Distinct Product and Box Topologies on R^N

Two Topologies on the Same Sequence Space

On a finite product, the product and box topologies agree. On \(\mathbb{R}^{\mathbb{N}}\), they do not: the product topology permits restrictions in only finitely many coordinates at a time, while the box topology permits restrictions in every coordinate. We will prove the difference by examining a specific box-open set and showing that none of its points has a product-open neighborhood contained in it.

Here \(\mathbb{R}\) has its standard topology, \(\mathbb{N}=\{1,2,\dots\}\), and \(\mathbb{R}^{\mathbb{N}}\) is the set of all real sequences \(x=(x_1,x_2,\dots)\). In each topology, a basic open set is formed by choosing an open set in each coordinate and taking their product. The difference is whether only finitely many coordinate sets may be proper subsets of \(\mathbb{R}\), or whether every coordinate may be restricted.

Definition: A basic open set in the product topology on \(\mathbb{R}^{\mathbb{N}}\) has the form \(\prod_{n\in\mathbb{N}} U_n\), where every \(U_n\) is open in \(\mathbb{R}\) and \(U_n=\mathbb{R}\) except for finitely many indices \(n\). A basic open set in the box topology has the same form, but each \(U_n\) may be any open subset of \(\mathbb{R}\), with no finite-restriction condition. The topologies consist of arbitrary unions of their respective basic open sets.

A basic open set containing \(x\) is a basic neighborhood of \(x\). In the product topology such a neighborhood can constrain only finitely many coordinates; all the other coordinates remain unrestricted. In the box topology, a basic neighborhood can constrain every coordinate at once. Since every product basic open set is also a box basic open set, the product topology is contained in the box topology. The key question is whether that containment is strict.

A Set That Separates the Topologies

Theorem: The product topology on \(\mathbb{R}^{\mathbb{N}}\) is strictly coarser than the box topology.

Proof. First, let \(G\) be open in the product topology. By definition, \(G\) is a union of product basic open sets. Each such set is box basic open, because its coordinate sets are open and the box definition allows all the restrictions permitted by the product definition. Therefore \(G\) is a union of box-open sets, so \(G\) is open in the box topology. This proves that every product-open set is box-open.

To show the containment is strict, let \(U=\prod_{n\in\mathbb{N}}(-1,1)\). Each interval \((-1,1)\) is open in \(\mathbb{R}\), so \(U\) is box-open. We show that no point \(x\in U\) has a product-open neighborhood contained in \(U\). Suppose \(G\) is product-open and \(x\in G\). There is a product basic open set \(B=\prod_{n\in\mathbb{N}}V_n\) such that \(x\in B\subseteq G\). There is a finite set \(F\subseteq\mathbb{N}\) outside which \(V_n=\mathbb{R}\). Because \(\mathbb{N}\) is infinite, choose \(k\in\mathbb{N}\setminus F\). Define \(y\in\mathbb{R}^{\mathbb{N}}\) by \(y_k=2\) and \(y_n=x_n\) for every \(n\ne k\). For \(n\in F\), \(y_n=x_n\in V_n\); for \(n\notin F\), \(V_n=\mathbb{R}\), so \(y_n\in V_n\) as well. Thus \(y\in B\subseteq G\). But \(y_k=2\notin(-1,1)\), so \(y\notin U\). Hence \(G\) is not contained in \(U\). Since this holds for every \(x\in U\) and every product-open \(G\) containing \(x\), \(U\) is not product-open. The inclusion of the product topology in the box topology is therefore strict. \(\square\)

The argument establishes more than non-openness: the interior of \(U\) in the product topology is empty. Indeed, if a point of \(U\) belonged to a product-open subset of \(U\), that subset would be a product-open neighborhood contained in \(U\), which the proof rules out.

Worked Example: The Box Around the Zero Sequence

Let \(0=(0,0,\dots)\) and \(U=\prod_{n\in\mathbb{N}}(-1,1)\). The point \(0\) lies in \(U\), since \(0\in(-1,1)\) in every coordinate. We will test a product basic neighborhood of \(0\).

For example, take \(B=(-1/2,1/2)\times\mathbb{R}\times(-1,1)\times\prod_{n\ge4}\mathbb{R}\). Its restrictions occur only in coordinates \(1\) and \(3\), and each restricted interval contains the corresponding coordinate of \(0\). Thus \(0\in B\), and \(B\) is product-open. Let \(y\) be the sequence with \(y_4=2\) and \(y_n=0\) for \(n\ne4\). Then \(y_1=0\in(-1/2,1/2)\), \(y_3=0\in(-1,1)\), and all unrestricted coordinates satisfy their factor condition \(\mathbb{R}\). Consequently \(y\in B\). However, \(y_4=2\notin(-1,1)\), so \(y\notin U\). Therefore this product neighborhood of \(0\) is not contained in \(U\).

Check: The escape coordinate is unrestricted because \(B\) restricts only coordinates \(1\) and \(3\). The value \(2\) is outside the interval \((-1,1)\), whose membership condition is strict. This is exactly the feature used in the theorem’s argument.

The Same Obstruction at Other Points

The failure is not tied to the zero sequence. A product neighborhood of any point has only finitely many restricted coordinates, so one can change an unrestricted coordinate to leave the box. The next example makes this point explicit at a sequence whose coordinates are not all equal.

Worked Example: A Nonzero Point and a Product Neighborhood

Define \(a\in\mathbb{R}^{\mathbb{N}}\) by \(a_n=n/(n+1)\). For every \(n\ge1\), \(0<n/(n+1)<1\), so \(a\in U=\prod_{n\in\mathbb{N}}(-1,1)\).

Consider the product basic open set whose coordinate \(1\) factor is \((1/4,3/4)\), whose coordinate \(3\) factor is \((13/20,17/20)\), and whose other factors are \(\mathbb{R}\). It contains \(a\): \(a_1=1/2\), and \(1/4<1/2<3/4\); also \(a_3=3/4=15/20\), and \(13/20<15/20<17/20\). Now define \(y\) by \(y_4=2\) and \(y_n=a_n\) for \(n\ne4\). The restricted coordinates of \(y\) equal those of \(a\), so \(y\) belongs to this product basic open set. But \(y_4=2\) is not in \((-1,1)\), so \(y\notin U\).

Check: The two restricted coordinates satisfy their open interval inequalities, and coordinate \(4\) is unrestricted. Thus the set is indeed a product neighborhood of \(a\), while the computed fourth coordinate shows it is not contained in \(U\).

The theorem proves the corresponding statement for every product neighborhood of \(a\), not just the particular one calculated here. If a product-open set contains \(a\), it contains a product basic open set containing \(a\); since that basic set restricts only finitely many coordinates, one can change a coordinate outside that finite set and obtain a point outside \(U\).

Coordinate-Dependent Restrictions

Box neighborhoods can also shrink at different rates in different coordinates. This gives another direct way to see why finite-coordinate reasoning cannot describe every box neighborhood.

Worked Example: Intervals Shrinking with the Coordinate Index

Let \(W=\prod_{n\in\mathbb{N}}(-1/n,1/n)\). Each factor is open in \(\mathbb{R}\), so \(W\) is box-open. It contains the zero sequence because \(0\) satisfies \(-1/n<0<1/n\) for every positive integer \(n\).

Take any product basic neighborhood \(B=\prod_{n\in\mathbb{N}}V_n\) of \(0\). Let \(F\) be the finite set of coordinates where \(V_n\ne\mathbb{R}\). Choose \(k\notin F\), and form \(y\) with \(y_k=2/k\) and \(y_n=0\) for \(n\ne k\). The point \(y\) belongs to \(B\): its coordinates in \(F\) equal the zero coordinates of \(0\), and its other coordinates belong to the unrestricted factors \(\mathbb{R}\). But \(y\notin W\), because \(y_k=2/k>1/k\), so \(y_k\notin(-1/k,1/k)\). Thus no product basic neighborhood of \(0\) is contained in \(W\), and consequently \(W\) is not product-open.

Check: The failure is at coordinate \(k\), where the interval condition is strict: \(2/k\) is larger than \(1/k\), not equal to it. The other coordinates are zero and satisfy their respective interval conditions.

A Common Mistake

Common pitfall: “Every coordinate factor is open, so the product of those factors is product-open.” This reasoning confuses the two definitions. For the set \(U=\prod_{n\in\mathbb{N}}(-1,1)\), every factor is indeed open, but infinitely many factors are proper subsets of \(\mathbb{R}\). That makes \(U\) box-open, not necessarily product-open. The product topology requires a basic open neighborhood to use \(\mathbb{R}\) in all but finitely many coordinates.

A related incorrect argument is to take a point \(x\in U\), choose an open interval around each \(x_n\) contained in \((-1,1)\), and then declare the product of all those intervals a product neighborhood. That construction does give a box basic neighborhood, but it imposes restrictions in infinitely many coordinates. To obtain a product basic neighborhood, all but finitely many factors must instead be \(\mathbb{R}\). The escaping-coordinate construction shows why those unrestricted coordinates matter.

Takeaway: The product topology on \(\mathbb{R}^{\mathbb{N}}\) allows finitely many coordinate restrictions in a basic open set; the box topology allows restrictions in every coordinate. Therefore every product-open set is box-open, but \(\prod_{n\in\mathbb{N}}(-1,1)\) is box-open and not product-open. At each point of this set, every product neighborhood contains a point outside it.

Check Your Understanding

For each question, distinguish carefully between finitely many restricted coordinates and restrictions in every coordinate.

Why is \(\prod_{n\in\mathbb{N}}(-1,1)\) box-open? Which part of the product basic-open-set definition prevents the same argument from proving it product-open?
Given a product basic neighborhood of a point in \(\prod_{n\in\mathbb{N}}(-1,1)\), explain how to choose an unrestricted coordinate and construct a point outside this set.
Is \(\prod_{n\in\mathbb{N}}(-1/n,1/n)\) box-open? Use the zero sequence to show that no product basic neighborhood of it is contained in this set.
Explain why every product-open set on \(\mathbb{R}^{\mathbb{N}}\) is box-open.
In the second worked example, verify the inequalities placing \(a_1=1/2\) and \(a_3=3/4\) in their specified open intervals.