From Coordinates to a Product
In a product space, a neighborhood can restrict only finitely many coordinates at a time. This is why convergence in the product topology can be checked one coordinate at a time. The box topology allows a neighborhood to restrict every coordinate simultaneously, so the same test does not suffice there. Nets make the distinction precise, including for products with infinitely many factors.
Let \(I\) be an index set, and for each \(\alpha\in I\) let \((X_\alpha,\mathcal{T}_\alpha)\) be a topological space. The product set is \(X=\prod_{\alpha\in I}X_\alpha\). A point \(x\in X\) has coordinates \(x_\alpha\in X_\alpha\). The product topology has a basis consisting of sets that restrict finitely many coordinates: each such basic open set has the form \(\prod_{\alpha\in I}U_\alpha\), where every \(U_\alpha\) is open in \(X_\alpha\), and \(U_\alpha=X_\alpha\) except for finitely many \(\alpha\). In the box topology, the same form is allowed, but there is no requirement that only finitely many \(U_\alpha\) differ from \(X_\alpha\).
The order on \(D\) records which indices count as later. Directedness is important: it lets us find one index that is later than finitely many indices, even when those indices arise from different coordinate convergence tests.
Proof. Suppose first that \(x_d\to x\) in the product topology. Fix \(\alpha\in I\) and an open neighborhood \(U_\alpha\) of \(x_\alpha\) in \(X_\alpha\). The set of product points whose \(\alpha\)-coordinate belongs to \(U_\alpha\), with no restrictions on other coordinates, is a basic open neighborhood of \(x\). The net is eventually in this set, so its \(\alpha\)-coordinates are eventually in \(U_\alpha\). This holds for every such \(U_\alpha\), hence the coordinate net converges to \(x_\alpha\).
Conversely, suppose every coordinate net converges to its corresponding coordinate of \(x\). Let \(W=\prod_{\alpha\in I}U_\alpha\) be a basic product neighborhood of \(x\), and let \(F=\{\alpha\in I:U_\alpha\ne X_\alpha\}\), which is finite. For each \(\alpha\in F\), coordinate convergence gives an index \(d_\alpha\) such that \(d\geq d_\alpha\) implies \((x_d)_\alpha\in U_\alpha\). If \(F\ne\varnothing\), directedness, applied finitely many times, gives an index \(d_0\) later than every \(d_\alpha\). Thus every \(d\geq d_0\) has \(x_d\in W\). If \(F=\varnothing\), then \(W=X\), and every term is already in \(W\). Finally, every open neighborhood of \(x\) contains a basic product neighborhood of \(x\), so the net converges to \(x\). \(\square\)
If \(I\) is empty, the product has one point and there are no coordinate conditions; every net in that one-point space converges to its point. If the product has no points, there is no net in it to which the theorem would apply.
Finite Coordinate Restrictions in Practice
Worked Example: A Sequence in \(\mathbb{R}^2\)
Give \(\mathbb{R}\) its standard topology and \(\mathbb{R}^2\) the product topology. Consider \(x_n=(1/n,3/n)\), for \(n\in\mathbb{N}\), and test whether it converges to \((0,0)\).
The first coordinate converges to \(0\): given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For \(n\geq N\), \(0<1/n\leq1/N<\varepsilon\). The second coordinate also converges to \(0\): choose \(N\in\mathbb{N}\) with \(N>3/\varepsilon\); then \(0<3/n\leq3/N<\varepsilon\) for \(n\geq N\). Taking \(N>3/\varepsilon\) works for both coordinates, since \(3/\varepsilon>1/\varepsilon\). The product convergence criterion gives \(x_n\to(0,0)\).
Check: a basic neighborhood of \((0,0)\) has the form \(U_1\times U_2\), where each \(U_i\) is an open neighborhood of \(0\) in \(\mathbb{R}\). Choose \(\varepsilon_1,\varepsilon_2>0\) with \((-\varepsilon_i,\varepsilon_i)\subseteq U_i\). Taking \(N> \max\{1/\varepsilon_1,3/\varepsilon_2\}\) ensures both coordinates lie in their respective intervals for every \(n\geq N\).
The key step in the theorem’s proof was finding one index that works for a finite list of coordinates. There need not be a single index that works for infinitely many independently chosen coordinate thresholds. The next example makes that difference explicit.
Worked Example: A Net in an Infinite Product
Give each factor \(\mathbb{R}\) its standard topology and give \(X=\mathbb{R}^{\mathbb{N}}\) the product topology. Let \(D\) be the set of finite subsets \(F\) of \(\mathbb{N}\), ordered by inclusion. This is directed: for finite \(F,G\), the finite set \(F\cup G\) contains both. Define \(x_F\in X\) by \[ (x_F)_k= \begin{cases} 0,&k\in F,\\ 1,&k\notin F. \end{cases} \] We claim that the net \((x_F)\) converges to the zero sequence.
Fix a coordinate \(k\). Once \(F\) contains \(\{k\}\), every larger finite set also contains \(k\), so the \(k\)-th coordinate is \(0\) thereafter. Thus the \(k\)-th coordinate net converges to \(0\). By the product convergence criterion, \(x_F\) converges to the zero sequence.
Check with a basic neighborhood: let a basic neighborhood of zero restrict a finite set \(K\subseteq\mathbb{N}\) of coordinates. At index \(F=K\), all coordinates in \(K\) equal \(0\). For every \(G\supseteq K\), those coordinates remain \(0\), so the net is eventually in the neighborhood. This verifies directly that the finitely many restrictions are satisfied together.
What Changes in the Box Topology?
Proof. A box neighborhood of \(x\) is a product \(\prod_{\alpha\in I}U_\alpha\) in which each \(U_\alpha\) is open and contains \(x_\alpha\). If the net converges in the box topology, it is eventually in every such neighborhood, which is exactly the stated condition. Conversely, if the stated condition holds, the net is eventually in every basic box neighborhood of \(x\). Every open neighborhood contains a basic box neighborhood, so the net converges to \(x\). \(\square\)
Convergence in the box topology implies coordinatewise convergence: to test one coordinate, take any open neighborhood in that coordinate and use the whole factor \(X_\beta\) in every other coordinate. But the converse can fail, because coordinatewise convergence gives separate thresholds, whereas the box test asks for one threshold that handles all coordinates at once.
Worked Example: Coordinatewise Convergence Without Box Convergence
Keep \(X=\mathbb{R}^{\mathbb{N}}\), but now give it the box topology. For \(n\in\mathbb{N}\), define \(y_n\in X\) by \[ (y_n)_k= \begin{cases} 1,&n<k,\\ 0,&n\geq k. \end{cases} \] For each fixed \(k\), the \(k\)-th coordinate equals \(0\) for every \(n\geq k\), so it converges to \(0\). The product convergence criterion therefore says that \(y_n\) converges to zero in the product topology.
Consider instead the box neighborhood \(W=\prod_{k\in\mathbb{N}}(-1/2,1/2)\) of zero. For each \(n\), take \(k=n+1\). Then \(n<k\), so \((y_n)_k=1\), and \(1\notin(-1/2,1/2)\). Hence \(y_n\notin W\) for every \(n\), and the sequence is not eventually in this box neighborhood. It does not converge to zero in the box topology.
Check: each coordinate really is eventually zero: for fixed \(k\), every \(n\geq k\) satisfies \(n\geq k\), so the second line of the definition applies. But for every \(n\), the coordinate \(k=n+1\) takes the value \(1\), which is outside the corresponding interval in \(W\). Thus coordinatewise convergence and failure of box convergence both follow from the displayed definition.
Worked Example: A Nonconstant Net That Converges in the Box Topology
Give \(X=\mathbb{R}^{\mathbb{N}}\) the box topology and define \(z_n\) by \((z_n)_1=1/n\) and \((z_n)_k=0\) for every \(k\geq2\). We show that \(z_n\) converges to zero in the box topology, not just coordinatewise.
Let \(W=\prod_{k\in\mathbb{N}}U_k\) be any box neighborhood of zero. Since \(U_1\) is open in the standard topology on \(\mathbb{R}\) and contains \(0\), there is \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq U_1\). Choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). If \(n\geq N\), then \(0<1/n\leq1/N<\varepsilon\), so \((z_n)_1\in U_1\). For each \(k\geq2\), \((z_n)_k=0\in U_k\) for every \(n\). Therefore \(z_n\in W\) for every \(n\geq N\).
Check: the argument works for an arbitrary choice of all the neighborhoods \(U_k\); no uniform radius across coordinates was assumed. Only the first coordinate varies, and every other coordinate equals the center of its neighborhood.
A Common Pitfall
Common pitfall. From coordinatewise convergence, one obtains a threshold \(N_k\) for each coordinate \(k\). It is incorrect to say “take \(N=\max_{k\in\mathbb{N}}N_k\)” unless that maximum exists. For the sequence \(y_n\) above, coordinate \(k\) is zero whenever \(n\geq k\), so one may take \(N_k=k\). The set \(\{k:k\in\mathbb{N}\}\) has no largest element, so these thresholds do not give a single finite \(N\). The product topology avoids this problem because each basic neighborhood mentions only finitely many coordinates; the box topology can mention all of them.
This distinction is useful whenever convergence is tested in an infinite product. For the product topology, check each coordinate separately and then combine the finitely many thresholds required by a basic neighborhood. For the box topology, test arbitrary coordinate neighborhoods simultaneously; separate coordinate convergence alone is not enough.
Check Your Understanding
Use the definitions and the two convergence tests to answer the following.
- For \(X=\mathbb{R}^{\mathbb{N}}\) with the product topology, explain why a net whose \(k\)-th coordinate is eventually in \(U_k\) for each \(k\) need not be eventually in \(\prod_k U_k\).
- In the finite-subset net \(x_F\), why does directedness allow the restrictions on a finite set \(K\) of coordinates to hold at once?
- For the sequence \(y_n\), identify a box neighborhood of zero that the sequence never enters.
- For \(z_n\), why do the neighborhoods in coordinates \(k\geq2\) impose no additional threshold?
- State the product convergence criterion and the box convergence test in your own words, making explicit which one requires only finitely many coordinate restrictions at a time.