What a Neighborhood Controls
In a finite product, a familiar open rectangle restricts each coordinate. In an infinite product with the product topology, a basic open set works differently: it restricts only finitely many coordinates and leaves every other coordinate unrestricted. This finite-support condition is the practical way to recognize neighborhoods, even when the product has infinitely many factors.
Let \((X_n,\mathcal{T}_n)\), for \(n\in\mathbb{N}\), be topological spaces, and let \(X=\prod_{n\in\mathbb{N}}X_n\) have the product topology. A point \(x\in X\) is a tuple \(x=(x_n)_{n\in\mathbb{N}}\), with \(x_n\in X_n\). A basic open set is a product \(\prod_{n\in\mathbb{N}}V_n\) in which each \(V_n\) is open in \(X_n\) and \(V_n=X_n\) for all but finitely many indices. A neighborhood of \(x\) means an open set containing \(x\).
This notation is equivalent to the basic open product above: set \(V_n=U_n\) for \(n\in F\), and \(V_n=X_n\) otherwise. The set \(F\) records exactly which coordinates are being controlled. It may be empty, in which case the basic neighborhood is all of \(X\).
The Neighborhood-Basis Test
Proof. By definition of the product topology, its basic open sets form a basis: every open set is a union of basic open sets. Since \(x\in O\), at least one of the basic open sets in such a union contains \(x\) and is contained in \(O\). Write that basic open set as \(\prod_n V_n\), and let \(F\) be the finite set of indices at which \(V_n\ne X_n\). For each \(n\in F\), take \(U_n=V_n\). Because \(x\in\prod_n V_n\), each \(U_n\) contains \(x_n\), and the resulting basic neighborhood is exactly \(\prod_n V_n\). This proves the first assertion. For the consequence, an open \(G\) has the stated property by the first assertion. Conversely, if every \(x\in G\) lies in a basic neighborhood contained in \(G\), then \(G\) is the union of those basic open neighborhoods, so \(G\) is open. \(\square\)
The criterion is local: to show that \(G\) is open, it is enough to find a suitable finite-coordinate neighborhood around each of its points. To show that a proposed set is not open, it is enough to find one point in it for which no such neighborhood fits inside the set.
Worked Example: Restricting Two Coordinates in \(\mathbb{R}^{\mathbb{N}}\)
Give \(\mathbb{R}\) its standard topology and \(X=\mathbb{R}^{\mathbb{N}}\) the product topology. Let \(x_n=n\), so \(x=(1,2,3,\ldots)\). Restrict the second coordinate to \((1,3)\) and the fifth to \((4,6)\). The resulting basic neighborhood is $$ C=\{y\in\mathbb{R}^{\mathbb{N}}:1<y_2<3\text{ and }4<y_5<6\}. $$
Here \(F=\{2,5\}\), a finite set. The intervals contain the corresponding coordinates of \(x\), since \(1<2<3\) and \(4<5<6\); therefore \(x\in C\). Every other coordinate is unrestricted. For instance, define \(y_2=5/2\), \(y_5=11/2\), and \(y_n=100+n\) for \(n\ne2,5\). Then \(1<5/2<3\) and \(4<11/2<6\), so \(y\in C\), regardless of the large values in the unrestricted coordinates.
The inequalities are strict. If instead \(z_2=2\), \(z_5=6\), and \(z_n=0\) for every other \(n\), then \(z\notin C\): the fifth-coordinate condition requires \(z_5<6\), and \(6<6\) is false. This checks both the membership rule and the endpoint convention.
Combining Finite Lists of Restrictions
When two basic neighborhoods overlap, their restrictions combine coordinate by coordinate. Only indices appearing in at least one of the two finite lists can be restricted in the intersection. If both neighborhoods restrict the same coordinate, that coordinate must satisfy both conditions.
Proof. Write the two basic open sets as \(\prod_n V_n\) and \(\prod_n W_n\), where \(V_n=X_n\) except at finitely many indices and \(W_n=X_n\) except at finitely many indices. Their intersection is \(\prod_n(V_n\cap W_n)\): a tuple belongs to both products exactly when each coordinate belongs to both corresponding factor sets. The set \(V_n\cap W_n\) is open, since it is the intersection of two open sets, and it equals \(X_n\) except possibly at indices where \(V_n\ne X_n\) or \(W_n\ne X_n\). There are only finitely many such indices. Thus this product is basic, unless one of its factors is empty, in which case the whole product is empty. If both original sets contain \(x\), every factor intersection contains \(x_n\), so none is empty, and their intersection is a basic neighborhood of \(x\). \(\square\)
Worked Example: Intersecting Restrictions
In \(\mathbb{R}^{\mathbb{N}}\), consider $$ C_1=\{x:-2<x_1<2,\ 0<x_4<2\} $$ and $$ C_2=\{x:-2<x_4<2,\ -4<x_7<2\}. $$ The first set restricts coordinates \(1\) and \(4\); the second restricts \(4\) and \(7\). Their intersection requires all the listed inequalities at once.
At coordinate \(1\), only \(C_1\) imposes a condition, so the restriction remains \(-2<x_1<2\). At coordinate \(7\), only \(C_2\) imposes a condition, so the restriction remains \(-4<x_7<2\). At coordinate \(4\), both conditions apply, and their intersection is $$ (0,2)\cap(-2,2)=(0,2). $$ Thus $$ C_1\cap C_2=\{x:-2<x_1<2,\ 0<x_4<2,\ -4<x_7<2\}. $$
This intersection is nonempty: take \(x_1=0\), \(x_4=1\), \(x_7=0\), and \(x_n=0\) for all other indices. The values satisfy \(-2<0<2\), \(0<1<2\), and \(-4<0<2\), so the tuple belongs to both sets. The result has three restricted coordinates, the union \(\{1,4\}\cup\{4,7\}=\{1,4,7\}\), and all other coordinates remain unrestricted.
Why Restrictions at Every Coordinate Are Different
In an infinite product, choosing an open set in every factor does not automatically produce an open set in the product topology. A basic open set requires that all but finitely many factor sets be the whole factor space. This is not a cosmetic condition: it determines which sets actually contain open neighborhoods.
Worked Example: A Coordinatewise Restriction That Is Not Open
Give \(X=\mathbb{R}^{\mathbb{N}}\) the product topology, and define $$ W=\{x\in X:|x_n|<1/n\text{ for every }n\in\mathbb{N}\}. $$ The zero tuple \(0=(0,0,\ldots)\) belongs to \(W\), since \(|0|=0<1/n\) for every positive integer \(n\). We show that \(W\) is not open by proving that no basic neighborhood of \(0\) is contained in \(W\).
Take any basic neighborhood \(C(0;F,(U_n))\). Its restriction set \(F\) is finite, while \(\mathbb{N}\) is infinite, so choose \(k\in\mathbb{N}\setminus F\). Define \(y\) by \(y_k=1/k\) and \(y_n=0\) for every \(n\ne k\). At each restricted coordinate \(n\in F\), \(y_n=0\in U_n\), because \(U_n\) contains the zero coordinate. Hence \(y\in C(0;F,(U_n))\). But \(y\notin W\), because its \(k\)-th coordinate satisfies \(|y_k|=1/k\), not the required strict inequality \(|y_k|<1/k\).
Thus every basic neighborhood of \(0\) contains a point outside \(W\). By the neighborhood-basis criterion, \(W\) is not open, despite imposing an open interval condition at each coordinate.
Common pitfall. One might argue, “Each interval \((-1/n,1/n)\) is open, so their product \(W\) is open.” This overlooks the finite-support requirement: the factors \((-1/n,1/n)\) differ from \(\mathbb{R}\) at infinitely many indices. The calculation above corrects the claim: for any proposed basic neighborhood, choose an unrestricted coordinate \(k\), set that coordinate equal to the boundary value \(1/k\), and set the other coordinates to zero. The resulting tuple is inside the basic neighborhood but outside \(W\). An infinite collection of open coordinate conditions needs a separate openness test; factorwise openness alone does not settle it.
Worked Example: Neighborhoods in a Product of Discrete Spaces
Let each factor be \(D_n=\{0,1\}\) with the discrete topology, in which every subset is open, and give \(D=\prod_{n\in\mathbb{N}}D_n\) the product topology. At the zero tuple, the set $$ C=\{x\in D:x_1=0,\ x_3=0\} $$ is a basic neighborhood: it restricts the finite set \(F=\{1,3\}\), using the open singleton \(\{0\}\) in each restricted factor.
For example, the tuple \(y\) with \(y_1=0\), \(y_3=0\), and \(y_n=1\) for every other \(n\) belongs to \(C\), since its first and third coordinates meet the restrictions. In contrast, the set requiring \(x_n=0\) for every \(n\) is not a neighborhood of the zero tuple. Given any basic neighborhood of zero with finite restriction set \(F\), choose \(k\notin F\), put \(y_k=1\), and put all other coordinates equal to zero. Then \(y\) belongs to that basic neighborhood but does not have every coordinate zero.
This example has the same finite-support rule as the real product. The factor spaces and their open sets differ, but the form of a basic neighborhood does not: only finitely many coordinates are restricted.
A Practical Test
For a point \(x\) in an infinite product, test a proposed neighborhood by asking which coordinates are actually restricted. If only finitely many are restricted by open sets containing the corresponding coordinates, the resulting set is a basic neighborhood. If infinitely many coordinates are constrained, do not assume the set is open: use the neighborhood-basis criterion at each point. In particular, when testing openness at a point, any proposed basic neighborhood leaves infinitely many coordinates unrestricted if the index set is \(\mathbb{N}\); those coordinates can be used to test whether the neighborhood stays inside the proposed set.
This finite-coordinate viewpoint makes infinite products workable. It lets us build local neighborhoods without controlling an entire infinite tuple, combine neighborhoods by merging finite lists, and detect when an all-coordinate condition is too restrictive to be open.
Check Your Understanding
For each question, identify the finite set of restricted coordinates when there is a basic neighborhood, and pay attention to whether interval endpoints are included.
- In \(\mathbb{R}^{\mathbb{N}}\), describe a basic neighborhood of \(x\) that requires \(x_2-1<y_2<x_2+1\) and \(x_6<y_6<x_6+3\). Which coordinates are unrestricted?
- In \(\mathbb{R}^{\mathbb{N}}\), intersect the conditions \(-1<x_3<2\) and \(0<x_3<4\). What is the resulting restriction at coordinate \(3\)?
- In the product of discrete spaces \(\{0,1\}^{\mathbb{N}}\), is the set requiring \(x_1=1\) and \(x_4=0\) a basic open set? Explain using the factor topologies.
- Show that \(\{x\in\mathbb{R}^{\mathbb{N}}:|x_n|<1\text{ for every }n\in\mathbb{N}\}\) is not a neighborhood of the zero tuple.
- State the neighborhood-basis criterion for openness in an arbitrary product, and explain why it involves only finitely many restricted coordinates at a time.