What the Basis Criterion Checks
A basis describes a topology by giving a collection of basic open sets from which every open set is formed by taking unions. In a product space, open rectangles are the basic candidates. To verify that a collection really works as a basis, it is not enough that its members are open and cover the space: whenever two basic sets overlap, the collection must also provide a basic set around each point of that overlap that fits inside both.
Let \((X,\mathcal{T})\) and \((Y,\mathcal{S})\) be topological spaces. Their product \(X\times Y\) consists of ordered pairs \((x,y)\), and the product topology is the topology whose basic open sets are rectangles \(U\times V\), where \(U\in\mathcal{T}\) and \(V\in\mathcal{S}\). A rectangle is nonempty exactly when both factors are nonempty. Empty rectangles can be omitted from a basis without changing the unions it produces.
- Every point of \(Z\) belongs to at least one member of \(\mathcal{B}\).
- Whenever \(B_1,B_2\in\mathcal{B}\) and \(z\in B_1\cap B_2\), there is \(B_3\in\mathcal{B}\) such that \(z\in B_3\subseteq B_1\cap B_2\).
Proof. Suppose first that \(\mathcal{B}\) satisfies the two conditions, and define \(\mathcal{T}_{\mathcal{B}}\) to be all unions of members of \(\mathcal{B}\). The empty set is the empty union, and \(Z\) is a union of basis members by the coverage condition. Any union of sets in \(\mathcal{T}_{\mathcal{B}}\) is again a union of members of \(\mathcal{B}\).
To check finite intersections, let \(G,H\in\mathcal{T}_{\mathcal{B}}\). Write \(G=\bigcup_{i\in I}B_i\) and \(H=\bigcup_{j\in J}C_j\), with all \(B_i,C_j\in\mathcal{B}\). If \(z\in G\cap H\), then \(z\in B_i\cap C_j\) for some \(i,j\). The second condition supplies \(D\in\mathcal{B}\) with \(z\in D\subseteq B_i\cap C_j\subseteq G\cap H\). Thus \(G\cap H\) is the union of all basis members that it contains, and hence belongs to \(\mathcal{T}_{\mathcal{B}}\). This proves that \(\mathcal{T}_{\mathcal{B}}\) is a topology and that \(\mathcal{B}\) is a basis for it.
Conversely, suppose \(\mathcal{B}\) is a basis for a topology on \(Z\). The union of the basis members is \(Z\), so the coverage condition holds. If \(z\in B_1\cap B_2\), then this intersection is open. By the definition of a basis, every point in an open set belongs to a basis member contained in that open set. Applying this at \(z\) gives \(B_3\in\mathcal{B}\) with \(z\in B_3\subseteq B_1\cap B_2\). Both conditions follow. \(\square\)
Why Open Rectangles Form a Basis
Proof. If \(X\times Y=\varnothing\), then there are no points to cover and there are no nonempty open rectangles. The basis conditions hold vacuously. Now suppose \(X\times Y\ne\varnothing\). For every \((x,y)\in X\times Y\), the sets \(X\in\mathcal{T}\) and \(Y\in\mathcal{S}\) are open, and \((x,y)\in X\times Y\). Thus the rectangles cover the product.
Take two such rectangles \(U_1\times V_1\) and \(U_2\times V_2\). Their intersection is $$ (U_1\times V_1)\cap(U_2\times V_2) =(U_1\cap U_2)\times(V_1\cap V_2). $$ If \((x,y)\) lies in the intersection, then \(x\in U_1\cap U_2\) and \(y\in V_1\cap V_2\). Finite intersections of open sets are open, so this displayed intersection is itself an open rectangle containing \((x,y)\). It therefore supplies the refinement required by the basis criterion. \(\square\)
This verifies both basis conditions directly: every product point is covered, and overlaps have basic neighborhoods inside them. The proof also identifies the key advantage of rectangles: intersection is calculated one coordinate at a time.
Worked Example: Intersecting Two Rectangles
In \(\mathbb{R}^2=\mathbb{R}\times\mathbb{R}\), with each copy of \(\mathbb{R}\) carrying its standard topology, let \(R_1=(-2,3)\times(1,5)\) and \(R_2=(0,4)\times(-1,2)\). Find their intersection and check the refinement condition at \((1,3/2)\).
In the first coordinate, \((-2,3)\cap(0,4)=(0,3)\): the larger lower endpoint is \(0\), and the smaller upper endpoint is \(3\). In the second coordinate, \((1,5)\cap(-1,2)=(1,2)\): the larger lower endpoint is \(1\), and the smaller upper endpoint is \(2\). Therefore $$ R_1\cap R_2=(0,3)\times(1,2). $$ The point \((1,3/2)\) belongs to this intersection because \(0<1<3\) and \(1<3/2<2\). The intersection itself is a basic open rectangle containing the point and contained in both \(R_1\) and \(R_2\).
Check: the coordinate inequalities for \(R_1\) are \(-2<1<3\) and \(1<3/2<5\); for \(R_2\) they are \(0<1<4\) and \(-1<3/2<2\). Thus the point lies in both original rectangles as well as their computed intersection.
Testing a Proposed Rectangle Family
For a proposed collection \(\mathcal{R}\) of open rectangles in \(X\times Y\), coverage and intersection refinement show that it is a basis for some topology. To show it is a basis for the product topology, there is an additional requirement: it must provide enough rectangles to fit inside every product-open neighborhood. Equivalently, for every open rectangle \(U\times V\) and every \((x,y)\in U\times V\), there must be \(R\in\mathcal{R}\) with \((x,y)\in R\subseteq U\times V\). This local refinement condition ensures that every product-open set is a union of members of \(\mathcal{R}\).
Proof. Suppose \(\mathcal{R}\) is a basis for the product topology. Each \(U\times V\) is open in that topology, so the basis property gives the stated refinement around each of its points. Conversely, suppose the refinement condition holds. Each \(R\in\mathcal{R}\) is product-open by assumption. Every product-open set is a union of open rectangles by the definition of the product topology; applying the refinement condition at every point of each such rectangle expresses that open set as a union of members of \(\mathcal{R}\). Hence the unions of members of \(\mathcal{R}\) are exactly the product-open sets, so \(\mathcal{R}\) is a basis for the product topology. \(\square\)
Worked Example: Rectangles with Rational Endpoints
In \(\mathbb{R}^2\) with the product of the standard topologies, consider all rectangles \((a,b)\times(c,d)\) whose four endpoints are rational numbers, with \(a<b\) and \(c<d\). We show this family is a basis for the product topology.
We use the density of \(\mathbb{Q}\) in \(\mathbb{R}\): if \(a<x<b\), there is a rational \(r\) with \(a<r<x\) and a rational \(s\) with \(x<s<b\). One way to obtain them is to choose \(n\in\mathbb{N}\) large enough that \(2/n<x-a\) and \(1/n<b-x\), and then let \(m\) be the greatest integer not exceeding \(nx\). The rationals \(r=(m-1)/n\) and \(s=(m+1)/n\) satisfy \(r<x<s\); the bounds on \(1/n\) place both strictly between \(a\) and \(b\).
Now take any open rectangle \(U\times V\) containing \((x,y)\). Since \(U\) and \(V\) are open in the standard topology, there are real numbers \(a_0<x<b_0\) and \(c_0<y<d_0\) such that \((a_0,b_0)\subseteq U\) and \((c_0,d_0)\subseteq V\). By the density argument, choose rationals \(a<x<b\) within \((a_0,b_0)\), and rationals \(c<y<d\) within \((c_0,d_0)\). Then \((a,b)\times(c,d)\) belongs to the proposed family, contains \((x,y)\), and is contained in \(U\times V\). The product-topology test proves the claim.
Check at a specific point: take \(U=(1.4,1.5)\), \(V=(3.1,3.2)\), and \((x,y)=(\sqrt{2},\pi)\). The inequalities \(1.4^2=1.96<2<2.25=1.5^2\) give \(1.4<\sqrt{2}<1.5\), and \(3.1<\pi<3.2\). The rational-endpoint rectangle \((7/5,3/2)\times(31/10,16/5)\) is exactly \(U\times V\), since \(7/5=1.4\), \(3/2=1.5\), \(31/10=3.1\), and \(16/5=3.2\). It contains the point and is contained in the chosen neighborhood.
Worked Example: Equal-Width Rectangles
In \(\mathbb{R}^2\) with the standard product topology, let \(\mathcal{R}\) consist of rectangles \((a,b)\times(c,d)\) for which \(b-a=d-c>0\). These equal-width rectangles still form a basis.
To verify local refinement, take any open rectangle \(U\times V\) containing \((x,y)\). There are \(r,s>0\) such that \((x-r,x+r)\subseteq U\) and \((y-s,y+s)\subseteq V\). Choose \(\delta=\min(r,s)>0\). The rectangle \[ (x-\delta/2,x+\delta/2)\times(y-\delta/2,y+\delta/2) \] has width \(\delta\) in each coordinate, contains \((x,y)\), and is contained in \(U\times V\). The product-topology test applies.
Check: the first interval has width \((x+\delta/2)-(x-\delta/2)=\delta\), and the second has width \((y+\delta/2)-(y-\delta/2)=\delta\). Since \(\delta/2\le r/2<r\) and \(\delta/2\le s/2<s\), both intervals lie inside the indicated neighborhoods.
A Family That Fails the Criterion
Worked Example: Rectangles Whose Widths Are at Least One
In \(\mathbb{R}^2\) with the standard product topology, let \(\mathcal{R}\) consist of all open rectangles \((a,b)\times(c,d)\) satisfying \(b-a\ge1\) and \(d-c\ge1\). This family covers \(\mathbb{R}^2\): for any \((x,y)\), the rectangle \((x-1,x+1)\times(y-1,y+1)\) belongs to it and contains \((x,y)\). Its members are open. Nevertheless, it is not a basis for the product topology.
Consider the product-open neighborhood \((-1/4,1/4)\times(-1/4,1/4)\) of \((0,0)\). Each of its coordinate intervals has width \(1/2\). A rectangle in \(\mathcal{R}\) has width at least \(1\) in each coordinate, so neither of its coordinate intervals can fit inside an interval of width \(1/2\). There is therefore no member of \(\mathcal{R}\) containing \((0,0)\) and contained in this neighborhood.
The intersection condition also fails directly. Set \(R_1=(-1/2,1/2)\times(-1/2,1/2)\) and \(R_2=(0,1)\times(0,1)\). Both belong to \(\mathcal{R}\), and \(R_1\cap R_2=(0,1/2)\times(0,1/2)\), which contains \((1/4,1/4)\). No rectangle in \(\mathcal{R}\) can be contained in this intersection, because each of its coordinate intervals would need width at least \(1\), while each interval of the intersection has width \(1/2\).
Check: \(0<1/4<1/2\), so the chosen point really is in the intersection. The intersection calculation uses \((-1/2,1/2)\cap(0,1)=(0,1/2)\) in both coordinates. The failure is not about coverage or openness; it is the absence of sufficiently small basis rectangles.
Common Pitfall: Coverage Is Not Enough
A common mistake is to argue that a family is a basis merely because its members are open and their union is the whole space. The family of rectangles with both widths at least \(1\) has exactly those properties, as the example shows, but it cannot refine small product neighborhoods. The basis criterion has a second, essential condition: overlaps must admit basis refinements. For a basis of a topology already specified, one must also check that the family refines every open set of that topology.
When testing a proposed rectangle family, keep the tasks separate: first check coverage; next check the intersection-refinement condition; finally, if the target is specifically the product topology, check local refinement inside arbitrary open rectangles. This prevents confusing “a basis for some topology” with “a basis for the product topology.”
Check Your Understanding
Use the basis criterion to decide each question, and state which condition supports your conclusion.
- For open rectangles \(U_1\times V_1\) and \(U_2\times V_2\), write their intersection as a product of two sets.
- In \(\mathbb{R}^2\), do rectangles with rational endpoints form a basis for the product topology? Explain the local refinement step.
- Does the family \(\{(a,b)\times(c,d):b-a=d-c>0\}\) form a basis for the standard product topology? Give the construction inside an arbitrary open rectangle.
- Why do open rectangles with both widths at least \(1\) fail to be a basis, despite covering \(\mathbb{R}^2\)?
- State the two conditions in the basis criterion for a collection of subsets of a set \(Z\).