Why Coordinates Determine Product Convergence
A point in a product space is a collection of coordinates, one in each factor. The product topology is designed so that an open neighborhood may constrain finitely many coordinates at a time, while leaving all other coordinates unrestricted. This finite-coordinate structure is exactly what lets us test convergence using the coordinate projections.
Let \(J\) be an index set, let \((X_j,\mathcal{T}_j)\) be a topological space for each \(j\in J\), and write \(P=\prod_{j\in J}X_j\) with the product topology. The \(j\)-th projection is \(\pi_j:P\to X_j\), defined by \(\pi_j((x_i)_{i\in J})=x_j\). A basic open set in \(P\) has the form \[ \bigcap_{j\in F}\pi_j^{-1}(U_j), \] where \(F\subseteq J\) is finite and each \(U_j\in\mathcal{T}_j\). When \(F=\varnothing\), this intersection is all of \(P\).
For an ultrafilter \(\mathcal{U}\) on \(P\), its \(j\)-th coordinate image is the collection \[ \pi_{j*}\mathcal{U}=\{B\subseteq X_j:\pi_j^{-1}(B)\in\mathcal{U}\}. \] This is an ultrafilter on \(X_j\). Indeed, preimages preserve inclusions and finite intersections, and \(\pi_j^{-1}(X_j)=P\), while \(\pi_j^{-1}(\varnothing)=\varnothing\). Also, for every \(B\subseteq X_j\), an ultrafilter contains either \(\pi_j^{-1}(B)\) or its complement \(\pi_j^{-1}(X_j\setminus B)\); this decision property follows from maximality. Thus the coordinate image is a proper filter that decides every subset and is itself maximal.
Proof. Suppose first that \(\mathcal{U}\) converges to \(x\). Fix \(j\in J\), and let \(U_j\) be any open neighborhood of \(x_j\) in \(X_j\). The set \(\pi_j^{-1}(U_j)\) is open in \(P\) and contains \(x\), so convergence of \(\mathcal{U}\) gives \(\pi_j^{-1}(U_j)\in\mathcal{U}\). By the definition of the coordinate image, \(U_j\in\pi_{j*}\mathcal{U}\). This holds for every open neighborhood \(U_j\), so \(\pi_{j*}\mathcal{U}\) converges to \(x_j\).
Conversely, suppose that \(\pi_{j*}\mathcal{U}\) converges to \(x_j\) for every \(j\in J\). Let \(V\) be any open neighborhood of \(x\) in \(P\). By the definition of the product topology, there is a basic open set \(N=\bigcap_{j\in F}\pi_j^{-1}(U_j)\) such that \(x\in N\subseteq V\), where \(F\) is finite and each \(U_j\) is open with \(x_j\in U_j\). Coordinate convergence gives \(U_j\in\pi_{j*}\mathcal{U}\), hence \(\pi_j^{-1}(U_j)\in\mathcal{U}\), for each \(j\in F\). Closure of a filter under finite intersections gives \(N\in\mathcal{U}\). Since \(N\subseteq V\), closure under supersets gives \(V\in\mathcal{U}\). Every open neighborhood of \(x\) therefore belongs to \(\mathcal{U}\), so \(\mathcal{U}\) converges to \(x\). If \(F\) is empty, then \(N=P\in\mathcal{U}\), and the same conclusion follows. \(\square\)
The proof uses only the neighborhood basis of the product topology and the filter axioms. In fact, the same equivalence holds for any filter on \(P\); the ultrafilter hypothesis is not needed for this particular result. It is useful in applications because ultrafilters can be pushed forward to the factors and tested there.
Worked Applications
Worked Example: A Principal Ultrafilter in a Two-Factor Product
Give \(\mathbb{R}\) its standard topology and \([0,1]\) its subspace topology from \(\mathbb{R}\). Let \(P=\mathbb{R}\times[0,1]\), let \(p=(2,1/3)\), and take the principal ultrafilter \(\mathcal{U}_p=\{A\subseteq P:p\in A\}\).
The first coordinate image consists of the subsets \(B\subseteq\mathbb{R}\) for which \(\pi_1^{-1}(B)\) contains \(p\). This is equivalent to \(2\in B\), so \(\pi_{1*}\mathcal{U}_p\) is the principal ultrafilter at \(2\). Likewise, \(\pi_{2*}\mathcal{U}_p\) is the principal ultrafilter at \(1/3\).
Each coordinate image converges to its indicated point: every open neighborhood of that point contains the point, and hence belongs to its principal ultrafilter. The product criterion now gives convergence of \(\mathcal{U}_p\) to \(p\). Directly, the basic neighborhood \((1,3)\times(1/4,1/2)\) contains \(p\), so it belongs to \(\mathcal{U}_p\). This also checks the conclusion using a specific product neighborhood.
Worked Example: A Nonprincipal Ultrafilter in an Infinite Product
Give each factor \(\{0,1\}\) the discrete topology, and set \(P=\{0,1\}^{\mathbb{N}}\) with the product topology. Fix a nonprincipal ultrafilter \(\mathcal{V}\) on \(\mathbb{N}\); in particular, it contains every cofinite subset. Such an ultrafilter exists by the ultrafilter extension principle. For each \(n\in\mathbb{N}\), define \(x_n\in P\) by \(x_n(k)=1\) if \(n\ge k\), and \(x_n(k)=0\) if \(n<k\). Let \(\mathcal{U}\) on \(P\) be the image of \(\mathcal{V}\) under \(f(n)=x_n\); that is, \(A\in\mathcal{U}\) exactly when \(f^{-1}(A)\in\mathcal{V}\).
Fix a coordinate \(k\). The set of indices for which the \(k\)-th coordinate equals \(1\) is \(\{n:n\ge k\}\), which is cofinite and therefore belongs to \(\mathcal{V}\). The set of indices for which that coordinate equals \(0\) is \(\{1,\ldots,k-1\}\), which is finite and does not belong to the nonprincipal ultrafilter. Thus the \(k\)-th coordinate image contains \(\{1\}\), and it converges to \(1\), since the only open neighborhoods of \(1\) in the discrete factor are \(\{1\}\) and \(\{0,1\}\).
For a direct product check, take a basic neighborhood of the all-ones point that restricts a finite set \(F\subseteq\mathbb{N}\) of coordinates to \(1\). If \(F\ne\varnothing\), let \(m=\max F\). Whenever \(n\ge m\), all the restricted coordinates of \(x_n\) equal \(1\). The preimage of this basic neighborhood therefore contains the cofinite set \(\{n:n\ge m\}\), so it belongs to \(\mathcal{V}\). If \(F=\varnothing\), the neighborhood is all of \(P\), whose preimage is \(\mathbb{N}\). Hence \(\mathcal{U}\) converges to the all-ones point. As a coordinate check, for \(k=3\) the values are \(0,0,1,1,\ldots\), and the indices with value \(1\) form the cofinite set \(\{n:n\ge3\}\).
Worked Example: Coordinate Limits Need Not Be Unique
Let \(S=\{a,b\}\) have topology \(\{\varnothing,\{a\},S\}\), and give \(P=S\times S\) the product topology. Let \(\mathcal{U}\) be the principal ultrafilter at \((a,a)\). Its two coordinate images are both the principal ultrafilter at \(a\).
The principal ultrafilter at \(a\) converges to \(a\), because every open neighborhood of \(a\) contains \(a\). It also converges to \(b\): the only open neighborhood of \(b\) is \(S\), which belongs to every proper filter on \(S\). Thus each coordinate image converges to both points of \(S\).
In the product, every open neighborhood of any point \(y\in S\times S\) contains \((a,a)\). To verify this, take a basic neighborhood of \(y\). In either coordinate, an open set containing \(a\) is \(\{a\}\) or \(S\), both of which contain \(a\); an open set containing \(b\) must be \(S\), which also contains \(a\). Their product therefore contains \((a,a)\). Every neighborhood of \(y\) belongs to the principal ultrafilter at \((a,a)\), so \(\mathcal{U}\) converges to all four points of \(P\). The coordinate criterion permits this: it asserts convergence to a specified point when all the coordinates converge to its specified coordinates, not that limits must be unique.
A Common Pitfall: Requiring Every Coordinate at Once
For the sequence in the infinite-product example, each fixed coordinate is eventually \(1\). A tempting but incorrect argument is to say that convergence requires all coordinates to be \(1\) at once, and then to demand a single index after which this holds for every \(k\in\mathbb{N}\). For any fixed \(n\), the coordinate \(k=n+1\) of \(x_n\) is \(0\), so no such index exists. The calculation is correct, but the requirement is not: the condition that every coordinate equal \(1\) describes the singleton containing the all-ones point, and that singleton is not a basic open neighborhood in the product topology.
For example, a basic neighborhood may require coordinates \(2\) and \(5\) to equal \(1\) and leave every other coordinate unrestricted. For the displayed sequence, both requirements hold whenever \(n\ge5\), so the preimage contains \(\{5,6,7,\ldots\}\), a cofinite set in \(\mathcal{V}\). The correction is to check each finite set of coordinate restrictions, not infinitely many restrictions simultaneously. This is precisely the step in the theorem’s proof where finiteness of \(F\) allows the filter’s finite-intersection property to apply.
The result itself makes no compactness assumption and supplies no limit when none has been specified. It says that once a candidate point \(x\) is given, convergence of the ultrafilter to \(x\) can be checked through its coordinate images. In compactness arguments, one uses this coordinate test together with separate results that supply coordinate limits; the test then converts those limits into a product limit.
Check Your Understanding
For each question, use the definition of convergence and the finite-coordinate basis of the product topology.
- For a filter \(\mathcal{U}\) on \(P\), write the definition of its coordinate image under \(\pi_j\).
- Why does the converse direction of the theorem require only finitely many applications of the filter’s intersection property?
- In \(\{0,1\}^{\mathbb{N}}\), with discrete factors, what is the coordinatewise limit of the sequence \(x_n(k)=1\) when \(n\ge k\) and \(x_n(k)=0\) otherwise?
- In the topology \(\{\varnothing,\{a\},\{a,b\}\}\), which points are limits of the principal ultrafilter at \(a\)?
- Does the coordinate criterion imply that a convergent ultrafilter has a unique limit? Explain using the two-point example.