Two Topologies on the Same Cartesian Product
For a finite list of spaces, it is natural to describe an open rectangle by choosing an open set in each coordinate. With an arbitrary index set, there are two choices: restrict only finitely many coordinates, or allow restrictions in every coordinate. These choices define the product topology and the box topology. Their basic sets look similar, but the allowed number of coordinate restrictions differs.
Let \(J\) be an index set, and for each \(i\in J\) let \((X_i,\mathcal{T}_i)\) be a topological space. Write \(X=\prod_{i\in J}X_i\) for the Cartesian product. A point \(x\in X\) has coordinates \(x_i\in X_i\). If \(U_i\subseteq X_i\) for each \(i\), the notation \(\prod_{i\in J}U_i\) means the set of points \(x\in X\) satisfying \(x_i\in U_i\) for every \(i\in J\).
Here “all but finitely many” means that the set \(\{i\in J:U_i\ne X_i\}\) is finite. Thus a product basic set can impose restrictions on finitely many coordinates, while a box basic set may impose restrictions on infinitely many. Every product basic set is also a box basic set: the product definition merely adds the requirement that only finitely many coordinates are restricted.
Why the Box Sets Form a Basis
To specify the box topology by its basic sets, we check the basis conditions. A basis is a family of subsets whose union covers the space and such that, whenever a point lies in the intersection of two family members, some family member contains that point and lies inside the intersection.
Proof. First suppose \(X\ne\varnothing\). The set \(X=\prod_{i\in J}X_i\) is a box basic set, since each \(X_i\) is open in itself, so the family covers \(X\). Now let \(B=\prod_{i\in J}U_i\) and \(C=\prod_{i\in J}V_i\) be box basic sets. Their intersection is \[ B\cap C=\prod_{i\in J}(U_i\cap V_i). \] Each \(U_i\cap V_i\) is open in \(X_i\), so \(B\cap C\) is a box basic set. In particular, whenever a point belongs to \(B\cap C\), this same box basic set contains the point and is contained in both \(B\) and \(C\). If \(X=\varnothing\), every subset of \(X\) is empty, and the family of box basic sets consists of \(\varnothing\); it covers \(X\) and satisfies the intersection condition. Thus the family is a basis in either case. \(\square\)
The product basic sets also form a basis: intersecting two of them intersects the coordinate-open sets, and the union of their two finite sets of restricted coordinates is finite. The product topology is the topology generated by this basis. The difference between the two definitions is not whether coordinate sets are open; they are open in both. It is whether restrictions may occur at infinitely many coordinates.
Every Product-Open Set Is Box-Open
Proof. Let \(P=\prod_{i\in J}U_i\) be any product basic set. Each \(U_i\) is open in \(X_i\), so \(P\) satisfies the definition of a box basic set as well. Hence every product basic set is open in the box topology. Every open set in the product topology is a union of product basic sets. That union is therefore a union of box-open sets, and is open in the box topology. This proves that every product-open set is box-open, which is the asserted containment. \(\square\)
The direction of containment follows directly from the definitions: allowing restrictions at any number of coordinates gives at least all the basic sets available when only finitely many restrictions are allowed. The theorem does not say that the topologies are equal. It establishes the inclusion that follows from the basic-set comparison.
Worked Example: A Finite-Coordinate Rectangle
Take \(J=\mathbb{N}\) and \(X_i=\mathbb{R}\) with its standard topology for every \(i\). Consider \[ P=\{x\in\mathbb{R}^{\mathbb{N}}:x_2\in(0,1),\ x_5\in(-2,2)\}. \] To write this as a product of coordinate-open sets, set \(U_2=(0,1)\), \(U_5=(-2,2)\), and \(U_i=\mathbb{R}\) for every \(i\notin\{2,5\}\). Then \(P=\prod_{i\in\mathbb{N}}U_i\). Only coordinates \(2\) and \(5\) are restricted, so \(P\) is product basic and therefore also box basic.
For example, the point \(x\) with \(x_2=\tfrac12\), \(x_5=1\), and \(x_i=0\) for every other \(i\) belongs to \(P\): \(\tfrac12\in(0,1)\), \(1\in(-2,2)\), and the remaining coordinates lie in their unrestricted factors \(\mathbb{R}\). If instead \(y_2=0\), then \(y\notin P\), because \(0\notin(0,1)\). This checks the strict endpoint condition in the second coordinate restriction.
Check: The factor description gives precisely the stated membership conditions: \(x\in\prod_i U_i\) if and only if \(x_i\in U_i\) for every \(i\). At coordinates other than \(2\) and \(5\), this condition is automatic because \(U_i=\mathbb{R}\).
Worked Example: Restrictions in Every Coordinate
Again take \(X=\mathbb{R}^{\mathbb{N}}\), and define \[ B=\prod_{n\in\mathbb{N}}\left(-\frac1n,\frac1n\right). \] Each factor is open in \(\mathbb{R}\), so \(B\) is box basic and hence box-open. Its membership test is coordinatewise: \(x\in B\) exactly when \(-1/n<x_n<1/n\) for every \(n\in\mathbb{N}\).
Consider the point \(a\) given by \(a_n=1/(2n)\) for every \(n\). For each positive integer \(n\), \(1/(2n)>0\) and \(1/(2n)<1/n\), so \(-1/n<a_n<1/n\); therefore \(a\in B\). By contrast, let \(b_4=1/4\) and \(b_n=0\) for \(n\ne4\). At coordinate \(4\), the required interval is \((-1/4,1/4)\), and \(b_4=1/4\) is not in that open interval. Thus \(b\notin B\).
This example uses a restriction at every coordinate. It is box basic by definition; the definition of product basic set, in contrast, requires that all but finitely many factors be the whole coordinate space. The distinction in these descriptions is exactly what the two topologies record.
Check: The point \(a\) satisfies the strict inequalities at every coordinate, whereas \(b\) fails the strict inequality at coordinate \(4\). No endpoint is included in any of the displayed intervals.
Intersections and Degenerate Products
Worked Example: Intersecting Two Box Basic Sets
In \(\mathbb{R}^{\mathbb{N}}\), let \(B=\prod_{n\in\mathbb{N}}U_n\) and \(C=\prod_{n\in\mathbb{N}}V_n\), where \(U_n=(-2,2)\) for every \(n\), \(V_1=(0,3)\), \(V_3=(-1,1)\), and \(V_n=\mathbb{R}\) for \(n\ne1,3\). Both are box basic. At each coordinate, intersection requires satisfying both coordinate conditions. Consequently, \[ B\cap C=(0,2)\times(-2,2)\times(-1,1)\times\prod_{n\ge4}(-2,2), \] where the first displayed factor is coordinate \(1\), the second coordinate \(2\), and the third coordinate \(3\). Indeed, \(U_1\cap V_1=(-2,2)\cap(0,3)=(0,2)\); \(U_2\cap V_2=(-2,2)\cap\mathbb{R}=(-2,2)\); \(U_3\cap V_3=(-2,2)\cap(-1,1)=(-1,1)\); and for \(n\ge4\), \(U_n\cap V_n=(-2,2)\cap\mathbb{R}=(-2,2)\). The result is again a box basic set.
If a coordinate intersection were empty, the whole product intersection would be empty, since no point could satisfy that coordinate condition. This is consistent with the basis condition: the empty set is open, and there is no point in the intersection for which a basis neighborhood must be found.
There are two harmless edge cases in the definitions. If some factor \(X_k\) is empty, then \(X=\prod_{i\in J}X_i\) is empty, because no tuple can have its \(k\)-th coordinate in \(X_k\). Both topologies on \(X\) then consist only of \(\varnothing\), so the claimed containment holds. If \(J=\varnothing\), the Cartesian product has one element, the empty tuple. There is just one choice of coordinate factors, and its product is that singleton; both generated topologies consist of the empty set and the singleton. These cases do not require any nonemptiness assumption on the factors for the containment theorem.
A Common Pitfall
Common pitfall: A frequent error is to define a box basic set using only finitely many coordinate restrictions. That is the definition of a product basic set, not a box basic set. For a concrete check, consider the set \(B\) from the previous example and suppose a finite set \(F\subseteq\mathbb{N}\) is chosen as the restricted coordinates. Since \(F\) is finite and \(\mathbb{N}\) is infinite, there is an index \(m\notin F\). Define \(z_i=0\) for \(i\ne m\) and \(z_m=2\). At every coordinate \(i\in F\), \(z_i=0\in(-1/i,1/i)\). Thus \(z\) satisfies those finitely many restrictions. But \(z\notin B\), because at coordinate \(m\), \(2\notin(-1/m,1/m)\); indeed \(1/m\le1\), so \(2>1/m\). Therefore those finite restrictions do not describe \(B\). The correction is to keep the quantifiers straight: a box basic set chooses an open factor for every index, while a product basic set chooses factors that differ from the whole space at only finitely many indices.
This comparison is useful whenever a Cartesian product has infinitely many coordinates: before deciding that a rectangle is basic open, identify which topology is being used and check how many coordinates the definition permits it to restrict. The inclusion proved here is one-way: a union of product basic sets is also a union of box basic sets. No reverse inclusion follows merely from that argument.
Check Your Understanding
Use the definitions to answer the following questions.
- For a product basic set \(\prod_{i\in J}U_i\), what condition must hold for all but finitely many indices?
- In \(\mathbb{R}^{\mathbb{N}}\), is \(\prod_{n\in\mathbb{N}}(-1,1)\) box basic, product basic, both, or neither? Explain using the coordinate restrictions.
- Explain why the intersection of two box basic sets is a box basic set, including the case where one coordinate intersection is empty.
- Give the one-step argument that every product basic set is box basic.
- What is the Cartesian product when the index set is empty, and what are its open sets in either topology?