Coordinate Restrictions in a Product
In a product space, a set can be specified by asking that one coordinate lie in a chosen open set, while leaving every other coordinate unrestricted. Such a set is a cylinder: its defining condition concerns one coordinate, but its points are tuples from the whole product. Combining finitely many coordinate conditions gives the basic open sets of the product topology. The purpose here is to make that description precise and prove that these cylinders generate the topology.
Let \(J\) be an index set, let \((X_i,\mathcal{T}_i)\) be a topological space for each \(i\in J\), and write \(X=\prod_{i\in J}X_i\) for their Cartesian product. A point \(x\in X\) is a tuple with \(i\)-th coordinate \(x_i\in X_i\). For each \(i\in J\), the coordinate projection is the function \(\pi_i:X\to X_i\) defined by \(\pi_i(x)=x_i\). The product topology has as a basis the sets that restrict finitely many coordinates to open sets and leave all remaining coordinates unrestricted.
The phrase “finite-coordinate cylinder” allows a restriction on several coordinates at once. A single coordinate cylinder is the special case \(F=\{i\}\). If \(X\) is empty, these definitions still apply: every coordinate preimage is empty, and the product topology on \(X=\varnothing\) is \(\{\varnothing\}\). If \(J=\varnothing\), the product consists of the unique empty tuple; the finite-coordinate cylinder for \(F=\varnothing\) is \(X\).
Finite Intersections of Coordinate Cylinders
Proof. A point \(x\in X\) belongs to the left-hand side exactly when \(x_{i_k}\in U_k\) for every \(k\in\{1,\ldots,m\}\). Fix a coordinate \(i\in F\). The conditions among these that concern \(i\) hold exactly when \(x_i\) belongs to every \(U_k\) with \(i_k=i\), which is equivalent to \(x_i\in V_i\). Requiring this for every \(i\in F\) is precisely membership in the right-hand side. Each \(V_i\) is open in \(X_i\), since it is a finite intersection of open sets. If \(m=0\), both sides are \(X\) by the empty-intersection convention. \(\square\)
This calculation also identifies two useful cases. If different coordinates are restricted, the restrictions remain side by side. If the same coordinate is restricted more than once, the conditions combine into a single open set in that factor. If that intersection of factor-open sets is empty, the resulting cylinder is empty.
Worked Example: Intersecting Cylinders in \(\mathbb{R}^3\)
Give \(\mathbb{R}^3\) the product topology from three copies of \(\mathbb{R}\) with their standard topologies. Let \(C_1=\pi_2^{-1}((-1,2))\), \(C_2=\pi_1^{-1}((0,3))\), and \(C_3=\pi_2^{-1}((0,1))\). Compute \(C_1\cap C_2\cap C_3\).
The first and third conditions both restrict the second coordinate. Their combined restriction is \((-1,2)\cap(0,1)=(0,1)\). The second condition restricts the first coordinate to \((0,3)\), and no condition restricts the third coordinate. Therefore \[ C_1\cap C_2\cap C_3=(0,3)\times(0,1)\times\mathbb{R}. \]
Check: For \(x=(2,\tfrac12,7)\), the second coordinate \(\tfrac12\) lies in both \((-1,2)\) and \((0,1)\), and the first coordinate \(2\) lies in \((0,3)\), so \(x\) belongs to all three cylinders and to the displayed product. For \(y=(2,\tfrac32,7)\), the second coordinate \(\tfrac32\) lies in \((-1,2)\) but not in \((0,1)\), so \(y\notin C_3\) and \(y\notin(0,3)\times(0,1)\times\mathbb{R}\).
Worked Example: Two Restrictions on One Finite Factor
Let \(X_1=\{a,b,c\}\) and \(X_2=\{0,1\}\), each with the discrete topology, and give \(X_1\times X_2\) the product topology. Compute \(\pi_1^{-1}(\{a,b\})\cap\pi_1^{-1}(\{b,c\})\cap\pi_2^{-1}(\{1\})\).
The two restrictions on the first coordinate combine as \(\{a,b\}\cap\{b,c\}=\{b\}\). The restriction on the second coordinate is \(\{1\}\). Thus the intersection is \(\{b\}\times\{1\}=\{(b,1)\}\).
For a direct check, the product has the six points \((a,0),(a,1),(b,0),(b,1),(c,0),(c,1)\). Of these, the first-coordinate conditions retain \((b,0)\) and \((b,1)\); the second-coordinate condition retains only \((b,1)\). The answer agrees with the cylinder calculation. If instead the first-coordinate restrictions were \(\{a\}\) and \(\{c\}\), their intersection would be empty, so the entire intersection of those cylinders would be empty.
Worked Example: A Cylinder in a Countable Product
Give \(X=\{0,1\}^{\mathbb{N}}\) the product topology, using the discrete topology on each factor \(\{0,1\}\). Consider \(C=\pi_2^{-1}(\{1\})\cap\pi_5^{-1}(\{0\})\). Describe membership and test two points.
By definition, \(C\) consists of all sequences \(x=(x_1,x_2,\ldots)\) whose second coordinate is \(1\) and whose fifth coordinate is \(0\). There is no condition on any other coordinate. For \(x=(0,1,0,1,0,1,1,\ldots)\), we have \(x_2=1\) and \(x_5=0\), so \(x\in C\). For \(y=(0,1,0,1,1,1,1,\ldots)\), we have \(y_2=1\) but \(y_5=1\ne0\), so \(y\notin C\).
Check: The membership test uses exactly the two specified coordinates: \(x_2=1\), \(x_5=0\); \(y_2=1\), \(y_5=1\). The values in the other coordinates do not affect either conclusion.
The Projection Subbasis Generates the Product Topology
A basis describes open sets as unions of its members. A subbasis is a family from which a basis can be formed by taking finite intersections. For the product topology, the coordinate cylinders form a subbasis. The next result proves this claim, including why the generated topology is exactly the product topology rather than merely a topology contained in it.
Proof. By the finite-intersection proposition, every finite intersection of members of \(\mathcal{S}\) is a set of the form \(\bigcap_{i\in F}\pi_i^{-1}(U_i)\), with \(F\subseteq J\) finite and each \(U_i\in\mathcal{T}_i\). Such sets are exactly the standard basic open sets for the product topology: they impose open restrictions on finitely many coordinates and leave all other coordinates unrestricted. Conversely, every standard basic open set has this form, so it is a finite intersection of members of \(\mathcal{S}\). The product topology consists of unions of its basic open sets. Therefore it consists exactly of arbitrary unions of finite intersections of members of \(\mathcal{S}\), which is the topology generated by \(\mathcal{S}\). If \(J=\varnothing\), there are no members of \(\mathcal{S}\); its only finite intersection is \(X\), and its arbitrary unions are \(\varnothing\) and \(X\), the topology on the one-point product. \(\square\)
The theorem is a statement about the open sets of the product topology: every open set is a union of finite-coordinate cylinders, and every such union is open. It does not say that every open set is itself one cylinder. A union can combine many different coordinate restrictions, while a finite intersection combines only finitely many of them at a time.
Worked Example: Recovering an Open Rectangle from the Subbasis
In \(\mathbb{R}^2\) with the product topology, let \(R=(1,4)\times(-2,0)\). Write \(R\) using coordinate projection inverse images, and verify membership for two points.
The first-coordinate condition is \(x_1\in(1,4)\), so its cylinder is \(\pi_1^{-1}((1,4))\). The second-coordinate condition is \(x_2\in(-2,0)\), giving \(\pi_2^{-1}((-2,0))\). Hence \[ R=\pi_1^{-1}((1,4))\cap\pi_2^{-1}((-2,0)). \] For the point \((2,-1)\), the inequalities \(1<2<4\) and \(-2<-1<0\) both hold, so the point belongs to both cylinders and to \(R\). For \((2,0)\), the first inequality holds, but the second-coordinate condition fails because \(0\notin(-2,0)\); thus \((2,0)\notin R\).
Check: The intersection imposes both coordinate conditions: \((2,-1)\) meets both, whereas \((2,0)\) fails the strict upper bound in the second coordinate.
A Common Mistake
Common pitfall. A finite intersection of coordinate cylinders need not be a cylinder restricting only one coordinate. For example, in \(\mathbb{R}^2\), \[ \pi_1^{-1}((0,1))\cap\pi_2^{-1}((0,1))=(0,1)\times(0,1). \] It is incorrect to treat this intersection as a restriction on just the first coordinate. The point \((\tfrac12,2)\) satisfies the first-coordinate restriction but is not in the intersection, since \(2\notin(0,1)\). The correct description retains both conditions, one on each coordinate. By contrast, two restrictions on the same coordinate do combine into one: \(\pi_1^{-1}((0,2))\cap\pi_1^{-1}((1,3))=\pi_1^{-1}((1,2))\).
Thus the practical rule is to collect all restrictions coordinate by coordinate. Intersect the factor-open sets whenever a coordinate repeats, keep restrictions on different coordinates together, and use the empty-intersection convention when there are no restrictions. This is the calculation behind the subbasis description of the product topology.
Check Your Understanding
Use the coordinate conditions directly, and distinguish restrictions on the same coordinate from restrictions on different coordinates.
- In \(\mathbb{R}^3\), rewrite \(\pi_1^{-1}((0,2))\cap\pi_3^{-1}((-1,1))\) as a product of three sets.
- For any \(i\in J\) and open sets \(U,V\in\mathcal{T}_i\), simplify \(\pi_i^{-1}(U)\cap\pi_i^{-1}(V)\).
- In \(\mathbb{R}^2\), determine whether \((1,0)\) belongs to \(\pi_1^{-1}((0,2))\cap\pi_2^{-1}((0,1))\).
- Explain why the empty intersection of coordinate cylinders is \(X\), not \(\varnothing\).
- For \(X=\{0,1\}^{\mathbb{N}}\), describe the sequences in \(\pi_1^{-1}(\{0\})\cap\pi_3^{-1}(\{1\})\).