Products as Lists of Coordinate Choices
A Cartesian product collects choices made from several sets, keeping track of which set each choice came from. For two sets, an element of \(X\times Y\) is an ordered pair: its first coordinate belongs to \(X\), and its second belongs to \(Y\). The same idea works for any set of labels, including an infinite set of labels. The labels matter: a product is not merely a collection of values but a collection of values assigned to specified coordinates.
In this tutorial, products are considered as sets. If the factors are topological spaces, each factor retains its specified topology, but no topology on the Cartesian product is needed here. In particular, when we identify the underlying set of \(\mathbb{R}^n\) with a Cartesian product, we use its usual Euclidean topology only as the standard topology on \(\mathbb{R}^n\); the identification is a statement about sets.
The function description makes the membership rule precise: to belong to the product, a tuple must assign a value to every index, and the value at index \(i\) must lie in \(X_i\). It is not enough for its coordinates to come from the union of the factors. For a finite index set \(J=\{1,\dots,n\}\), the familiar notation \(X_1\times\cdots\times X_n\) means this same set of functions, usually written as ordered \(n\)-tuples \((x_1,\dots,x_n)\).
Proof. The first statement is exactly the defining membership condition for the product. For the second, if \(x=y\) as functions, then their values agree at each input \(i\in J\), so \(x_i=y_i\) for every \(i\). Conversely, suppose \(x_i=y_i\) for every \(i\in J\). Then \(x(i)=y(i)\) for every element \(i\) of their common domain \(J\). Functions with the same domain and equal values at every input are equal, so \(x=y\). If \(J\) is empty, there are no coordinates to compare; both functions in the product, if present, are the unique empty function, so the conclusion still holds. \(\square\)
Reading Membership Coordinate by Coordinate
Worked Example: A Product of Three Intervals and Sets
Let \(J=\{1,2,3\}\), with \(X_1=[-1,2]\), \(X_2=\{0,2\}\), and \(X_3=(1,4)\). The intervals have the subspace topology inherited from the standard topology on \(\mathbb{R}\); \(X_2\) is being used just as a set. Test whether \(a=(1,2,3/2)\) and \(b=(0,1,4)\) belong to \(X_1\times X_2\times X_3\).
For \(a\), check each coordinate against its own factor: $$ a_1=1\in[-1,2],\qquad a_2=2\in\{0,2\},\qquad a_3=\frac32\in(1,4), $$ where \(1<3/2<4\). All three conditions hold, so \(a\in X_1\times X_2\times X_3\).
For \(b\), the first coordinate satisfies \(0\in[-1,2]\). But \(b_2=1\notin\{0,2\}\), so this coordinate already fails its membership test. Also, \(b_3=4\notin(1,4)\), since the right endpoint is excluded. Thus \(b\notin X_1\times X_2\times X_3\) for two separate reasons.
Check: The tuple \(a\) has exactly three coordinates, one for each index, and each lies in the corresponding factor. The tuple \(b\) fails at coordinates 2 and 3, in particular respecting the strict endpoint in \(X_3=(1,4)\).
For products of intervals, the same rule becomes a list of coordinate inequalities. For example, a tuple \(x=(x_1,\dots,x_n)\) belongs to \(\prod_{i=1}^n[a_i,b_i]\) precisely when \(a_i\leq x_i\leq b_i\) for each index \(i\). If a factor is open at an endpoint, the corresponding inequality is strict there. Membership in the product is a conjunction: every coordinate condition must hold.
Worked Example: A Coordinate Description in \(\mathbb{R}^3\)
Give a coordinate test for the set \(C=[-1,2]\times[0,3]\times\{-1,1\}\), and decide whether \(u=(2,1,-1)\) and \(v=(-2,3,1)\) belong to \(C\). Here each interval has the subspace topology inherited from standard \(\mathbb{R}\); the set \(C\) is considered as a subset of \(\mathbb{R}^3\) with its standard Euclidean topology.
By the product membership rule, \((x,y,z)\in C\) exactly when all three conditions hold: $$ -1\leq x\leq 2,\qquad 0\leq y\leq 3,\qquad z\in\{-1,1\}. $$ For \(u\), substituting its coordinates gives \(-1\leq2\leq2\), \(0\leq1\leq3\), and \(-1\in\{-1,1\}\). Thus \(u\in C\).
For \(v\), the first condition would require \(-1\leq-2\), which is false because \(-2<-1\). Its second coordinate does satisfy \(0\leq3\leq3\), and its third satisfies \(1\in\{-1,1\}\), but those two successes do not fix the failed first condition. Therefore \(v\notin C\).
Check: The coordinate test applied to \(u=(2,1,-1)\) gives \(2\in[-1,2]\), \(1\in[0,3]\), and \(-1\in\{-1,1\}\). For \(v=(-2,3,1)\), its first coordinate is outside \([-1,2]\), so the product membership test rejects it.
Finite Products and Euclidean Space
The notation \(\mathbb{R}^n\) refers to \(n\)-tuples of real numbers, with the usual Euclidean topology. As a set, it is the Cartesian product of \(n\) copies of \(\mathbb{R}\). Thus an element of \(\mathbb{R}^n\) has one real coordinate for each index from 1 to \(n\); for \(n=2\), these are the familiar points \((x_1,x_2)\), and for \(n=3\), they are \((x_1,x_2,x_3)\). This interpretation specifies the points of \(\mathbb{R}^n\), without making any claim here about how a topology on a product is defined.
Worked Example: Coordinates in \(\mathbb{R}^3\)
Let \(A=[0,1]\), \(B=[-2,2]\), and \(C=\{0\}\), each regarded as a subset of \(\mathbb{R}\). Decide whether \(w=(1/2,-2,0)\) and \(z=(0,3,0)\) lie in \(A\times B\times C\).
For \(w\), the coordinate checks are $$ \frac12\in[0,1],\qquad -2\in[-2,2],\qquad 0\in\{0\}. $$ The first check holds because \(0\leq1/2\leq1\), and the second holds at its included left endpoint. All three hold, so \(w\in A\times B\times C\).
For \(z\), its first coordinate satisfies \(0\in[0,1]\), and its third satisfies \(0\in\{0\}\). Its second coordinate does not satisfy \(3\in[-2,2]\), because \(3>2\). Hence \(z\notin A\times B\times C\).
Check: Substitution into the three factor tests accepts \(w=(1/2,-2,0)\) and rejects \(z=(0,3,0)\) at its second coordinate. In particular, the endpoints \(-2\) and \(2\) belong to the closed interval \([-2,2]\), but \(3\) does not.
Proof. If the product contains a tuple \(x\), then \(x_i\in X_i\) for every \(i\in J\), so every factor is nonempty. For the converse, we prove the claim by induction on the number of indices. If \(J\) is empty, the unique empty function belongs to the product, as detailed below. If \(J\) has one element \(i\), nonemptiness of \(X_i\) supplies an element \(x_i\), and assigning it to \(i\) gives a tuple in the product. Suppose the claim holds for families with \(n\) indices and consider a family with \(n+1\) indices, all of whose factors are nonempty. Choose one index \(j\). The other \(n\) factors have a tuple by the induction hypothesis; since \(X_j\) is nonempty, choose \(x_j\in X_j\). Combining that tuple with \(x_j\) gives a function on all \(n+1\) indices, and each of its coordinates lies in its factor. It belongs to the product by the membership test. This proves the claim for every finite \(J\). \(\square\)
This finite argument must not be applied without care to an arbitrary infinite family. A tuple in the product is a choice function, selecting an element of every factor. The statement that every family of nonempty sets has such a function is the axiom of choice; it is not a consequence of the finite argument alone. For an arbitrary index set, the product is nonempty exactly when such a choice function exists.
Empty Factors and the Product with No Factors
The empty-factor rule follows directly from the definition: a tuple in the product would need a coordinate \(x_j\in X_j\), but no such value exists when \(X_j=\varnothing\). The empty-index case is different. A function with empty domain exists, and there is exactly one such function, because there are no inputs at which two functions could differ. Its coordinate conditions are vacuously satisfied: there is no index at which a condition can fail.
Worked Example: An Empty Factor Versus No Factors
Let \(J=\{1,2,3\}\), \(X_1=\{a\}\), \(X_2=\varnothing\), and \(X_3=\{c,d\}\). Then \(X_1\times X_2\times X_3=\varnothing\): any putative tuple would need a second coordinate in \(X_2=\varnothing\), which is impossible. In contrast, the product over the empty index set contains the empty function and is \(\{\varnothing\}\), not the empty set.
Check: A tuple for the three-factor product would have to satisfy \(x_2\in\varnothing\), which no value can satisfy. The no-factor product has no coordinate requirements and has exactly one element, the empty function. Thus these two cases give different sets.
A Common Notation Pitfall
A tuple is ordered: each entry belongs to the factor with the matching position or label. It is not an unordered set of values. For instance, in the product \([0,1]\times[2,3]\), the tuple \((1/2,5/2)\) belongs because \(0\leq1/2\leq1\) and \(2\leq5/2\leq3\). Reversing its entries gives a different tuple, \((5/2,1/2)\), which does not belong: \(5/2\notin[0,1]\) and \(1/2\notin[2,3]\). Treating the entries as an unordered pair would miss both failures.
Another frequent error is to check only one coordinate. In a product, membership requires every coordinate test to pass. The coordinate functions make this systematic: for a candidate tuple \(x\), check \(\pi_i(x)=x_i\in X_i\) separately for each index \(i\). The coordinate test is also useful for comparing tuples: equality requires agreement at every coordinate, not just at most of them.
Check Your Understanding
For each question, apply the coordinate condition at every index and pay attention to which endpoints are included.
- For \(X_1=[-1,1]\) and \(X_2=(0,2]\), does \((1,2)\) belong to \(X_1\times X_2\)? Does \((1,0)\) belong?
- Describe membership in \([0,1]\times\{-1,1\}\) using inequalities and a condition on the second coordinate.
- What is the Cartesian product of \(\{a\}\), \(\varnothing\), and \(\{b\}\)? Explain using the coordinate test.
- How many elements does the product over the empty index set have, and what function is its element?
- If \(x\) and \(y\) belong to the same indexed product and \(x_i=y_i\) for every index, what can be concluded about \(x\) and \(y\)?