Open Sets in a Product Space
A point in a product space is a tuple whose coordinates belong to different spaces. To describe a neighborhood of that point, we can impose open conditions on some of its coordinates while leaving the others unrestricted. The product topology formalizes this idea: a basic open set restricts only finitely many coordinates, even when the product has infinitely many factors.
This construction extends the topology on a finite product that has appeared in metric-space arguments. It also makes a crucial distinction visible: in a finite product, restricting every coordinate is still a finite collection of conditions; in an infinite product, it generally is not. We will define the topology through coordinate conditions, identify a basis for it, and compare it with the maximum-metric topology in the finite case.
A set \(\pi_i^{-1}(U)\) consists of the tuples whose \(i\)-th coordinate belongs to \(U\); it makes no restriction on any other coordinate. The product topology is the smallest topology on \(X\) containing all these sets. When \(I\) is empty, the product is understood to be a one-point set, and the generated topology is its unique topology. If one of the factors is empty, the product itself is empty.
Finite-Coordinate Cylinders Form a Basis
The basis generated by a subbasis consists of finite intersections of subbasic sets, along with the whole space. Thus in this setting the natural candidate basis consists of sets that impose open conditions on only finitely many coordinates. The result follows from the basis-generation theorem in “Generating a Topology From a Basis”; we verify the basis conditions directly to see how the product construction works.
Proof. First consider the basis covering condition. If \(x\in X\), take \(F=\varnothing\). There are no coordinate conditions to check, so \(B(\varnothing,(U_i))=X\) and \(x\in X\). If \(X\) is empty, the covering condition holds vacuously.
Now take two such sets \(B(F,(U_i))\) and \(B(G,(V_i))\). Their intersection imposes both conditions at every coordinate involved. More precisely, it is the set obtained by using the finite coordinate set \(F\cup G\), with the coordinate condition \(U_i\) when \(i\in F\setminus G\), \(V_i\) when \(i\in G\setminus F\), and \(U_i\cap V_i\) when \(i\in F\cap G\). Each condition is open in its factor, since finite intersections of open sets are open. Therefore the intersection is again a member of the stated family. In particular, whenever a point belongs to the intersection, it belongs to a basis member contained in that intersection. These are the basis conditions, so the family is a basis for the topology generated by the coordinate sets \(\pi_i^{-1}(U)\). \(\square\)
This description gives a practical test for openness. A subset \(O\subseteq X\) is open exactly when, for every \(x\in O\), there is a finite-coordinate basis set \(B\) with \(x\in B\subseteq O\). Some coordinates in \(B\) may be left unrestricted; the finite set of coordinates used can vary with the point and with the basis neighborhood.
Worked Example: Rectangles Give the Usual Topology on the Plane
Consider \(X=\mathbb{R}\times\mathbb{R}\), with the usual topology on each factor. The basis sets are products \(U\times V\), where \(U\) and \(V\) are open in \(\mathbb{R}\); either coordinate can also be unrestricted. For example, $$ (0,2)\times(-3,0) $$ is a basic open neighborhood of \((1,-1)\), since \(1\in(0,2)\) and \(-1\in(-3,0)\).
Equip \(\mathbb{R}^2\) with the maximum metric \(D((x_1,x_2),(y_1,y_2))=\max\{|x_1-y_1|,|x_2-y_2|\}\). The \(D\)-ball of radius \(1/2\) about \((1,-1)\) lies in this rectangle: if \(D((x_1,x_2),(1,-1))<1/2\), then \(1/2<x_1<3/2\) and \(-3/2<x_2<-1/2\). Conversely, the product \((1-r,1+r)\times(-1-r,-1+r)\) is the \(D\)-ball of radius \(r\) about \((1,-1)\). Thus these rectangles describe the same neighborhoods as the maximum metric. The agreement is not accidental; it holds for every finite product of metric spaces.
Finite Products and the Maximum Metric
Suppose \(X_1,\ldots,X_n\) are metric spaces with metrics \(d_1,\ldots,d_n\), where \(n\geq1\). On their Cartesian product, define $$ D(x,y)=\max_{1\leq i\leq n}d_i(x_i,y_i). $$ The maximum metric measures the largest coordinate distance. The next theorem shows that its open sets are precisely the open sets of the product topology.
Proof. Let \(O\) be open in the product topology, and take \(x\in O\). By the basis characterization, there is a finite-coordinate basis set containing \(x\) and contained in \(O\). Because there are only \(n\) coordinates, we can regard this as a condition in every coordinate: use the given open set for a restricted coordinate and the whole factor \(X_i\) for an unrestricted coordinate. For each \(i\), choose \(\varepsilon_i>0\) such that $$ B_{\varepsilon_i}^{X_i}(x_i)\subseteq U_i, $$ where \(U_i\) is the chosen open set in coordinate \(i\). For an unrestricted coordinate, we may take \(U_i=X_i\) and, for example, \(\varepsilon_i=1\). Since there are finitely many positive numbers \(\varepsilon_i\), their minimum \(\varepsilon=\min_{1\leq i\leq n}\varepsilon_i\) is positive.
If \(D(x,y)<\varepsilon\), then \(d_i(x_i,y_i)\leq D(x,y)<\varepsilon\leq\varepsilon_i\) for every \(i\). Hence \(y_i\in U_i\) for every coordinate, so \(y\) belongs to the chosen basis set and therefore to \(O\). This proves that every product-open set is open in the maximum-metric topology.
For the reverse inclusion, take a maximum-metric ball \(B_\varepsilon^D(x)\), with \(\varepsilon>0\). The product set $$ \prod_{i=1}^n B_\varepsilon^{X_i}(x_i) $$ is a basic open set in the product topology. It equals \(B_\varepsilon^D(x)\): membership in the product means \(d_i(x_i,y_i)<\varepsilon\) for every \(i\), which, because the maximum is over finitely many coordinates, is equivalent to \(\max_i d_i(x_i,y_i)<\varepsilon\). Thus every maximum-metric ball is product-open. Every open set in a metric topology is a union of open balls, so every maximum-metric open set is product-open. The two topologies are equal. \(\square\)
Finiteness matters in this argument. It ensures that the minimum of the finitely many neighborhood radii is positive, and that the maximum of finitely many coordinate distances is strictly less than \(\varepsilon\) when every coordinate distance is strictly less than \(\varepsilon\). An infinite product has a different neighborhood structure.
Worked Example: A Basic Neighborhood in a Countable Product
Let \(X=\mathbb{R}^{\mathbb{N}}\), the set of real sequences with the product topology. Consider the point \(a\) whose coordinates are \(a_i=i\). A basic neighborhood of \(a\) is $$ B=\{x\in\mathbb{R}^{\mathbb{N}}: |x_2-2|<1\text{ and }|x_5-5|<1\}. $$ This set is open because it restricts coordinates \(2\) and \(5\) to open intervals and leaves every other coordinate unrestricted.
For example, the sequence \(x\) with \(x_2=2\), \(x_5=5\), \(x_{100}=10^6\), and \(x_i=a_i\) at all other coordinates belongs to \(B\). The large value at coordinate \(100\) does not affect membership, because coordinate \(100\) was not restricted. In fact, any finite-coordinate basis neighborhood of \(a\) leaves all but finitely many coordinates unrestricted.
Worked Example: A Coordinatewise Restriction That Is Not Open
In the same space \(\mathbb{R}^{\mathbb{N}}\), consider $$ C=\prod_{i=1}^{\infty}\left(-\frac{1}{i+1},\frac{1}{i+1}\right). $$ The zero sequence belongs to \(C\), since \(0\) lies in every interval. Nevertheless, \(C\) is not open in the product topology.
To verify this, take any basic neighborhood \(B\) of the zero sequence. It restricts only a finite set \(F\) of coordinates. Choose an index \(j\notin F\), which is possible because the positive integers are infinite. Define \(y_j=1\) and \(y_i=0\) for \(i\neq j\). At every restricted coordinate \(i\in F\), \(y_i=0\), so \(y\in B\). But \(y\notin C\), because \(1\notin(-1/(j+1),1/(j+1))\). Thus no basic neighborhood of the zero sequence is contained in \(C\), and the openness test fails. This illustrates why imposing a separate open condition at every coordinate does not, in general, give an open set in the product topology.
Worked Example: A Singleton in a Product of Discrete Spaces
Let \(X=\{0,1\}^{\mathbb{N}}\), where each factor \(\{0,1\}\) has the discrete topology, and let \(p\) be the sequence whose coordinates are all \(0\). The set $$ U=\{x\in X:x_1=0,\ x_4=1\} $$ is a basic open neighborhood of any point satisfying those two coordinate conditions. The coordinate sets \(\{0\}\) and \(\{1\}\) are open in the discrete factors.
The singleton \(\{p\}\), however, is not open. If it were open, it would contain a basic neighborhood \(B\) of \(p\). Such a neighborhood restricts only finitely many coordinates, say those in \(F\). Choose \(j\notin F\) and let \(q_j=1\), with \(q_i=0\) for all \(i\neq j\). Then \(q\neq p\), but \(q\) agrees with \(p\) on all coordinates in \(F\), so \(q\in B\). This contradicts \(B\subseteq\{p\}\). The example shows that specifying every coordinate of an infinite tuple is not a finite-coordinate neighborhood condition.
Why the Finite-Coordinate Rule Matters
In an infinite product, it is tempting to treat a neighborhood as though it must constrain every coordinate. That intuition can produce a strictly finer topology than the product topology. The defining rule is instead local and finite: around each point of an open set, one must be able to find a basic neighborhood that imposes conditions on only finitely many coordinates.
The same rule explains why finite products behave more familiarly. When there are finitely many factors, all coordinates can be restricted at once, and the maximum metric captures exactly those product neighborhoods. When there are infinitely many factors, the product topology still permits any finite collection of coordinate restrictions, but it does not generally permit infinitely many independent restrictions in a single basic neighborhood. Keeping this distinction in view is essential when working with products of topological spaces.
Check Your Understanding
Use the finite-coordinate basis description and the results proved above to answer the following questions.
- What form do the basic open sets in a product \(\prod_{i\in I}X_i\) take?
- Why does the intersection of two finite-coordinate basis sets remain a finite-coordinate basis set?
- Why do the product topology and the maximum-metric topology agree for a finite product of metric spaces?
- In \(\mathbb{R}^{\mathbb{N}}\) with the product topology, why can a basic neighborhood leave every coordinate outside a finite set unrestricted?
- Why is the set \(\prod_{i=1}^{\infty}(-1/(i+1),1/(i+1))\) not open in the product topology?
- Why is a singleton generally not open in \(\{0,1\}^{\mathbb{N}}\) when each factor is discrete?