[proofplan] The correspondence is the analytic Nullstellensatz sheafified. A reduced analytic set is recovered from the ideal of functions vanishing on it, and a radical coherent ideal is recovered from its zero set. Inclusion of sets reverses inclusion of ideals, giving the order-reversing property. [/proofplan] [step:From a reduced analytic set to a radical coherent ideal] Let $V\subseteq X$ be a closed reduced analytic subset. Its ideal sheaf $\mathcal{I}(V)$ consists of germs of holomorphic functions vanishing on $V$. Cartan's coherence theorem for analytic ideals gives coherence. For each $p\in X$, the quotient $\mathcal{O}_{X,p}/\mathcal{I}(V)_p$ is the local ring of the reduced analytic germ $V_p$, so it has no nilpotents. Hence $\mathcal{I}(V)_p$ is radical for every $p$, and $\mathcal{I}(V)$ is a radical coherent ideal sheaf. [/step] [step:From a radical coherent ideal to a reduced analytic set] Let $\mathfrak a\subset\mathcal{O}_X$ be a radical coherent ideal sheaf. Coherence implies that locally $\mathfrak a$ is generated by finitely many holomorphic functions $f_1,\dots,f_r$. Their common zero set defines a closed analytic subset $V(\mathfrak a)$. Since each stalk $\mathfrak a_p$ is radical, the local quotient $\mathcal{O}_{X,p}/\mathfrak a_p$ is reduced; therefore $V(\mathfrak a)$ is a reduced analytic set. [/step] [step:Use the analytic Nullstellensatz] For any coherent ideal sheaf $\mathfrak a$, the analytic Nullstellensatz gives \begin{align*} \mathcal{I}(V(\mathfrak a))=\sqrt{\mathfrak a}. \end{align*} If $\mathfrak a$ is radical, this becomes $\mathcal{I}(V(\mathfrak a))=\mathfrak a$. Conversely, for a reduced analytic set $V$, the zero set of its vanishing ideal is $V$ itself and the ideal is radical, so $V(\mathcal{I}(V))=V$. [/step] [step:Check order reversal] If $V_1\subseteq V_2$, then every holomorphic germ vanishing on $V_2$ also vanishes on $V_1$, so $\mathcal{I}(V_2)\subseteq\mathcal{I}(V_1)$. Similarly, if $\mathfrak a\subseteq\mathfrak b$, then $V(\mathfrak b)\subseteq V(\mathfrak a)$. Thus both maps reverse inclusions, and the previous step shows they are inverse bijections. [/step]