Products Indexed by an Arbitrary Set
A product of two spaces can be described by open rectangles: one open condition on each coordinate. With an arbitrary family of spaces, it is important to specify how many coordinates may be restricted at once. In the product topology, a basic open set imposes conditions on only finitely many coordinates; every other coordinate is unrestricted.
Let \(I\) be any set, and for each \(i\in I\) let \((X_i,\mathcal{T}_i)\) be a topological space. The set underlying their product consists of all choices of one point from each factor. Write \(x_i\) for the \(i\)-th coordinate of a point \(x\).
The finite set \(F\) records the coordinates where conditions are imposed. If \(j\notin F\), membership in \(B(F;(U_i))\) places no condition on \(x_j\) other than \(x_j\in X_j\), as already required for every point of the product. In particular, a basic open set need not be a finite product: its points still have a coordinate in every factor, even though only finitely many coordinates are restricted.
Why These Sets Form a Basis
Proof. We check the two basis conditions. First, every point of the product belongs to a member of the collection: take \(F=\varnothing\). The condition on coordinates is then vacuous, so \(B(\varnothing;())=\prod_{i\in I}X_i\). If the product has no points, the condition that every point be covered holds vacuously.
Second, suppose \(x\) belongs to two basic sets \(B(F;(U_i))\) and \(B(G;(V_i))\). The set \(F\cup G\) is finite. For each \(i\in F\cup G\), define \(W_i\) to be the intersection of the restrictions on coordinate \(i\): use \(U_i\) if \(i\in F\), and \(X_i\) otherwise; use \(V_i\) if \(i\in G\), and \(X_i\) otherwise; then intersect these two sets. Since each factor \(X_i\) and each \(U_i,V_i\) used here is open in \(X_i\), \(W_i\) is open in \(X_i\). A point satisfies the combined restrictions exactly when it belongs to both original sets. Therefore \(B(F;(U_i))\cap B(G;(V_i))=B(F\cup G;(W_i))\). This is a basic set containing \(x\), so the basis intersection condition holds.
For completeness, the unions of basic sets form a topology. The empty set is the union of no basic sets, and the whole product is the basic set with \(F=\varnothing\). Arbitrary unions of unions of basic sets are again unions of basic sets. For two such unions, their intersection is the union of the pairwise intersections of their basic sets; each pairwise intersection is a basic set by the argument above. Thus finite intersections are also unions of basic sets. These are the topology axioms, and this topology is, by definition, the product topology. \(\square\)
The intersection argument gives a useful coordinate rule. On a coordinate restricted by both sets, intersect the two open restrictions. On a coordinate restricted by just one, keep that restriction. On all other coordinates, impose no condition.
Proof. Let the two coordinate sets be \(F\) and \(G\). On each coordinate in \(F\cup G\), regard a missing restriction as the whole factor \(X_i\), and intersect the resulting restrictions. A point belongs to the intersection of the original sets if and only if it satisfies every one of these combined coordinate conditions. Hence the intersection is the basic set described by those conditions. If a combined restriction is empty, no point can satisfy it, so the intersection is empty. \(\square\)
Reading Basic Sets in Examples
Worked Example: Two Restrictions on a Space of Real Sequences
Take \(I=\mathbb{N}\) and \(X_n=\mathbb{R}\) with the standard topology for every \(n\). Thus \(\prod_{n\in\mathbb{N}}\mathbb{R}\) is the set of all real sequences. Let \(F=\{2,5\}\), \(U_2=(-1,1)\), and \(U_5=(3,4)\). The associated basic open set is \(B=\{x\in\mathbb{R}^{\mathbb{N}}:-1<x_2<1\text{ and }3<x_5<4\}\).
Consider the sequence \(x\) defined by \(x_2=0\), \(x_5=7/2\), and \(x_n=0\) for every other \(n\). The first restriction holds because \(-1<0<1\). For the second, \(3=6/2<7/2<8/2=4\). Hence \(x\in B\).
Now define \(y\) by changing only coordinate \(1\) of \(x\), setting \(y_1=100\). Since \(1\notin F\), this change affects neither required inequality, so \(y\in B\) as well. This checks directly that coordinate \(1\) is unrestricted: its value may be any real number. Coordinates other than \(2\) and \(5\) are unrestricted for the same reason.
Check: the only conditions defining \(B\) are the two displayed strict inequalities. The values \(x_2=0\) and \(x_5=7/2\) satisfy them, while replacing \(x_1\) by \(100\) leaves both values unchanged.
Worked Example: An Uncountable Index Set
Let \(I=\mathbb{R}\), and for each \(t\in\mathbb{R}\) let \(X_t=\{0,1\}\) with the discrete topology, in which every subset is open. Restrict the coordinates in \(F=\{-1,2,\pi\}\) by \(U_{-1}=\{1\}\), \(U_2=\{0\}\), and \(U_\pi=\{1\}\). The basic set is the collection of all functions \(x:\mathbb{R}\to\{0,1\}\) satisfying \(x_{-1}=1,\quad x_2=0,\quad x_\pi=1\).
Define \(x\) to have those three specified values and set \(x_t=0\) for every \(t\notin F\). Then \(x\) belongs to the basic set because it meets all three coordinate conditions. The index \(0\) is not in \(F\), so define \(y\) by changing just \(x_0\) from \(0\) to \(1\). The three restricted coordinates of \(y\) are still \(1,0,1\), respectively, and thus \(y\) also belongs. This example has an uncountable index set, but its basic set still imposes conditions on only three coordinates.
Check: the singleton restrictions are open because each factor has the discrete topology. The changed coordinate \(0\) is distinct from \(-1\), \(2\), and \(\pi\), so changing its value does not violate a restriction.
Worked Example: Open Restrictions in Different Factors
Let \(I=\{a,b,c\}\). Give \(X_a=\mathbb{R}\) the standard topology, \(X_b=[0,1]\) the subspace topology from \(\mathbb{R}\), and \(X_c=\{p,q\}\) the discrete topology. Choose \(U_a=(1,2)\), \(U_b=[0,1/2)\), and \(U_c=\{q\}\). Each is open in its own factor: for \(U_b\), the equality \([0,1/2)=[0,1]\cap(-1,1/2)\) exhibits it as relatively open.
Here \(F=I\), so the basic set consists of triples \((x_a,x_b,x_c)\) with \(1<x_a<2\), \(0\leq x_b<1/2\), and \(x_c=q\). The triple \((3/2,1/4,q)\) belongs: \(1<3/2<2\), \(0\leq1/4<1/2\), and its last coordinate is \(q\). The triple \((3/2,3/4,q)\) does not belong, because \(3/4\geq1/2\). The triple \((3/2,1/4,p)\) does not belong, because its last coordinate is not \(q\).
Check: the first triple satisfies all three restrictions. Each of the other two triples fails a specific restriction, while its other coordinates meet theirs.
Empty Cases and a Common Mistake
The definition also covers the edge cases. If \(I=\varnothing\), there is exactly one function with domain \(I\), the empty function. The product is therefore a singleton. The only finite coordinate set is \(F=\varnothing\), giving the whole singleton as a basic open set; its topology is \(\{\varnothing,\{\text{empty function}\}\}\). If any factor \(X_j\) is empty, there is no function choosing a point in every factor, so the product is empty. Its only subset is \(\varnothing\), and the product topology is \(\{\varnothing\}\). The basis argument does not require that the product contain a point.
Common pitfall. In \(\mathbb{R}^{\mathbb{N}}\), consider the set \(A=\{x:-1/n<x_n<1/n\text{ for every }n\in\mathbb{N}\}\). It is tempting to call this a basic open set by choosing the interval \((-1/n,1/n)\) at every coordinate. That choice makes restrictions at infinitely many coordinates, so it does not have the required finite form. In fact, \(A\) is not open in the product topology. To verify this, let \(0\) be the sequence whose coordinates are all zero; it belongs to \(A\). If \(A\) were open, some basic set containing \(0\) would be contained in \(A\). Let \(F\) be the finite set of coordinates restricted by that basic set. Choose \(n\geq2\) outside \(F\), which is possible because \(\mathbb{N}\) is infinite and \(F\) is finite. The sequence that is \(1\) at coordinate \(n\) and \(0\) elsewhere belongs to the basic set: it agrees with \(0\) on every restricted coordinate. But it does not belong to \(A\), since its \(n\)-th coordinate fails \(1<1/n\). This contradicts the assumed containment. The correction is to impose restrictions at finitely many coordinates only; the other coordinates must remain unrestricted.
Check Your Understanding
In each question, keep track of which coordinates are actually restricted.
- For \(I=\mathbb{N}\) and \(X_n=\mathbb{R}\), describe the basic open set with \(F=\{1,4\}\), \(U_1=(0,2)\), and \(U_4=(-3,-1)\).
- Why is the whole product a basic open set, including when the index set is empty?
- For two basic sets with coordinate sets \(F\) and \(G\), what restriction is imposed on a coordinate in \(F\cap G\)?
- In the product of discrete spaces \(X_t=\{0,1\}\) indexed by \(\mathbb{R}\), can a basic open set restrict every coordinate? Explain.
- What is the product set if one factor is empty, and what is its topology?