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Let $\mathcal{I}\subseteq {\mathcal{O}}_X$ be the ideal sheaf of $Y$. The formal completion of a scheme along $Y$ has underlying topological space $|Y|$ and topological structure sheafThe topology is the adic topology given by the kernels of the projections to the successive infinitesimal neighbourhoods $X_n=(|Y|,{\mathcal{O}}_X/{\mathcal{I}}^{n+1})$. In the locally Noetherian scheme setting of these examples this is a formal scheme. On a Noetherian affine scheme, with $Y=V(I)\subseteq \mathrm{Spec}A$, it is the formal spectrum $\mathrm{Spf}\hat{A}$, where $\hat{A}={\underleftarrow{\lim }}_{n}A/I^{n+1}$. Thus the completion retains all orders of infinitesimal information normal to $Y$.

Here $Z=D(t)=\mathrm{Spec}k[q,t,t^{-1}]$ is an open subscheme of $X$. The relevant affine completions arewith the $q$-adic topology. The complement of $t=0$ in the first formal scheme is its formal open $D(t)$. The sections on this open are the completed localization of an adic ring:The same calculation on every smaller principal open subset identifies the structure sheaves, not just their global sections. Equivalently, every infinitesimal neighbourhood of $Z\cap (q=0)$ is precisely the restriction to $D(t)$ of the corresponding neighbourhood in $X$. By formal completion commutes with restriction to an open subscheme,The isomorphism is canonical and respects the morphisms to $\mathrm{Spf}k[[q]]$.

It would give the wrong answer to replace the formal-open structure sheaf by the ordinary localization of a ring $k[t][[q]][t^{-1}]$. That ring has a uniform bound on the negative powers of $t$ across the coefficients of any one element, whereashas no such bound. This difference between ordinary and completed localization does not obstruct the isomorphism of formal schemes; it explains why the completed localization is essential.