Why Box-Product Continuity Needs a Stronger Test
In a product topology, a basic neighborhood restricts only finitely many coordinates. In a box topology, a basic neighborhood may restrict every coordinate at once. A map can therefore behave continuously in each coordinate and still fail to fit inside a single box neighborhood near a given input. The right test asks whether one domain neighborhood works for all the coordinate restrictions simultaneously.
Let \((X,\mathcal{T})\) be a topological space, and let \((Y_n,\mathcal{T}_n)\), \(n\in\mathbb{N}\), be topological spaces. The underlying set of their product is \(\prod_{n\in\mathbb{N}}Y_n\). In the box topology, a basis consists of all sets \(\prod_{n\in\mathbb{N}}U_n\), where each \(U_n\) is open in \(Y_n\); there is no requirement that only finitely many \(U_n\) differ from \(Y_n\). For a map \(f:X\to\prod_{n\in\mathbb{N}}Y_n\), write \(f_n=\pi_n\circ f\) for its \(n\)th coordinate map. A map is continuous when the preimage of every open set is open.
Proof. Suppose first that \(f\) is continuous. The set \(\prod_n U_n\) is a basic open neighborhood of \(f(x)\). Its preimage is open and contains \(x\), so it can serve as \(V\). For the equivalence in the statement, a point \(z\in X\) satisfies \(f(z)\in\prod_n U_n\) exactly when \(f_n(z)\in U_n\) for every \(n\); thus \(f^{-1}(\prod_n U_n)=\bigcap_n f_n^{-1}(U_n)\). Conversely, suppose the neighborhood condition holds. Let \(O\) be open in the box topology, and let \(x\in f^{-1}(O)\). By the basis definition, some basic box \(\prod_n U_n\) contains \(f(x)\) and is contained in \(O\). The assumed condition gives an open neighborhood \(V\) of \(x\) with \(f(V)\subseteq\prod_n U_n\subseteq O\). Hence every point of \(f^{-1}(O)\) has an open neighborhood contained in \(f^{-1}(O)\), so \(f^{-1}(O)\) is open. This proves continuity. \(\square\)
The criterion is simultaneous: the neighborhood \(V\) must satisfy every coordinate restriction in the chosen box at once. It does not assert that the intersection of infinitely many open sets is automatically open. The examples below show why that distinction matters.
Continuity Forces Coordinate Continuity
Proof. Fix \(n\), and let \(W\) be open in \(Y_n\). The set of all product points whose \(n\)th coordinate lies in \(W\), with no restrictions on the other coordinates, is a box-open set: it has \(n\)th factor \(W\) and every other factor \(Y_m\). Its preimage under \(f\) is \(f_n^{-1}(W)\). Continuity of \(f\) makes this preimage open. Since this holds for every open \(W\), \(f_n\) is continuous. \(\square\)
This direction is useful as a first check: if even one coordinate map is discontinuous, the full map cannot be continuous. But passing every coordinate check is only a necessary condition. The box criterion also requires all coordinate restrictions to be met on one common neighborhood.
Worked Example: A Map with Only Two Varying Coordinates
Give \(\mathbb{R}\) its standard topology and give \(\mathbb{R}^{\mathbb{N}}\) the box topology. Consider \(f:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) defined by \(f(t)=(t,t^2,0,0,\ldots)\). We check continuity by taking an arbitrary basic box \(B=\prod_{n\in\mathbb{N}}U_n\), with each \(U_n\) open in \(\mathbb{R}\).
If \(0\notin U_n\) for even one \(n\ge3\), then no value of \(f\) belongs to \(B\), because the \(n\)th coordinate of every \(f(t)\) is \(0\). In that case \(f^{-1}(B)=\varnothing\), which is open. If \(0\in U_n\) for every \(n\ge3\), the remaining conditions are exactly \(t\in U_1\) and \(t^2\in U_2\). Therefore \(f^{-1}(B)=U_1\cap\{t\in\mathbb{R}:t^2\in U_2\}\). The function \(t\mapsto t^2\) is continuous, so the second set is open; \(U_1\) is open as well. Their intersection is open. Thus every basic box has open preimage, and \(f\) is continuous.
Check with a particular box. Take \(U_1=(-2,2)\), \(U_2=(0,2)\), and \(U_n=(-1,1)\) for \(n\ge3\). This box contains \(f(1)\), since \(1\in(-2,2)\), \(1^2=1\in(0,2)\), and \(0\in(-1,1)\). Its preimage is the set of \(t\) satisfying \(-2<t<2\) and \(0<t^2<2\), namely \((-\sqrt{2},0)\cup(0,\sqrt{2})\). This is open, as the general calculation predicts.
The Diagonal Map: Continuous Coordinates, Discontinuous Map
Worked Example: The Diagonal into the Box Topology
Give \(\mathbb{R}\) its standard topology and \(\mathbb{R}^{\mathbb{N}}\) the box topology. Define \(d:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) by \(d(t)=(t,t,t,\ldots)\). Every coordinate map is the identity \(t\mapsto t\), hence is continuous. We show that \(d\) itself is not continuous.
Consider the box \(B=\prod_{n\in\mathbb{N}}(-1/n,1/n)\). Each interval is open and contains \(0\), so \(B\) is a basic box neighborhood of \(d(0)=(0,0,\ldots)\). A real number \(t\) belongs to \(d^{-1}(B)\) exactly when \(-1/n<t<1/n\) for every \(n\), or equivalently \(|t|<1/n\) for every \(n\). The value \(t=0\) satisfies all these strict inequalities. If \(t\ne0\), choose \(n\in\mathbb{N}\) with \(n>1/|t|\), which is possible by the Archimedean property of the real numbers. Then \(1/n<|t|\), so \(t\) fails the \(n\)th inequality. Consequently \(d^{-1}(B)=\{0\}\).
The singleton \(\{0\}\) is not open in the standard topology on \(\mathbb{R}\): every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and that interval contains nonzero real numbers. Thus the preimage of the open box \(B\) is not open, and \(d\) is discontinuous. This verifies directly that continuous coordinate maps do not suffice for continuity into a box product.
The obstruction is not a discontinuity in any one coordinate. The chosen box demands that the same input \(t\) satisfy narrower and narrower bounds in every coordinate. No neighborhood of \(0\) in \(\mathbb{R}\) can meet all those bounds.
Comparison with the Product Topology
Worked Example: The Same Diagonal into the Product Topology
Keep the standard topology on \(\mathbb{R}\), but now give \(\mathbb{R}^{\mathbb{N}}\) the product topology. Its basic open sets restrict only finitely many coordinates: a basic set has the form \(\prod_n U_n\), where every \(U_n\) is open and \(U_n=\mathbb{R}\) except for finitely many \(n\). For the diagonal map \(d(t)=(t,t,\ldots)\), the preimage of such a basic set is \(\bigcap_{n:U_n\ne\mathbb{R}}U_n\). This is a finite intersection of open subsets of \(\mathbb{R}\), so it is open; if there are no restricted coordinates, the preimage is all of \(\mathbb{R}\), also open.
Every open set in the product topology is a union of basic open sets, and the preimage of a union is the union of the preimages. Since unions of open sets are open, \(d\) is continuous into the product topology. The contrast with the previous example comes from the allowed basic neighborhoods: a product neighborhood imposes only finitely many coordinate conditions, whereas a box neighborhood can impose infinitely many.
A Common Pitfall and a Practical Test
Common pitfall. A tempting but incorrect argument is: “For any box \(\prod_n U_n\), its preimage is \(\bigcap_n f_n^{-1}(U_n)\). Each coordinate map \(f_n\) is continuous, so each set \(f_n^{-1}(U_n)\) is open; therefore their intersection is open.” The last implication is false for an infinite intersection. For the diagonal map and the box \(B=\prod_n(-1/n,1/n)\), each individual preimage is the open interval \((-1/n,1/n)\), but their intersection is \(\{0\}\), which is not open. The correction is to check the simultaneous neighborhood condition, or directly check that every basic box has open preimage; separate coordinate continuity alone does not settle the question.
In practice, first test whether all coordinate maps are continuous. If not, the map is not continuous. If they are, examine an arbitrary box neighborhood, including boxes with infinitely many restricted coordinates. At a point \(x\), the issue is whether there is one open neighborhood \(V\) of \(x\) such that every \(z\in V\) satisfies all the conditions \(f_n(z)\in U_n\). The distinction between “one coordinate at a time” and “all coordinates at once” is the central feature of continuity into a box product.
Check Your Understanding
For each question, keep track of whether the target has the box topology or the product topology.
- State the neighborhood criterion for a map into a box product using the coordinate maps.
- Why does continuity into the box topology imply continuity of each coordinate map?
- For the diagonal map \(d(t)=(t,t,\ldots)\), compute the preimage of \(\prod_n(-2/n,2/n)\) and decide whether that particular preimage is open.
- Give a basic open neighborhood in the product topology on \(\mathbb{R}^{\mathbb{N}}\) that restricts exactly the first three coordinates.
- In the proof for \(f(t)=(t,t^2,0,0,\ldots)\), why is the preimage empty if \(0\notin U_n\) for some \(n\ge3\)?