Why Finite Coordinates Are Enough
In a product, a basic open neighborhood restricts only finitely many coordinates, even when the product has infinitely many factors. This is the key to the regularity proof: separate a point from a closed set by first finding such a basic neighborhood, then shrinking each of its finitely many coordinate restrictions. The result applies to products of any size, not just finite products.
For a family of spaces \((X_i,\mathcal{T}_i)\), indexed by a set \(I\), write \(X=\prod_{i\in I}X_i\) with the product topology. A basic open set has the form \(\prod_{i\in I}U_i\), where each \(U_i\) is open in \(X_i\), and \(U_i=X_i\) for all but finitely many \(i\). Equivalently, it is an intersection of finitely many sets \(\pi_i^{-1}(U_i)\), where \(\pi_i:X\to X_i\) is the coordinate projection.
The point-closed-set formulation of regularity will be used directly. A useful consequence is that open neighborhoods can be shrunk while keeping their closures inside the original neighborhood.
Proof. The set \(C_i=X_i\setminus U_i\) is closed and does not contain \(x_i\). By regularity there are disjoint open sets \(V_i,W_i\) with \(x_i\in V_i\) and \(C_i\subseteq W_i\). Since \(W_i\) is open and disjoint from \(V_i\), its complement is a closed set containing \(V_i\). The closure of \(V_i\), being contained in every closed set that contains \(V_i\), therefore satisfies \(\overline{V_i}\subseteq X_i\setminus W_i\). But \(C_i\subseteq W_i\), so \(X_i\setminus W_i\subseteq X_i\setminus C_i=U_i\). Thus \(\overline{V_i}\subseteq U_i\). \(\square\)
The Product Theorem
Proof. First suppose \(X\) is nonempty. To prove regularity, let \(F\) be closed in \(X\) and let \(x=(x_i)_{i\in I}\in X\setminus F\). The complement \(X\setminus F\) is open and contains \(x\). By the definition of the product topology, it contains a basic open set \(B=\bigcap_{i\in J}\pi_i^{-1}(U_i)\) with \(x\in B\subseteq X\setminus F\), where \(J\subseteq I\) is finite and each \(U_i\) is open in \(X_i\). In particular, \(x_i\in U_i\) for \(i\in J\).
If \(J\) is empty, then \(B=X\), so \(X\subseteq X\setminus F\) and \(F=\varnothing\). The open sets \(X\) and \(\varnothing\) separate \(x\) from \(F\). Now suppose \(J\) is nonempty. For each \(i\in J\), apply the closure-shrinking lemma to choose an open \(V_i\) with \(x_i\in V_i\) and \(\overline{V_i}\subseteq U_i\). Set \(V=\bigcap_{i\in J}\pi_i^{-1}(V_i)\). This is an open neighborhood of \(x\).
For each \(i\in J\), \(\pi_i^{-1}(\overline{V_i})\) is closed: its complement is the preimage of the open set \(X_i\setminus\overline{V_i}\) under the continuous coordinate projection. It contains \(V\), so it also contains \(\overline{V}\). Therefore
The middle inclusion follows from \(\overline{V_i}\subseteq U_i\) in each coordinate. Since \(B\subseteq X\setminus F\), the set \(\overline{V}\) is disjoint from \(F\). Now \(W=X\setminus\overline{V}\) is open, contains \(F\), and is disjoint from \(V\). This separates \(x\) and \(F\), proving regularity.
To prove the \(T_1\) condition, take distinct points \(x,y\in X\). They differ in some coordinate \(i\), so \(x_i\ne y_i\). Since \(X_i\) is \(T_1\), \(X_i\setminus\{y_i\}\) is open. Its preimage under \(\pi_i\) is open in \(X\), contains \(x\), and excludes \(y\). Thus \(X\setminus\{y\}\) is open: indeed, it is the union, over all \(i\in I\), of the open sets \(\pi_i^{-1}(X_i\setminus\{y_i\})\). Hence \(\{y\}\) is closed. This holds for every \(y\), so \(X\) is \(T_1\).
If \(X\) is empty, it is \(T_1\), and the regularity condition holds because it has no point \(x\) to test. This also covers a product with an empty factor. If \(I\) is empty, the product is a singleton; its only subsets are open and closed, so it is regular and \(T_1\). \(\square\)
Worked Separations
Worked Example: A Closed Set in the Plane
Take \(X=\mathbb{R}\times\mathbb{R}\) with the product of the standard topologies, \(x=(0,0)\), and \(F=\{(u,v):u\geq1\text{ or }v\geq1\}\). The set \(F\) is closed: its complement is the open rectangle \((-\infty,1)\times(-\infty,1)\). We construct disjoint open neighborhoods of \(x\) and \(F\).
The open rectangle \(B=(-1,\tfrac12)\times(-1,\tfrac12)\) contains \((0,0)\). Each coordinate of every point in \(B\) is less than \(\tfrac12<1\), so \(B\cap F=\varnothing\). Shrink the coordinate intervals to obtain \(V=(-\tfrac12,\tfrac14)\times(-\tfrac12,\tfrac14)\). Its closure is the rectangle \(\overline V=[-\tfrac12,\tfrac14]\times[-\tfrac12,\tfrac14]\), which is disjoint from \(F\): both coordinates are at most \(\tfrac14<1\). Thus \(W=\mathbb{R}^2\setminus\overline V\) is open, contains \(F\), and is disjoint from \(V\). In particular, the coordinate shrinkings have kept the closure of the neighborhood away from the closed set.
Check: \(0\in(-\tfrac12,\tfrac14)\) in each coordinate, so \(x\in V\); and \(\overline V\cap F=\varnothing\), so \(F\subseteq W\) and \(V\cap W=\varnothing\).
Worked Example: An Infinite Product with One Restricted Coordinate
Let \(X=\mathbb{R}^{\mathbb{N}}\) with the product topology, let \(x=(0,0,\ldots)\), and let \(F=\{y=(y_n):y_3\geq2\}\). The coordinate projection \(\pi_3\) is continuous, and \([2,\infty)\) is closed in \(\mathbb{R}\), so \(F=\pi_3^{-1}([2,\infty))\) is closed.
Consider the basic open neighborhood \(B=\{y:y_3\in(-1,2)\}\) of \(x\). It misses \(F\), since \(y_3<2\) for every \(y\in B\). Shrink the third-coordinate restriction to \(V=\{y:y_3\in(-1,1)\}\). The set \(C=\{y:y_3\in[-1,1]\}\) is closed and contains \(V\), so \(\overline V\subseteq C\). Since every member of \(F\) has third coordinate at least \(2\), \(C\cap F=\varnothing\). Therefore \(W=X\setminus C\) is open, contains \(F\), and is disjoint from \(V\).
This construction controls only coordinate \(3\). All the other coordinates remain unrestricted, as required for a basic open set in the product topology.
Check: \(0\in(-1,1)\), so \(x\in V\); every \(y\in F\) has \(y_3\geq2\), so \(y\notin C\) and \(y\in W\); and \(V\subseteq C\), so \(V\cap W=\varnothing\).
Worked Example: A Real Factor and a Discrete Factor
Give \(\mathbb{R}\) its standard topology and \(\mathbb{Z}\) the discrete topology, in which every subset is open. These factors are regular and \(T_1\): \(\mathbb{R}\) is a metric space (and every metric space is normal, hence regular and \(T_1\)); in \(\mathbb{Z}\), every subset is both open and closed. In \(X=\mathbb{R}\times\mathbb{Z}\), take \(x=(0,0)\) and \(F=[2,\infty)\times\mathbb{Z}\). The set \(F\) is closed because its complement \((-\infty,2)\times\mathbb{Z}\) is open.
The open set \(B=(-1,2)\times\{0\}\) contains \(x\) and misses \(F\), since its real coordinate is less than \(2\). Shrink it to \(V=(-1,1)\times\{0\}\). The set \(C=[-1,1]\times\{0\}\) is closed: its complement is \((\mathbb{R}\setminus[-1,1])\times\mathbb{Z}\,\cup\,[-1,1]\times(\mathbb{Z}\setminus\{0\})\), which is open in the product topology. It contains \(V\) and is disjoint from \(F\). Consequently \(W=X\setminus C\) is open, contains \(F\), and is disjoint from \(V\).
Check: the real coordinates in \(C\) lie between \(-1\) and \(1\), while those in \(F\) are at least \(2\); hence \(C\cap F=\varnothing\). Also \(V\subseteq C\), so \(V\cap W=\varnothing\).
What the Finite-Coordinate Condition Does
The proof needs only finitely many applications of regularity in the factors: the basic neighborhood \(B\) restricts coordinates in a finite set \(J\), and each of those restrictions can be shrunk. The set \(\overline V\) is then controlled by finitely many closed coordinate conditions. This is why the proof works even when the index set is infinite.
Common pitfall. In \(\mathbb{R}^{\mathbb{N}}\), it is incorrect to treat \((-1,1)^{\mathbb{N}}\) as a basic open neighborhood in the product topology. It restricts every coordinate, not finitely many. In fact it is not open in the product topology: any basic neighborhood of the zero sequence restricts only a finite set \(J\) of coordinates. Choose \(k\in\mathbb{N}\setminus J\). The sequence whose \(k\)-th coordinate is \(2\) and whose other coordinates are \(0\) lies in that basic neighborhood, but not in \((-1,1)^{\mathbb{N}}\). The correction is to use finite-coordinate neighborhoods, as in the theorem; arbitrary restrictions in every coordinate belong to the box topology, not the product topology.
Regularity and the \(T_1\) property are established separately here. The shrinking argument proves regularity by constructing open sets separating a point from a closed set. The \(T_1\) argument uses the \(T_1\) property of a factor in a coordinate where two product points differ. Keeping those two claims distinct makes explicit which hypothesis supplies each part of the conclusion.
| Goal | Construction or test | Where the hypothesis is used |
|---|---|---|
| Find an initial neighborhood | Choose a basic product neighborhood inside \(X\setminus F\) | Product topology |
| Keep a point-neighborhood away from \(F\) | Shrink each of the finitely many restricted coordinate neighborhoods | Regularity of the factors |
| Separate the point and the closed set | Use \(V\) and \(X\setminus\overline V\) | Closure containment in the basic neighborhood |
| Show singletons are closed | Separate distinct points in a coordinate where they differ | \(T_1\) property of the factors |
Check Your Understanding
Use the product topology in each question, and identify where regularity or the \(T_1\) hypothesis is needed.
- In \(\mathbb{R}^2\), construct disjoint open sets separating \((0,0)\) from \(F=\{(u,v):u\geq3\}\).
- In \(\mathbb{R}^{\mathbb{N}}\), explain why a basic neighborhood of a point restricts only finitely many coordinates.
- In a product of \(T_1\) spaces, prove that the complement of any singleton is open.
- In the proof of the theorem, why does \(\overline V\subseteq\bigcap_{i\in J}\pi_i^{-1}(\overline{V_i})\) hold?
- What changes in the proof when the finite coordinate set \(J\) is empty?