Countability in Products
A neighborhood in a product space can restrict several coordinates at once. For a finite product, this gives a direct way to combine countable local bases, or countable bases, from the factor spaces. The key is that a finite tuple of choices from countable collections still has only countably many possibilities.
We use the product topology throughout. In a finite product, its basic open sets are rectangles \(U_1\times\cdots\times U_n\), where each \(U_i\) is open in its factor. The number of factors matters: an open set in the product topology is built using restrictions on only finitely many coordinates, a distinction that becomes important for infinite products.
The Product Basis
Let \((X_i,\mathcal{T}_i)\), for \(1\le i\le n\), be topological spaces, with \(n\) a positive integer. Write \(\pi_i:\prod_{j=1}^n X_j\to X_i\) for the coordinate projection. By definition, the product topology is generated by the sets \(\pi_i^{-1}(U_i)\), where \(U_i\in\mathcal{T}_i\). Finite intersections of these generating sets give rectangles: coordinates not restricted by the intersection are assigned the whole factor \(X_i\). Thus every open neighborhood of a point contains an open rectangle about that point.
Proof. Each set \(B_1\times\cdots\times B_n\) is open in the product topology, since it is the finite intersection \(\bigcap_{i=1}^n\pi_i^{-1}(B_i)\) of open sets. Now let \(W\) be open in the product and let \(x=(x_1,\ldots,x_n)\in W\). The description of the product topology gives open sets \(U_i\subseteq X_i\), with \(x_i\in U_i\), such that \(x\in\prod_{i=1}^n U_i\subseteq W\). For every \(i\), the basis property of \(\mathcal{B}_i\) gives a set \(B_i\in\mathcal{B}_i\) with \(x_i\in B_i\subseteq U_i\). Consequently \(x\in\prod_{i=1}^n B_i\subseteq W\). This is the basis condition. \(\square\)
We will also use the elementary countability fact that a finite product of countable sets is countable. For two copies of \(\mathbb{N}\), for instance, the map \((a,b)\mapsto 2^{a-1}(2b-1)\) is an injection into \(\mathbb{N}\): every positive integer has a unique expression as a power of \(2\) times an odd integer. Repeating this pairing argument shows that \(\mathbb{N}^n\) is countable for every finite \(n\). Finite sets and finite collections can be included by indexing or padding them with unused indices.
Finite Products of First-Countable Spaces
Proof. Let \(x=(x_1,\ldots,x_n)\in\prod_{i=1}^n X_i\). For each \(i\), choose a countable local base \(\mathcal{U}_i\) at \(x_i\). Consider the collection \(\mathcal{U}=\{U_1\times\cdots\times U_n:U_i\in\mathcal{U}_i\}\). Every member is an open neighborhood of \(x\). The collection is countable because it is indexed by the finite product of the countable collections \(\mathcal{U}_1,\ldots,\mathcal{U}_n\). To verify the local-base condition, let \(W\) be any open neighborhood of \(x\). It contains a rectangle \(\prod_{i=1}^n O_i\), where each \(O_i\) is open in \(X_i\) and \(x_i\in O_i\). Since \(\mathcal{U}_i\) is a local base at \(x_i\), choose \(U_i\in\mathcal{U}_i\) with \(U_i\subseteq O_i\). Then \(x\in\prod_{i=1}^n U_i\subseteq W\). Hence \(\mathcal{U}\) is a countable local base at \(x\). This holds for every \(x\), proving first countability. \(\square\)
Worked Example: A Local Base in the Plane
Give an explicit countable local base at \((1,-2)\) in \(\mathbb{R}\times\mathbb{R}\), where each copy of \(\mathbb{R}\) has its standard topology and the product has the product topology.
For each \(n\in\mathbb{N}\), define \(U_n=(1-1/n,1+1/n)\times(-2-1/n,-2+1/n)\). Each factor interval is open and contains the corresponding coordinate, so \(U_n\) is an open neighborhood of \((1,-2)\).
To check the local-base condition, let \(W\) be any open neighborhood of \((1,-2)\). It contains a rectangle \(O_1\times O_2\) with \(1\in O_1\) and \(-2\in O_2\), where \(O_1,O_2\) are open in \(\mathbb{R}\). There are positive numbers \(\delta_1,\delta_2\) such that \((1-\delta_1,1+\delta_1)\subseteq O_1\) and \((-2-\delta_2,-2+\delta_2)\subseteq O_2\). Choose \(n\in\mathbb{N}\) large enough that \(1/n<\delta_1\) and \(1/n<\delta_2\). Then \(U_n\subseteq O_1\times O_2\subseteq W\). Thus \(\{U_n:n\in\mathbb{N}\}\) is a countable local base.
Check: for \(n=2\), the set is \((1/2,3/2)\times(-5/2,-3/2)\), which contains \((1,-2)\). If \(1/n<\delta_i\) for both \(i=1,2\), each of its coordinate intervals is contained in the corresponding interval of radius \(\delta_i\), as required.
Worked Example: The Sorgenfrey Line Times the Real Line
Let \(\mathbb{R}_\ell\) denote the real line with the lower limit topology, whose basis consists of intervals \([a,b)\) with \(a<b\). Give a countable local base at \((0,\pi)\) in \(\mathbb{R}_\ell\times\mathbb{R}\), where the second factor has its standard topology.
For \(n\in\mathbb{N}\), take \(V_n=[0,1/n)\times(\pi-1/n,\pi+1/n)\). The first interval is a basic open neighborhood of \(0\) in \(\mathbb{R}_\ell\), and the second is an open neighborhood of \(\pi\) in the standard real line. Thus each \(V_n\) is an open neighborhood of \((0,\pi)\).
Any open neighborhood \(W\) of \((0,\pi)\) contains a product \(O_1\times O_2\) with \(0\in O_1\) and \(\pi\in O_2\). The basis definition for \(\mathbb{R}_\ell\) gives some \(\delta>0\) with \([0,\delta)\subseteq O_1\). The standard topology gives some \(\eta>0\) with \((\pi-\eta,\pi+\eta)\subseteq O_2\). Choose \(n\) such that \(1/n<\delta\) and \(1/n<\eta\). Then \(V_n\subseteq O_1\times O_2\subseteq W\). This explicit local base illustrates the product construction even when the two factors use different topologies.
Check: taking \(n=2\) gives \([0,1/2)\times(\pi-1/2,\pi+1/2)\). It contains \((0,\pi)\), since \(0\in[0,1/2)\) and \(\pi\in(\pi-1/2,\pi+1/2)\).
Finite Products of Second-Countable Spaces
Proof. For each \(i\), let \(\mathcal{B}_i\) be a countable basis for \(X_i\). By the product-basis proposition, the collection \(\mathcal{B}=\{\prod_{i=1}^n B_i:B_i\in\mathcal{B}_i\}\) is a basis for the product. It is countable because it is indexed by the finite product \(\prod_{i=1}^n\mathcal{B}_i\) of countable collections. Therefore the product has a countable basis and is second countable. \(\square\)
Worked Example: A Countable Basis for the Plane
Use the standard topology on each copy of \(\mathbb{R}\). For \(a,b\in\mathbb{Q}\) with \(a<b\), let \((a,b)\) be a rational-endpoint interval. The collection of all such intervals is a countable basis for \(\mathbb{R}\). The product-basis construction gives the collection \(\mathcal{C}=\{(a,b)\times(c,d):a,b,c,d\in\mathbb{Q},\ a<b,\ c<d\}\).
This collection is countable: it is indexed by a subset of \(\mathbb{Q}^4\), and \(\mathbb{Q}\) is countable, so the finite-product countability argument applies. To check that it is a basis, take an open set \(W\subseteq\mathbb{R}^2\) and a point \((x,y)\in W\). The product topology gives open intervals \(O_1,O_2\) such that \((x,y)\in O_1\times O_2\subseteq W\). By density of \(\mathbb{Q}\) in \(\mathbb{R}\), choose rational numbers \(a<x<b\) with \((a,b)\subseteq O_1\), and rational numbers \(c<y<d\) with \((c,d)\subseteq O_2\). Therefore \((x,y)\in(a,b)\times(c,d)\subseteq W\). This proves the basis condition directly.
Check: the rectangle \((-1,2)\times(3,5)\) belongs to \(\mathcal{C}\), since \(-1,2,3,5\) are rational and \(-1<2\), \(3<5\). It contains \((0,4)\), as \(-1<0<2\) and \(3<4<5\).
Why the Finite Hypothesis Matters
The proof for first countability cannot simply be repeated for an arbitrary product. In the product topology, a basic neighborhood restricts only finitely many coordinates. If the index set is uncountable, there can be too many different coordinates to capture with a countable local base. The following example verifies this failure directly.
Worked Example: An Uncountable Product That Is Not First Countable
Let \(J\) be an uncountable set, give \(\{0,1\}\) the discrete topology, and give \(X=\{0,1\}^{J}\) the product topology. Let \(p\in X\) be the point whose every coordinate is \(0\). We show that \(X\) is not first countable at \(p\).
For a finite set \(F\subseteq J\), a basic neighborhood of \(p\) can require specified values in coordinates in \(F\), while leaving all other coordinates unrestricted. Suppose, toward a contradiction, that \(\mathcal{B}\) is a countable local base at \(p\). For each \(B\in\mathcal{B}\), choose one basic neighborhood \(C_B\) satisfying \(p\in C_B\subseteq B\). Let \(F_B\) denote the finite set of coordinates restricted by this chosen \(C_B\). The union \(F=\bigcup_{B\in\mathcal{B}}F_B\) is countable, because it is a countable union of finite sets. Since \(J\) is uncountable, choose \(\alpha\in J\setminus F\).
Now \(U_\alpha=\{x\in X:x_\alpha=0\}\) is an open neighborhood of \(p\): it is the inverse image of the open singleton \(\{0\}\) under the coordinate projection at \(\alpha\). The local-base property supplies \(B\in\mathcal{B}\) with \(p\in B\subseteq U_\alpha\). Then \(C_B\subseteq B\subseteq U_\alpha\). But \(\alpha\notin F\), so \(\alpha\notin F_B\): the cylinder \(C_B\) does not restrict coordinate \(\alpha\). Set that coordinate to \(1\) and leave the coordinates in \(F_B\) as required by \(C_B\); assign arbitrary values, for example \(0\), to any remaining coordinates. The resulting point belongs to \(C_B\) but not to \(U_\alpha\), contradicting \(C_B\subseteq U_\alpha\). No countable local base at \(p\) exists.
Check: the contradiction uses a single fixed \(C_B\) for each basis member \(B\), not a separate choice of cylinder for each coordinate \(\alpha\). The union of the finite supports \(F_B\) is therefore countable, while the local-base condition forces every coordinate \(\alpha\in J\) to occur in that union. This is impossible for uncountable \(J\).
A Common Pitfall
Common pitfall. It is incorrect to use unrestricted coordinate boxes as a basis for an infinite product with the product topology. For example, in \(\{0,1\}^{\mathbb{N}}\), the set \(\prod_{n\in\mathbb{N}}\{0\}=\{p\}\), where \(p\) is the all-zero point, is not open in the product topology. Every basic open neighborhood of \(p\) restricts only finitely many coordinates. Choose a coordinate not among those restrictions and change its value to \(1\); the resulting point remains in that neighborhood but is not \(p\). Thus no basic neighborhood of \(p\) is contained in \(\{p\}\).
In the box topology, by contrast, products of open sets may restrict every coordinate, so \(\{p\}\) is open in this particular product of discrete spaces. That is a different topology. For finite products there is no difference between allowing restrictions on all coordinates and requiring restrictions on only finitely many, because there are only finitely many coordinates. The finite-product theorems above depend on that distinction being handled correctly.
Check Your Understanding
Use the product topology and the definitions of local base and basis in your answers.
- Write a countable local base at \((2,0)\) in \(\mathbb{R}\times\mathbb{R}\) with the standard topologies.
- Explain why the product of two countable local bases at \(x_1\) and \(x_2\) is a local base at \((x_1,x_2)\).
- If \(X_1,\ldots,X_n\) each have a countable basis, describe a countable basis for their product.
- In the uncountable-product example, why must a cylinder contained in \(U_\alpha\) restrict coordinate \(\alpha\)?
- In \(\{0,1\}^{\mathbb{N}}\) with the product topology, explain why the singleton consisting of the all-zero point is not open.