Example. A moving prime separates two topologies [fcap-004I]

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Use the hyperbolic plane \(Q(x,y)=xy\) and the integral orthogonal references of the example in Definition [fcap-004F]. Enumerate the rational primes without repetition as \(p_1,p_2,\ldots \). Define \(x_n\in O_f(U)\) by \[(x_n)_p=\begin {cases} \operatorname {diag}(p_n,p_n^{-1})&p=p_n,\\ 1&p\ne p_n. \end {cases}\] The exceptional diagonal matrix preserves \(xy\), and every other component is the identity. Thus \(x_n\) belongs to the principal stage \(O_{\{p_n\}}(U)\).

For each fixed prime \(p\), the coordinate \((x_n)_p\) is the identity for all but at most one \(n\). Hence \(x_n\to 1\) in the topology inherited from the unrestricted product \(\prod _pO_p\). Explicitly, a basic product neighborhood of the identity restricts only a finite set \(F\) of coordinates. Once the moving prime has passed every prime in \(F\), all those coordinates equal the identity, so the tuple belongs to that neighborhood.

In contrast, \(x_n\notin K(U)=\prod _pU_p^O\) for every \(n\): the entry \(p_n^{-1}\) is not in \(\mathbb Z_{p_n}\), so its matrix does not preserve the standard integral lattice. Since \(K(U)\) is an open identity neighborhood in the restricted-product topology, the sequence does not converge to the identity there. Coordinate continuity in Definition [fcap-004F] gives one topology inclusion; this sequence proves that the restricted-product topology is strictly finer in this example.

For the classical open-stage topology, see [conrad2011adelic, Section 4 opening, p. 9] and its finite-stage formula in Theorem 3.6. The calculation above isolates what that topology detects: the nonintegral coordinate keeps moving, although every fixed coordinate eventually becomes the identity.