Continuity Into a Product: The Coordinate Test
A function into a product space has one coordinate function for each factor. The product topology is designed so that these coordinate functions provide a complete test for continuity: continuity of the whole map is equivalent to continuity in every coordinate. The key point is that basic open sets in the product restrict only finitely many coordinates.
Let \((X,\mathcal{T})\) be a topological space, let \((X_\alpha,\mathcal{T}_\alpha)\) be topological spaces indexed by a set \(J\), and give \(\prod_{\alpha\in J}X_\alpha\) the product topology. A basic open set in this topology is a finite intersection of sets of the form \(\pi_\alpha^{-1}(U_\alpha)\), where \(U_\alpha\in\mathcal{T}_\alpha\) and \(\pi_\alpha\) is the projection onto the \(\alpha\)-coordinate. Such a set is called a cylinder restriction. Equivalently, a basic open set imposes open conditions on finitely many coordinates and leaves all other coordinates unrestricted.
Projections and Cylinder Sets
The cylinder sets are open by the definition of the product topology. In particular, each projection is continuous.
Proof. Let \(U\) be any open subset of \(X_\alpha\). By the definition of a cylinder restriction, \(\pi_\alpha^{-1}(U)\) is open in the product topology. Thus the preimage of every open set under \(\pi_\alpha\) is open, which is precisely the definition of continuity. \(\square\)
This proposition gives one direction of the coordinate test: if \(f\) is continuous, then composing \(f\) with a projection gives a continuous coordinate map. The converse depends on the finite-intersection structure of basic open sets.
Proof. First suppose \(f\) is continuous. Fix \(\alpha\in J\), and let \(U\) be open in \(X_\alpha\). The projection proposition shows that \(\pi_\alpha^{-1}(U)\) is open in the product. Therefore \(f_\alpha^{-1}(U)=f^{-1}(\pi_\alpha^{-1}(U))\) is open in \(X\), by continuity of \(f\). Since this holds for every open \(U\), \(f_\alpha\) is continuous.
Conversely, suppose every \(f_\alpha\) is continuous. Consider a basic open set \(B=\bigcap_{k=1}^{n}\pi_{\alpha_k}^{-1}(U_{\alpha_k})\), where \(n\) is finite and each \(U_{\alpha_k}\) is open in \(X_{\alpha_k}\). The preimage under \(f\) is \(f^{-1}(B)=\bigcap_{k=1}^{n}f_{\alpha_k}^{-1}(U_{\alpha_k})\). Each set in this intersection is open in \(X\), since \(f_{\alpha_k}\) is continuous. A finite intersection of open sets is open, so \(f^{-1}(B)\) is open. Every open set in the product is a union of basic open sets; its preimage is therefore a union of open sets and is open. Hence \(f\) is continuous.
If \(J\) is empty, the product is a one-point space. There are no coordinate conditions, and every map into a one-point space is continuous because its only open sets are the empty set and the whole space. Thus the theorem also holds in this case. If a factor is empty, a map into the product can exist only when its domain is empty; for any map that does exist, the proof above applies. \(\square\)
Applying the Theorem
Worked Example: A Map into the Plane
Give \(\mathbb{R}\) and both factors of \(\mathbb{R}\times\mathbb{R}\) their standard topologies. Define \(f:\mathbb{R}\to\mathbb{R}\times\mathbb{R}\) by \(f(t)=(t^2,3t-1)\). The coordinate maps \(t\mapsto t^2\) and \(t\mapsto 3t-1\) are continuous real functions, so the coordinatewise theorem proves that \(f\) is continuous.
We can see the finite-intersection calculation directly on the basic open rectangle \(U=(-1,4)\times(2,8)\). Its preimage is determined by both coordinate conditions:
For the first condition, \(t^2\geq 0>-1\), while \(t^2<4\) is equivalent to \(-2<t<2\). For the second, adding \(1\) gives \(3<3t<9\), and dividing by \(3\) gives \(1<t<3\). The intersection is therefore \((1,2)\). The endpoints are excluded: at \(t=1\), the second coordinate is \(2\), and at \(t=2\), the first coordinate is \(4\), neither of which belongs to the corresponding open interval. As a check, \(t=3/2\) gives \(f(t)=(9/4,7/2)\), and \(9/4\in(-1,4)\) and \(7/2\in(2,8)\).
Worked Example: The Same Rule for Many Factors
Let \(J\) be any set with at least two elements, give every factor \(\mathbb{R}\) its standard topology, and give \(\mathbb{R}^J=\prod_{j\in J}\mathbb{R}\) the product topology. Define \(f:\mathbb{R}\to\mathbb{R}^J\) by \(f(t)_j=t\) for every \(j\in J\). For each \(j\), the coordinate map is the identity \(t\mapsto t\), which is continuous. The theorem gives continuity of \(f\), regardless of how many factors \(J\) contains.
To calculate a basic preimage, choose distinct \(j_1,j_2\in J\) and restrict those coordinates to \((-2,3)\) and \((0,4)\). The resulting cylinder is \(B=\pi_{j_1}^{-1}((-2,3))\cap\pi_{j_2}^{-1}((0,4))\). Its preimage is \(f^{-1}(B)=(-2,3)\cap(0,4)=(0,3)\). The intersection is open. The strict endpoints matter: \(t=0\) fails the second restriction, and \(t=3\) fails the first. As a check, \(t=1\) has value \(1\) in both selected coordinates, which lies in both intervals. No conditions on the other coordinates are needed.
Worked Example: One Discontinuous Coordinate
Give \(\mathbb{R}\) its standard topology, and define \(h:\mathbb{R}\to\mathbb{R}\) by \(h(t)=0\) for \(t\leq 0\) and \(h(t)=1\) for \(t>0\). Define \(f:\mathbb{R}\to\mathbb{R}\times\mathbb{R}\) by \(f(t)=(t,h(t))\). The first coordinate is continuous, but the second is not: for the open interval \((-1/2,1/2)\), its preimage is \(h^{-1}((-1/2,1/2))=(-\infty,0]\). This set is not open, because it contains \(0\) but no open interval about \(0\) is contained in it. Thus \(h\) is discontinuous, and the coordinatewise theorem implies that \(f\) is not continuous.
Directly, the cylinder \(\mathbb{R}\times(-1/2,1/2)\) is open in the product, and its preimage under \(f\) is \((-\infty,0]\), confirming the conclusion. In particular, the inclusion of the endpoint \(0\) is essential: \(h(0)=0\) belongs to the target interval, while \(h(t)=1\) for every \(t>0\).
A Common Mistake: Restricting Every Coordinate
In an infinite product, it is tempting to think that an open neighborhood must restrict every coordinate. That is not the product topology. Its basic open sets restrict only finitely many coordinates; requiring restrictions on infinitely many coordinates can produce a set that is not open.
For a concrete calculation, give \(\mathbb{R}^{\mathbb{N}}\) the product topology and define \(d:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) by \(d(t)_n=t\) for every \(n\in\mathbb{N}\). The cylinder \(C=\pi_1^{-1}((-1,1))\) restricts just the first coordinate, so \(d^{-1}(C)=(-1,1)\). A mistaken calculation might add the conditions \(-1/n<t<1/n\) for every \(n\), giving \(\bigcap_{n\geq1}(-1/n,1/n)=\{0\}\). Indeed, \(0\) belongs to every interval, and for any \(t\ne0\) one can choose \(n>1/|t|\), so \(|t|>1/n\). But those infinitely many conditions describe a different set, not the cylinder \(C\). In the product topology they cannot be substituted for the single-coordinate restriction defining \(C\).
The coordinatewise theorem matters because it turns a potentially complicated question about open subsets of a product into separate continuity checks on the factors. When checking a particular product-open preimage, the practical rule is to write down the finitely many coordinate conditions that define a basic open set, take each coordinate preimage, and intersect those open sets.
Check Your Understanding
Use the definition of the product topology and the coordinatewise continuity theorem to answer these questions.
- For \(f:\mathbb{R}\to\mathbb{R}^2\) defined by \(f(t)=(2t+1,t^3)\), explain why continuity of each coordinate implies continuity of \(f\).
- For \(g:\mathbb{R}\to\mathbb{R}^2\) defined by \(g(t)=(t,h(t))\), where \(h\) is the step function in the third example, find an open set whose preimage proves that \(g\) is not continuous.
- Let \(J\) be nonempty and let \(c:\mathbb{R}\to\mathbb{R}^J\) be defined by \(c(t)_j=5\) for every \(j\). Find each coordinate map and determine whether \(c\) is continuous.
- For a basic open set restricting coordinates \(\alpha_1,\ldots,\alpha_n\), write its preimage under \(f\) in terms of the coordinate maps \(f_{\alpha_k}\).
- Why does the theorem’s converse use a finite intersection of open sets rather than an arbitrary intersection?