From Factor Bases to Product Bases
A product topology is described using open rectangles, but those rectangles can be refined using smaller, more manageable collections of open sets in each factor. If \(\mathcal{B}_X\) is a basis for \(X\) and \(\mathcal{B}_Y\) is a basis for \(Y\), the corresponding rectangles \(B\times C\) provide a basis for \(X\times Y\). This lets us build explicit bases from familiar ones, such as rational intervals in \(\mathbb{R}\) or singletons in a finite discrete space.
Throughout, \(X\) and \(Y\) are topological spaces with topologies \(\mathcal{T}_X\) and \(\mathcal{T}_Y\). The product \(X\times Y\) has the product topology: its basis consists of open rectangles \(U\times V\), where \(U\in\mathcal{T}_X\) and \(V\in\mathcal{T}_Y\). A basis for a topology is a collection of open sets that covers the space and has the following intersection property: if a point belongs to two members of the collection, some member of the collection contains the point and is contained in their intersection.
The Product-Basis Theorem
Proof. We check the two basis conditions. First, let \((x,y)\in X\times Y\). Since \(\mathcal{B}_X\) covers \(X\), there is \(B\in\mathcal{B}_X\) with \(x\in B\). Since \(\mathcal{B}_Y\) covers \(Y\), there is \(C\in\mathcal{B}_Y\) with \(y\in C\). Thus \((x,y)\in B\times C\), so the basis rectangles cover \(X\times Y\). If \(X\times Y\) is empty, this covering condition holds vacuously.
For the intersection condition, suppose \((x,y)\in(B_1\times C_1)\cap(B_2\times C_2)\), where \(B_1,B_2\in\mathcal{B}_X\) and \(C_1,C_2\in\mathcal{B}_Y\). Then \(x\in B_1\cap B_2\) and \(y\in C_1\cap C_2\). The basis property in \(X\) gives a set \(B_3\in\mathcal{B}_X\) such that \(x\in B_3\subseteq B_1\cap B_2\). The basis property in \(Y\) gives a set \(C_3\in\mathcal{B}_Y\) such that \(y\in C_3\subseteq C_1\cap C_2\). It follows that \((x,y)\in B_3\times C_3\subseteq(B_1\times C_1)\cap(B_2\times C_2)\). So the intersection condition holds.
Each basis rectangle is open in the product topology because \(B\) and \(C\) are open in their respective factors. Conversely, take any open rectangle \(U\times V\) in the product topology and any \((x,y)\in U\times V\). The factor basis properties give \(B\in\mathcal{B}_X\) and \(C\in\mathcal{B}_Y\) with \(x\in B\subseteq U\) and \(y\in C\subseteq V\). Hence \((x,y)\in B\times C\subseteq U\times V\). Thus every open rectangle is a union of basis rectangles, so the basis rectangles generate the product topology. \(\square\)
The proof separates into two simple tasks: cover each coordinate using its factor basis, and refine each coordinate intersection separately. The theorem does not require the same basis in both factors, and it does not require either space to be metric.
Rational Rectangles in \(\mathbb{R}^2\)
Proof. Every real number \(x\) lies in some interval with rational endpoints: by density of \(\mathbb{Q}\) in \(\mathbb{R}\), choose rationals \(p<x<q\). For the intersection condition, suppose \(x\in(p_1,q_1)\cap(p_2,q_2)\). This intersection is open and contains \(x\), so it contains an interval \((x-\varepsilon,x+\varepsilon)\) for some \(\varepsilon>0\). Density of the rationals supplies rational numbers \(p,q\) with \(x-\varepsilon<p<x<q<x+\varepsilon\). Then \(x\in(p,q)\subseteq(p_1,q_1)\cap(p_2,q_2)\). The interval family is therefore a basis. The product-basis theorem now gives the claimed basis of rational rectangles. \(\square\)
Worked Example: A Rational Rectangle Inside a Ball
Use the rational-rectangle basis to find a basis neighborhood of \((1/3,2/3)\) contained in the Euclidean open ball \(B((1/3,2/3),1/4)\) in \(\mathbb{R}^2\).
Consider \(R=(1/6,1/2)\times(1/2,5/6)\). All four endpoints are rational. Since \(1/6<1/3<1/2\) and \(1/2<2/3<5/6\), the point \((1/3,2/3)\) belongs to \(R\). For any \((u,v)\in R\), the coordinate differences satisfy \(|u-1/3|<1/6\) and \(|v-2/3|<1/6\). Therefore \(\|(u,v)-(1/3,2/3)\|^2< (1/6)^2+(1/6)^2=1/18\). Since \(1/18<1/16=(1/4)^2\), we have \(\|(u,v)-(1/3,2/3)\|<1/4\), so \(R\subseteq B((1/3,2/3),1/4)\).
Check: the center belongs to \(R\), and every point of \(R\) has squared distance less than \(1/18\), which is less than the squared radius \(1/16\). Thus this is indeed a basis neighborhood inside the specified ball.
A Factor with the Subspace Topology
For a subset \(A\) of a topological space \(X\), the subspace topology on \(A\) consists of the sets \(A\cap U\), where \(U\) is open in \(X\). In particular, a neighborhood of an endpoint of an interval can look different from an ordinary open interval in \(\mathbb{R}\): it is the intersection of such an interval with the subspace.
Proof. A subspace-open set in \([0,1]\) has the form \([0,1]\cap U\) for some open \(U\subseteq\mathbb{R}\). If \(x\in[0,1]\cap U\), the real-line basis gives rational \(p<q\) with \(x\in(p,q)\subseteq U\). Then \(x\in[0,1]\cap(p,q)\subseteq[0,1]\cap U\), which proves that the stated collection refines every subspace-open neighborhood. It covers \([0,1]\) by applying this argument to \(U=\mathbb{R}\). For an intersection of two members, write it as \([0,1]\cap((p_1,q_1)\cap(p_2,q_2))\). At any point \(x\) in the intersection, the real-line basis supplies a rational interval around \(x\) contained in \((p_1,q_1)\cap(p_2,q_2)\); intersecting that interval with \([0,1]\) gives a member of the stated collection containing \(x\) and contained in the intersection. This proves the basis condition. The product-basis theorem gives the final claim. \(\square\)
Worked Example: A Neighborhood at the Endpoint of the Interval
In \([0,1]\times\mathbb{R}\), consider the product-open neighborhood \(U=[0,1]\times(1,3)\) of \((0,2)\). Find a basis rectangle containing \((0,2)\) and contained in \(U\).
In the first factor, choose \([0,1]\cap(-1,1/2)=[0,1/2)\). The equality follows because the points of \([0,1]\) that are greater than \(-1\) and strictly less than \(1/2\) are exactly those in \([0,1/2)\). This set contains \(0\). In the second factor, choose the rational interval \((3/2,5/2)\), which contains \(2\) because \(3/2<2<5/2\). The resulting basis rectangle is \([0,1/2)\times(3/2,5/2)\).
Its first coordinate lies in \([0,1]\), and its second coordinate lies in \((1,3)\), since \(1<3/2\) and \(5/2<3\). Thus the rectangle is contained in \(U\), and it contains \((0,2)\).
Check: \(0\in[0,1/2)\) and \(2\in(3/2,5/2)\), so the chosen rectangle contains the stated point; both coordinate sets are contained in the corresponding factors of \(U\).
A Finite Discrete Factor
A space is discrete if every subset is open. In a discrete space, the collection of all singletons is a basis: each point belongs to its singleton, and two singletons that contain the same point must be the same singleton. This makes discrete factors particularly direct to combine with a basis for another space.
Worked Example: A Two-Point Discrete Space Times the Real Line
Let \(D=\{a,b\}\) have the discrete topology, and let \(\mathbb{R}\) have its standard topology. A basis for \(D\) is \(\{\{a\},\{b\}\}\), and a basis for \(\mathbb{R}\) is the rational open intervals. The product-basis theorem gives the basis \(\{\{a\}\times(p,q),\{b\}\times(p,q):p,q\in\mathbb{Q},\ p<q\}\) for \(D\times\mathbb{R}\).
For a direct neighborhood check, let \(G=(\{a\}\times(-1,2))\cup(\{b\}\times(0,3))\). This is open in the product topology because each displayed piece is a product of open sets. At \((b,1)\), the basis rectangle \(\{b\}\times(1/2,3/2)\) is available: \(1/2\) and \(3/2\) are rational, and \(1/2<1<3/2\). Also, \(0<1/2\) and \(3/2<3\), so this rectangle is contained in \(\{b\}\times(0,3)\), hence in \(G\).
At \((a,0)\), the rectangle \(\{a\}\times(-1/2,1/2)\) contains the point because \(-1/2<0<1/2\). Its interval lies in \((-1,2)\), since \(-1<-1/2\) and \(1/2<2\), so this rectangle is contained in \(G\) as well. These checks illustrate how the discrete coordinate remains fixed while the interval basis refines the other coordinate.
Check: Each selected rectangle uses a singleton open in \(D\) and a rational interval open in \(\mathbb{R}\); the inequalities above verify containment in the corresponding piece of \(G\).
A Common Pitfall
A basis element in a subspace need not be an open interval in the ambient real line. For example, \((0,1/2)\) does not contain \(0\), so it cannot be a neighborhood of \(0\) in \([0,1]\). A mistaken product-basis attempt might therefore say that no interval-basis rectangle can describe a neighborhood of \((0,2)\) at the left edge of \([0,1]\). The error is using ambient intervals without taking their intersections with the subspace.
The correction is to use relative basis elements. The ambient interval \((-1,1/2)\) gives the subspace-open set \([0,1]\cap(-1,1/2)=[0,1/2)\), which does contain \(0\). Thus \([0,1/2)\times(3/2,5/2)\) is a valid basis rectangle at \((0,2)\). The basis theorem applies to the topology actually placed on each factor, so the factor bases must match those topologies.
Once factor bases are known, the construction is systematic: choose one basis set around each coordinate, then take their product. The theorem guarantees that these rectangles both cover the product and refine every product-open neighborhood. In concrete work, the remaining task is to check coordinate membership and containment in each factor.
Check Your Understanding
Use the factor-basis construction and the definitions to answer the following.
- Describe a basis for \(\mathbb{R}\times\mathbb{R}\) using intervals with rational endpoints.
- In \([0,1]\) with its subspace topology, express \([0,1/3)\) as the intersection of \([0,1]\) with an open interval in \(\mathbb{R}\).
- Find a rational-rectangle basis neighborhood of \((1,1)\) contained in \(B((1,1),1/2)\) in \(\mathbb{R}^2\).
- If \(D=\{a,b,c\}\) has the discrete topology, describe a basis for \(D\times\mathbb{R}\).
- Explain which basis condition is used when refining the intersection of two product basis rectangles.