Where Product-Open-Set Errors Begin
In a product space, a basic open set can restrict only finitely many coordinates. This is the detail that makes product topology different from box topology, where every coordinate may be restricted at once. Many errors about product-open sets come from forgetting that difference, or from misreading what a projection inverse image actually restricts.
Let \(\{X_i:i\in J\}\) be a family of topological spaces, with topology \(\mathcal{T}_i\) on \(X_i\), and write \(X=\prod_{i\in J}X_i\). A point of \(X\) is a tuple \(x=(x_i)_{i\in J}\). The product topology has as a basis the sets \(\prod_{i\in J}U_i\), where each \(U_i\) is open in \(X_i\) and \(U_i=X_i\) for all but finitely many indices. A basis means that open sets are unions of these basic sets. In particular, unrestricted coordinates have factor \(X_i\), not an extra condition that the coordinate equal some chosen value.
These sets do form a basis: they cover \(X\), since \(X\) itself is one of them. The intersection of two such sets is either empty or another such set. Indeed, coordinate by coordinate, the intersection uses the intersection of the two open factors, and only the union of the two finite sets of restricted indices can have a factor different from \(X_i\). This gives the local test we will use: a set is open precisely when every one of its points lies in a basic set contained in it.
Worked Examples: Restrictions and Projections
Worked Example: Restricting Infinitely Many Coordinates
Give \(\mathbb{R}^{\mathbb{N}}=\prod_{n\in\mathbb{N}}\mathbb{R}\) the product topology, using the standard topology on each copy of \(\mathbb{R}\). Consider \(A=\prod_{n\in\mathbb{N}}(-1,1)\). Although every factor is open, \(A\) is not open in the product topology.
To test openness, consider the zero tuple \(0=(0,0,\dots)\), which belongs to \(A\). Any basic open set \(B\) containing \(0\) restricts only finitely many coordinates; call that finite set of indices \(F\). Choose \(k\in\mathbb{N}\setminus F\). Define \(y\) by \(y_k=2\) and \(y_n=0\) for \(n\ne k\). Every restricted coordinate of \(y\) agrees with \(0\), so \(y\in B\). But \(y_k=2\notin(-1,1)\), so \(y\notin A\). Thus no basic neighborhood of \(0\) is contained in \(A\), and \(A\) is not product-open.
Check: in the box topology, \(A\) is open because each factor \((-1,1)\) is open and the box basis permits all coordinates to be restricted. The same formula for \(A\) therefore gives an open set in the box topology, but not in the product topology.
Worked Example: A Projection Inverse Image
In \(X=\prod_{n\in\mathbb{N}}\mathbb{R}\) with the product topology, let \(\pi_3:X\to\mathbb{R}\) be the third-coordinate projection, so \(\pi_3(x)=x_3\). For \(U=(2,4)\),
This set has exactly one restricted coordinate. It is a basic product-open set. The infinitely many other coordinates are unrestricted: each has factor \(\mathbb{R}\). Therefore the claim that this inverse image is not open because it involves infinitely many coordinates is incorrect.
More generally, for any index \(i\in J\) and open \(U\subseteq X_i\), the inverse image \(\pi_i^{-1}(U)\) is the basic set with factor \(U\) at coordinate \(i\) and factor \(X_j\) at every other coordinate. This proves directly that each projection \(\pi_i\) is continuous: the inverse image of every open set in its target is open.
Proof. Suppose first that \(A\ne\varnothing\) and is open. Choose \(x\in A\). For each \(y_i\in U_i\), form a point of \(A\) by using \(y_i\) in coordinate \(i\) and \(x_j\) in every other coordinate. Since \(A\) is open, this point lies in a basic set contained in \(A\). The factor of that basic set at coordinate \(i\) is an open neighborhood of \(y_i\) contained in \(U_i\). This holds for every \(y_i\in U_i\), so \(U_i\) is open. Also, take a basic set containing \(x\) and contained in \(A\), with finite restricted-index set \(F\). If \(i\notin F\), its factor at \(i\) is \(X_i\). Replacing \(x_i\) by any point of \(X_i\) gives a point in the basic set and hence in \(A\). Therefore \(U_i=X_i\). This proves the required conditions. Conversely, if every \(U_i\) is open and only finitely many factors differ from \(X_i\), then \(A\) is itself a basic product-open set. The empty set is open by the definition of a topology, which handles the separate empty case. \(\square\)
The nonempty condition matters: an empty product of subsets can arise because some factor is empty, and the resulting set is open even when other factors fail the stated conditions.
Coordinatewise Continuity: What the Theorem Says
A second frequent error is to assume that continuity of each coordinate automatically proves continuity into every topology one might put on the product. For the product topology, it does. For the box topology, it need not. Here is the precise product-topology result.
Proof. Suppose \(f\) is continuous. For any open \(U_i\subseteq X_i\), the set \(\pi_i^{-1}(U_i)\) is open by the projection calculation above. Hence \(f_i^{-1}(U_i)=f^{-1}(\pi_i^{-1}(U_i))\) is open. This proves continuity of each \(f_i\).
Conversely, suppose each \(f_i\) is continuous. A basic open set in the product has the form \(B=\prod_{i\in J}U_i\), with only finitely many \(U_i\ne X_i\). Let \(F=\{i\in J:U_i\ne X_i\}\). Then \(f^{-1}(B)=\bigcap_{i\in F}f_i^{-1}(U_i)\). Each set in this intersection is open by continuity of \(f_i\), and a finite intersection of open sets is open. If \(F\) is empty, the intersection is \(Y\), also open. Every open set in the product is a union of basic open sets, and its inverse image is the corresponding union of open inverse images. Therefore \(f\) is continuous. \(\square\)
Worked Example: Coordinatewise Continuity Is Not Enough for the Box Topology
Let \(\Delta:\mathbb{R}\to\mathbb{R}^{\mathbb{N}}\) be the diagonal map \(\Delta(t)=(t,t,t,\dots)\). Each coordinate map is the identity map \(t\mapsto t\), which is continuous in the standard topology on \(\mathbb{R}\). By the theorem, \(\Delta\) is continuous when the target has the product topology.
Now give the target the box topology. The set \(V=\prod_{n\in\mathbb{N}}(-1/n,1/n)\) is a box-open neighborhood of the zero tuple because every factor is open. Its inverse image is \(\Delta^{-1}(V)=\bigcap_{n\in\mathbb{N}}(-1/n,1/n)=\{0\}\). To verify the equality, \(0\) belongs to every interval. If \(t\ne0\), choose \(n\in\mathbb{N}\) with \(n>1/|t|\). Then \(1/n<|t|\), so \(t\notin(-1/n,1/n)\). Thus no nonzero \(t\) belongs to the intersection.
The singleton \(\{0\}\) is not open in the standard topology on \(\mathbb{R}\), since every open neighborhood of \(0\) contains an interval around \(0\), and that interval contains nonzero points. Hence \(\Delta^{-1}(V)\) is not open, so \(\Delta\) is not continuous into the box topology. The coordinate maps remain continuous; what changed is the target topology.
A Common Mistake and a Reliable Test
Common pitfall: “Each coordinate factor is open, so their product is product-open.” The incorrect reasoning is: “Every \((-1,1)\) is open, therefore \(\prod_{n\in\mathbb{N}}(-1,1)\) is a basic open set in the product topology.” The error is in the last step: being open in every coordinate is not enough for a product-basic set. Infinitely many factors differ from \(\mathbb{R}\). The zero-tuple test above gives the correction: any product-basic neighborhood of zero leaves some coordinate \(k\) unrestricted, and setting that coordinate equal to \(2\) produces a point outside the proposed set.
When checking a product-space argument, ask which coordinates are actually restricted. A basic product-open neighborhood allows finitely many restrictions; a projection inverse image uses one; a box-open set may use infinitely many. For a map into a product, test continuity coordinate by coordinate using the theorem, but do not transfer that conclusion to the box topology without a separate argument.
Check Your Understanding
For each question, specify which topology is being used and justify the conclusion from the relevant definition or theorem.
- In \(\mathbb{R}^{\mathbb{N}}\) with the product topology, is \((-1,1)\times\mathbb{R}\times(-2,2)\times\mathbb{R}\times\cdots\) open?
- For a projection \(\pi_k:\prod_{i\in J}X_i\to X_k\) and open \(U\subseteq X_k\), describe every coordinate factor of \(\pi_k^{-1}(U)\).
- In \(\mathbb{R}^{\mathbb{N}}\), why is \(\prod_{n\in\mathbb{N}}(-1/n,1/n)\) box-open but not product-open?
- If all coordinate maps of \(f:Y\to\prod_{i\in J}X_i\) are continuous, what additional fact about the target topology is required to conclude that \(f\) is continuous?
- Why does the product-open-rectangle criterion include a separate case for an empty product of subsets?