Rectangles in a Product Space
A product space pairs points from two spaces: a point of \(X\times Y\) has the form \((x,y)\), with \(x\in X\) and \(y\in Y\). A rectangular subset has the form \(A\times B\), where \(A\subseteq X\) and \(B\subseteq Y\). Its membership rule is coordinate-by-coordinate: \((x,y)\in A\times B\) exactly when \(x\in A\) and \(y\in B\). When \(A\) and \(B\) are open in their respective spaces, the rectangle is open in the product topology.
Here is the product-topology convention we use. If \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) are topological spaces, their product topology on \(X\times Y\) is generated by the sets \(U\times Y\) and \(X\times V\), where \(U\in\mathcal{T}_X\) and \(V\in\mathcal{T}_Y\). In other words, these coordinate restrictions are subbasic open sets; finite intersections of them are open, as are arbitrary unions of open sets. A product rectangle \(U\times V\) is the intersection \((U\times Y)\cap(X\times V)\), so it is open whenever both factors are open.
The empty set requires care in statements about rectangular sets. If either factor is empty, then \(A\times B=\varnothing\), no matter what the other factor is. Thus the rectangle is open even if its nonempty factor is not open. If neither factor is empty, openness of the rectangle forces both factors to be open. The following result gives the exact condition.
Proof. Suppose first that \(A=\varnothing\) or \(B=\varnothing\). Then \(A\times B=\varnothing\), which is open. If both factors are open, then \(A\times B=(A\times Y)\cap(X\times B)\). Both sets in this intersection are subbasic open sets, so their intersection is open.
For the converse, suppose \(A\times B\) is open and both \(A\) and \(B\) are nonempty. Fix \(a\in A\) and \(b\in B\). By the definition of the topology generated by the coordinate restrictions, an open set containing \((a,b)\) contains a finite intersection of such restrictions that contains \((a,b)\). Combining all restrictions from the first coordinate gives an open set \(U\subseteq X\) containing \(a\); combining those from the second coordinate gives an open set \(V\subseteq Y\) containing \(b\). If there were no restrictions on one coordinate, use the whole space for that coordinate, which is open. The resulting intersection is \(U\times V\), and it is contained in \(A\times B\). Since \(b\in V\), each \(u\in U\) gives \((u,b)\in U\times V\subseteq A\times B\), so \(u\in A\). Therefore \(U\subseteq A\). For every \(a\in A\), this supplies an open neighborhood \(U\) contained in \(A\), so \(A\) is open. To show \(B\) is open, fix any \(b\in B\) and choose one \(a\in A\), which is possible because \(A\ne\varnothing\). The same argument supplies an open neighborhood \(V\) of \(b\) contained in \(B\). Hence \(B\) is open as well. \(\square\)
Rectangular Neighborhoods in the Plane
Take \(X=Y=\mathbb{R}\), each with its standard topology. The product topology on \(\mathbb{R}\times\mathbb{R}\), identified with \(\mathbb{R}^2\), has open rectangles \(U\times V\) for open subsets \(U,V\subseteq\mathbb{R}\). In particular, if \(a\in U\) and \(b\in V\), then \(U\times V\) is an open neighborhood of \((a,b)\). When \(U\) and \(V\) are intervals, this is the familiar rectangular shape, but the definition applies to any open factors.
Worked Example: A Neighborhood in \(\mathbb{R}^2\)
In the product topology on \(\mathbb{R}\times\mathbb{R}\), consider the rectangle \(R=(0,2)\times(1,3)\). Check that \(R\) is an open neighborhood of \((1,2)\), and compare it with a Euclidean ball around that point.
The factors \((0,2)\) and \((1,3)\) are open in the standard topology on \(\mathbb{R}\). Theorem 1 therefore gives that \(R\) is open in the product topology. Since \(0<1<2\) and \(1<2<3\), the point \((1,2)\) belongs to \(R\). Thus \(R\) is an open neighborhood of that point.
For a direct comparison using the Euclidean metric, let \((s,t)\in\mathbb{R}^2\) satisfy \(\|(s,t)-(1,2)\|<1\). Then \[ |s-1|\leq\sqrt{(s-1)^2+(t-2)^2}<1 \quad\text{and}\quad |t-2|\leq\sqrt{(s-1)^2+(t-2)^2}<1. \] It follows that \(0<s<2\) and \(1<t<3\). Hence \(B((1,2),1)\subseteq R\). Conversely, the rectangle \((1-\tfrac14,1+\tfrac14)\times(2-\tfrac14,2+\tfrac14)\) lies in \(B((1,2),1)\): for a point in this rectangle, each coordinate difference is less than \(\tfrac14\), so its distance from \((1,2)\) is less than \(\sqrt{(\tfrac14)^2+(\tfrac14)^2}=\tfrac{\sqrt{2}}4<1\). These inclusions verify how rectangles and Euclidean balls give neighborhoods at this point.
Check: the center satisfies \(1\in(0,2)\) and \(2\in(1,3)\), and the radius-one ball is contained in the rectangle by the coordinate bounds just computed.
A Line Times a Circle
Let \(X=\mathbb{R}\) with its standard topology, and let \(Y=S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\) with the subspace topology inherited from the Euclidean topology on \(\mathbb{R}^2\). An open subset of \(S^1\) is the intersection of \(S^1\) with an open subset of \(\mathbb{R}^2\). A rectangle \(U\times V\), with \(U\) open in \(\mathbb{R}\) and \(V\) open in \(S^1\), is open in the product topology on \(\mathbb{R}\times S^1\). The second factor need not be an interval in the real line: it is open relative to the circle.
Worked Example: A Neighborhood on the Cylinder
In \(\mathbb{R}\times S^1\), construct an open rectangular neighborhood of \((2,(0,1))\), and verify that its circle factor is open in \(S^1\).
Use \(U=(1,3)\subseteq\mathbb{R}\), and let \[ V=S^1\cap B_{\mathbb{R}^2}((0,1),\tfrac12). \] The Euclidean ball \(B_{\mathbb{R}^2}((0,1),\tfrac12)\) is open in \(\mathbb{R}^2\). By the definition of the subspace topology, its intersection \(V\) with \(S^1\) is open in \(S^1\). Also, \((0,1)\in S^1\), since \(0^2+1^2=1\), and its Euclidean distance from the center \((0,1)\) is \(0<\tfrac12\). Thus \((0,1)\in V\). The first coordinate \(2\) satisfies \(1<2<3\), so \((2,(0,1))\in U\times V\).
Both factors are open in their stated spaces. Therefore \(U\times V\) is open in \(\mathbb{R}\times S^1\), and the membership check shows that it is a neighborhood of the specified point. The radius condition also describes exactly which circle points are admitted: \((u,v)\in V\) precisely when \(u^2+v^2=1\) and \(u^2+(v-1)^2<\tfrac14\).
Check: substituting the center point gives \(0^2+(1-1)^2=0<\tfrac14\), as required for membership in \(V\).
A Discrete Space Times the Real Line
Let \(D=\{p,q\}\) have the discrete topology, in which every subset of \(D\) is open, and let \(\mathbb{R}\) have its standard topology. In \(D\times\mathbb{R}\), a set of the form \(\{p\}\times J\), with \(J\) open in \(\mathbb{R}\), is an open rectangle. It restricts the first coordinate to \(p\) and the second coordinate to an open set \(J\). The discrete factor allows either coordinate value to be selected as an open set on its own.
Worked Example: A Neighborhood in a Discrete Product
In \(D\times\mathbb{R}\), find an open rectangular neighborhood of \((p,0)\), then decide whether \(A\times B\) is open for \(A=\{p\}\) and \(B=[0,1]\).
Take \(U=\{p\}\) and \(V=(-1,1)\). The set \(U\) is open because \(D\) is discrete, and \(V\) is open in the standard topology on \(\mathbb{R}\). Also \(p\in U\) and \(-1<0<1\), so \((p,0)\in U\times V\). Thus \(\{p\}\times(-1,1)\) is an open rectangular neighborhood of \((p,0)\).
For the second question, \(A=\{p\}\) is open in \(D\), but \(B=[0,1]\) is not open in \(\mathbb{R}\): every open interval around \(0\) contains negative real numbers, so no open neighborhood of \(0\) is contained in \([0,1]\). Neither factor is empty, and \(B\) is not open. Theorem 1 shows that \(\{p\}\times[0,1]\) is not open in \(D\times\mathbb{R}\).
Check: the point \((p,0)\) belongs to \(\{p\}\times[0,1]\), but the rectangle fails the open-set condition because its second factor is not open.
Common Pitfall: Forgetting the Empty Case
A common error is to say that \(A\times B\) is open exactly when both \(A\) and \(B\) are open, without first checking whether either factor is empty. For a counterexample, use \(X=Y=\mathbb{R}\) with the standard topology, \(A=\varnothing\), and \(B=[0,1]\). The set \(B\) is not open, but \[ A\times B=\varnothing\times[0,1]=\varnothing, \] which is open. The theorem handles this case correctly because \(A=\varnothing\). When both factors are nonempty, however, the requirement that both be open is necessary, not just sufficient.
Rectangles are useful because they let us specify a neighborhood by choosing conditions on each coordinate separately. To verify an open rectangle, state the topology on each factor, check that each factor is open there, and check that the point under consideration belongs to both factors. These checks also prevent a relative-open set on a subspace, such as the circle, from being confused with an open subset of the ambient plane.
Check Your Understanding
For each question, keep track of the topology on both factors and consider separately the case in which a factor is empty.
- In the product topology on \(\mathbb{R}\times\mathbb{R}\), is \((2,5)\times(-1,1)\) open? State the reason.
- Let \(X\) and \(Y\) be topological spaces, \(A\subseteq X\), and \(B\subseteq Y\). If \(A\times B\) is open and both factors are nonempty, what can be concluded about \(A\) and \(B\)?
- In \(\mathbb{R}\times S^1\), explain why \(\mathbb{R}\times\bigl(S^1\cap B_{\mathbb{R}^2}((1,0),1)\bigr)\) is open.
- Let \(D=\{p,q\}\) have the discrete topology. Is \(\{q\}\times(2,4)\) open in \(D\times\mathbb{R}\)? Why?
- Give an example of a rectangle that is open although one of its factors is not open, and explain which condition makes this possible.