Basic Neighborhoods in the Box Topology

Neighborhoods Built Coordinate by Coordinate

In a product space, a point has one coordinate in each factor. The box topology allows a neighborhood to restrict every one of those coordinates at once. There is no requirement that only finitely many coordinates be restricted. The resulting basic neighborhoods are products of open sets, one chosen in every factor.

Let \(J\) be an index set, and for each \(\alpha\in J\) let \((X_\alpha,\mathcal{T}_\alpha)\) be a topological space. The underlying product set is \(X=\prod_{\alpha\in J}X_\alpha\). A point \(x\in X\) has coordinates \(x_\alpha\in X_\alpha\). Unless specified otherwise, each factor keeps its given topology \(\mathcal{T}_\alpha\).

Definition: A box in \(X=\prod_{\alpha\in J}X_\alpha\) is a set of the form \(\prod_{\alpha\in J}U_\alpha\), where \(U_\alpha\in\mathcal{T}_\alpha\) for every \(\alpha\in J\). The topology whose open sets are arbitrary unions of boxes is the box topology. A box containing \(x\) has \(x_\alpha\in U_\alpha\) for every \(\alpha\in J\).

Thus the choice of \(U_\alpha\) must be made at every index, but it need not differ from \(X_\alpha\) at every index. If \(U_\alpha=X_\alpha\), that coordinate places no restriction on membership in the box. If \(U_\alpha\ne X_\alpha\), it imposes a restriction. The definition allows such restrictions at finitely many, infinitely many, or all indices.

The Basis Test for Box-Open Sets

Theorem (Box basis criterion): The boxes form a basis for the box topology. Consequently, \(G\subseteq X\) is box-open if and only if for every \(x\in G\) there are open sets \(U_\alpha\in\mathcal{T}_\alpha\), one for each \(\alpha\in J\), such that \(x_\alpha\in U_\alpha\) for all \(\alpha\) and \(\prod_{\alpha\in J}U_\alpha\subseteq G\).

Proof. To verify the basis conditions, first take any \(x\in X\). For each \(\alpha\), choose \(U_\alpha=X_\alpha\), which is open in \(X_\alpha\). Then the box \(\prod_{\alpha\in J}X_\alpha=X\) contains \(x\), so the boxes cover \(X\). Next, take two boxes \(B=\prod_{\alpha\in J}U_\alpha\) and \(C=\prod_{\alpha\in J}V_\alpha\), and suppose \(x\in B\cap C\). Their intersection is \(\prod_{\alpha\in J}(U_\alpha\cap V_\alpha)\). Each \(U_\alpha\cap V_\alpha\) is open in \(X_\alpha\), since it is the intersection of two open sets, and it contains \(x_\alpha\). The intersection is therefore a box containing \(x\) and contained in \(B\cap C\). These two conditions show that the boxes form a basis. By the definition of the topology generated by a basis, \(G\) is open exactly when each of its points lies in a basis element contained in \(G\). Applying that statement to the boxes gives the stated criterion. \(\square\)

The criterion is local: to check that \(G\) is open, take each \(x\in G\) and find a box containing \(x\) that stays inside \(G\). A general open neighborhood need not itself be a single box. It is a union of boxes, and the criterion guarantees at least one suitable box around each of its points.

Corollary (Neighborhood form): If \(G\) is box-open and \(x\in G\), then there is a box \(\prod_{\alpha\in J}U_\alpha\) with \(x\in\prod_{\alpha\in J}U_\alpha\subseteq G\). Conversely, any set containing such a box around each of its points is box-open.

Proof. If \(G\) is open and \(x\in G\), the box basis criterion supplies the required box. Conversely, suppose every \(x\in G\) lies in a box \(B_x\subseteq G\). Then \(G=\bigcup_{x\in G}B_x\), because each \(B_x\) is contained in \(G\) and contains \(x\). A union of boxes is open in the box topology, so \(G\) is open. If \(G=\varnothing\), it is open by the definition of a topology, and the condition on its points holds vacuously. \(\square\)

Worked Neighborhoods

Worked Example: A Product with Different Kinds of Factors

Let \(X=\mathbb{R}\times\{p,q\}\times[0,1]\), where \(\mathbb{R}\) has its standard topology, \(\{p,q\}\) has the discrete topology, and \([0,1]\) has the subspace topology inherited from \(\mathbb{R}\). Construct a box neighborhood of \(x=(3,p,1/2)\).

Choose \(U_1=(2,4)\subseteq\mathbb{R}\), \(U_2=\{p\}\subseteq\{p,q\}\), and \(U_3=(1/4,3/4)\cap[0,1]\subseteq[0,1]\). Each is open in its stated factor: \(U_1\) is an open interval, \(U_2\) is open because the second factor is discrete, and \(U_3\) is the intersection of \([0,1]\) with an open subset of \(\mathbb{R}\). The resulting box is \(\displaystyle B=(2,4)\times\{p\}\times\bigl((1/4,3/4)\cap[0,1]\bigr).\)

Check the center coordinate by coordinate: \(2<3<4\), \(p\in\{p\}\), and \(1/4<1/2<3/4\), with \(1/2\in[0,1]\). Thus \(x\in B\). More generally, \((r,s,t)\in B\) exactly when \(2<r<4\), \(s=p\), and \(1/4<t<3/4\) with \(t\in[0,1]\). These conditions verify directly that the box restricts each coordinate using an open set in its own factor.

Check: the third coordinate set is \((1/4,3/4)\), since this interval already lies inside \([0,1]\). Its endpoints are excluded, so \(t=1/4\) and \(t=3/4\) do not belong to \(B\). The coordinate \(s=q\) also fails membership because \(q\notin\{p\}\).

Worked Example: Infinitely Many Restrictions in \(\mathbb{R}^{\mathbb{N}}\)

Give a box neighborhood of the zero sequence in \(\mathbb{R}^{\mathbb{N}}\), where every factor \(\mathbb{R}\) has its standard topology, and exhibit a point in that box with every coordinate nonzero.

For each \(n\in\mathbb{N}=\{1,2,\dots\}\), choose the open interval \(U_n=(-1/n,1/n)\). The box is \(\displaystyle B=\prod_{n\in\mathbb{N}}(-1/n,1/n).\) The zero sequence \(0=(0,0,\dots)\) belongs to \(B\), since for every \(n\ge1\), \(-1/n<0<1/n\).

Define \(y=(y_n)_{n\in\mathbb{N}}\) by \(y_n=1/(2n)\). For each \(n\ge1\), \(1/(2n)>0\) and \(1/(2n)<1/n\), so \(-1/n<y_n<1/n\). Hence \(y\in B\). Also \(y_n=1/(2n)\ne0\) for every \(n\). This verifies that the box imposes a genuine restriction in every coordinate, not merely in a finite list of coordinates.

Check: at \(n=1\), the interval is \((-1,1)\) and \(y_1=1/2\) lies inside it. At \(n=2\), the interval is \((-1/2,1/2)\) and \(y_2=1/4\) lies inside it. The general inequalities above establish membership at every remaining index as well.

Worked Example: Factor Topologies Determine the Available Restrictions

For each \(n\in\mathbb{N}\), let \(X_n=\{0,1\}\). Give \(X_n\) the discrete topology when \(n\) is even and the indiscrete topology when \(n\) is odd. The discrete topology consists of all subsets; the indiscrete topology consists only of \(\varnothing\) and \(X_n\). In \(X=\prod_{n\in\mathbb{N}}X_n\), construct a box containing the zero sequence.

For even \(n\), set \(U_n=\{0\}\). This is open because \(X_n\) is discrete. For odd \(n\), set \(U_n=X_n=\{0,1\}\), which is open in every topology on \(X_n\), including the indiscrete topology. Then \(\displaystyle B=\prod_{n\in\mathbb{N}}U_n\) is a box and contains the zero sequence, since \(0\in U_n\) at every index.

For a sequence \(z=(z_n)\in X\), membership in \(B\) means \(z_n=0\) whenever \(n\) is even, while odd coordinates may be either \(0\) or \(1\). For example, take \(z_n=0\) for even \(n\) and \(z_n=1\) for odd \(n\); this sequence belongs to \(B\). If instead \(w_2=1\), then \(w\notin B\), regardless of its other coordinates, because \(U_2=\{0\}\).

Check: the restrictions are open in their respective factors: \(\{0\}\) is open at every even index because that factor is discrete, and \(\{0,1\}\) is the whole factor at every odd index. The example shows why one must check openness in each factor's stated topology, rather than treating all coordinate spaces as if they had the same open sets.

A Common Mistake: Truncating the Restrictions

A frequent error is to assume that a box neighborhood may restrict only finitely many coordinates. That finite-support condition is not part of the box definition. Here is a concrete calculation showing why replacing infinitely many coordinate sets by the whole factor changes the set.

Start with \(B=\prod_{n\in\mathbb{N}}(-1/n,1/n)\subseteq\mathbb{R}^{\mathbb{N}}\). Fix any \(N\in\mathbb{N}\), and form the truncated set \(\displaystyle B_N=\left(\prod_{n=1}^{N}(-1/n,1/n)\right)\times\left(\prod_{n>N}\mathbb{R}\right).\) The sequence \(v\) defined by \(v_{N+1}=1/(N+1)\) and \(v_n=0\) for \(n\ne N+1\) belongs to \(B_N\): its first \(N\) coordinates are all zero, and the \((N+1)\)-st coordinate is unrestricted there. But \(v\notin B\), because \(v_{N+1}=1/(N+1)\) is an endpoint of the required interval \((-1/(N+1),1/(N+1))\), and the endpoint is excluded. Thus \(B_N\ne B\).

The correction is to check the box definition at every index: for each \(n\), the set \((-1/n,1/n)\) is open in the \(n\)-th factor \(\mathbb{R}\), and the box is their product. No finite cutoff is required. In general, a box is open by virtue of having an open coordinate set at every index; whether that coordinate set equals the whole factor is a separate choice.

Using the Criterion Efficiently

When constructing a box around \(x\), work coordinate by coordinate. At each index \(\alpha\), choose an open set \(U_\alpha\) in \(X_\alpha\) that contains \(x_\alpha\). Then the product contains \(x\). If the aim is to place the box inside a proposed open set \(G\), the choices must also ensure that every point satisfying all the coordinate restrictions lies in \(G\). In applications, this final containment is a condition to verify; coordinatewise openness alone does not prove it.

For a product over an arbitrary index set, the same definition and basis test apply. The index set need not be countable, and the coordinate neighborhoods need not come from intervals or metrics. What matters is that each \(U_\alpha\) is open in its own factor and contains the corresponding coordinate of the point. The neighborhoods in the examples illustrate three uses of the criterion: combining different factor topologies, imposing infinitely many interval restrictions in \(\mathbb{R}^{\mathbb{N}}\), and adapting restrictions to the topology of each individual factor.

Check Your Understanding

Use the box basis criterion and check openness in each coordinate space.

In \(\mathbb{R}^{\mathbb{N}}\), does \(\prod_{n\in\mathbb{N}}(-1/(n+1),1/(n+1))\) contain the zero sequence? Is each coordinate set open?
In \(\prod_{n\in\mathbb{N}}\{0,1\}\) with every factor discrete, describe the box \(\prod_{n\in\mathbb{N}}\{0\}\).
Explain why a box \(\prod_{\alpha\in J}U_\alpha\) containing \(x\) requires \(x_\alpha\in U_\alpha\) at every index.
Suppose \(G\) is box-open and \(x\in G\). What basis criterion guarantees a box neighborhood of \(x\) contained in \(G\)?
For the mixed-topology product in the third worked example, what happens to membership in the displayed box if an even coordinate equals \(1\)?