for which $t_1-c$ does not divide $d$. Such a choice exists because a nonzero polynomial has only finitely many distinct factors $t_1-c$. The element $t_1-c$ is a nonunit even after inverting $d$: an inverse would imply that $t_1-c$ divides a power of $d$. Thus $R$ would not be a <field>, a contradiction. Hence $r=0$, and $L$ is algebraic over the <algebraically closed field> $\mathbb C$, forcing $L=\mathbb C$. This proves the <Zariski lemma> in the form used here.
Now if $f\notin\sqrt I$, the <localization of a ring> $B=(A/I)[1/f]$ is nonzero: vanishing would mean $f^n\in I$ for some $n$. It is still a <finitely generated algebra> over $\mathbb C$, since one need only adjoin an inverse of $f$. Choose a <maximal ideal> $\mathfrak n$ of $B$. By the preceding argument, $B/\mathfrak n=\mathbb C$. The composite $A\to B\to\mathbb C$ is surjective because it contains the constant scalars. Its <kernel> is consequently a <maximal ideal> $\mathfrak m$ of $A$, contains $I$, and does not contain $f$, whose image is invertible. This proves that the displayed intersection is contained in $\sqrt I$. The reverse inclusion follows because every <maximal ideal> is prime and contains the <radical of an ideal> whenever it contains that ideal.
For a closed subset $Z\subseteq\operatorname{MaxSpec}A$, put $I(Z)=\bigcap_{\mathfrak m\in Z}\mathfrak m$. It is a <radical ideal>. The formulas now show
$$
I\longmapsto V(I)\cap\operatorname{MaxSpec}A,\qquad
Z\longmapsto I(Z)
$$
are inverse maps on <radical ideals> and closed sets. Indeed the first followed by the second gives $\sqrt I=I$, while if $Z=V(J)\cap\operatorname{MaxSpec}A$, the second gives $\sqrt J$ and $V(\sqrt J)=V(J)$. Thus \b[the correspondence is a bijection reversing inclusion]. This is the relation between <radical ideals and closed subsets of the maximal spectrum>.
For $A=\mathbb C[x]$, the <maximal ideals> are $\mathfrak m_c=(x-c)$, for $c\in\mathbb C$. Every nonzero proper <radical ideal> is generated by a squarefree polynomial, and hence has the form
$$
I_S=\left(\prod_{c\in S}(x-c)\right)
$$
for a finite nonempty subset $S\subset\mathbb C$. The associated closed set is $\{\mathfrak m_c:c\in S\}$. A repeated factor of a polynomial does not change its zero set, but passing to its <radical ideal> removes that repeated factor. The two endpoint cases are
$$
\boxed{(0)\longleftrightarrow\mathbb C,\qquad
(1)\longleftrightarrow\varnothing,\qquad
I_S\longleftrightarrow S\text{ for finite }S.}
$$
The empty-product convention makes $I_\varnothing=(1)$. Under $c\mapsto\mathfrak m_c$, the closed sets are exactly the finite subsets and the whole affine line.
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