Arbitrary Products of Hausdorff Spaces

Separation in a Product

To show that a space is Hausdorff, one must separate every pair of distinct points by disjoint open neighborhoods. In a product, points are tuples, so distinct points differ in at least one coordinate. The product topology is designed so that restricting a single coordinate gives an open set in the product. That observation proves the forward result, even when the product has arbitrarily many factors.

Let \(J\) be an index set, and for each \(j\in J\) let \((X_j,\mathcal{T}_j)\) be a topological space. The product set is \(X=\prod_{j\in J}X_j\), whose elements \(x\) have coordinates \(x_j\in X_j\). For each \(j\), the projection is the map \(\pi_j:X\to X_j\) defined by \(\pi_j(x)=x_j\). We give \(X\) the product topology.

In particular, for any \(i\in J\) and any open \(U\subseteq X_i\), the set \(\pi_i^{-1}(U)\) is open: it is a basic open set with the single restricted coordinate \(i\). We will use precisely this one-coordinate form of openness.

The Product of Hausdorff Spaces Is Hausdorff

Proof. Let \(x,y\in X\) be distinct. Since they are unequal tuples, there is an index \(i\in J\) with \(x_i\ne y_i\). The factor \(X_i\) is Hausdorff, so there are disjoint open sets \(U_i,V_i\subseteq X_i\) with \(x_i\in U_i\) and \(y_i\in V_i\). The sets \(\pi_i^{-1}(U_i)\) and \(\pi_i^{-1}(V_i)\) are open neighborhoods of \(x\) and \(y\). They are disjoint: a tuple in their intersection would have its \(i\)-th coordinate in \(U_i\cap V_i=\varnothing\). Thus every distinct pair in \(X\) has disjoint open neighborhoods, as required. \(\square\)

The argument requires just one coordinate restriction, regardless of how many coordinates the product has. It does not require constructing a neighborhood by making infinitely many independent restrictions.

Recovering Hausdorffness of Each Factor

The converse needs care. To compare two points of a factor using points of the product, we first need a tuple in the product whose other coordinates can be kept fixed. The following argument makes that requirement explicit.

Proof. Fix \(i\in J\). Since \(X\) is nonempty, choose a tuple \(z\in X\). Take distinct \(a,b\in X_i\). Define \(p,q\in X\) by \(p_i=a\), \(q_i=b\), and \(p_j=q_j=z_j\) for every \(j\ne i\). These are distinct product points, so Hausdorffness of \(X\) gives disjoint open neighborhoods \(O\) of \(p\) and \(P\) of \(q\).

By the definition of the product topology, there is a basic open set \(B\) with \(p\in B\subseteq O\), and a basic open set \(C\) with \(q\in C\subseteq P\). Write \(B=\bigcap_{j\in F}\pi_j^{-1}(U_j)\) and \(C=\bigcap_{j\in G}\pi_j^{-1}(V_j)\), with \(F,G\) finite and all restricted coordinate sets open. Define \(U_i=X_i\) if \(i\notin F\), and \(V_i=X_i\) if \(i\notin G\). The resulting \(U_i,V_i\) are open neighborhoods of \(a,b\) in \(X_i\), respectively.

We show they are disjoint. If \(w\in U_i\cap V_i\), form the tuple \(r\) by setting \(r_i=w\) and \(r_j=z_j\) for \(j\ne i\). At coordinate \(i\), \(r\) satisfies the restrictions defining both \(B\) and \(C\). At every coordinate \(j\ne i\), it has the same coordinate as \(p\) and \(q\). Since \(p\in B\) and \(q\in C\), those coordinates satisfy the restrictions defining \(B\) and \(C\) as well. Hence \(r\in B\cap C\subseteq O\cap P\), contradicting \(O\cap P=\varnothing\). Thus \(U_i\cap V_i=\varnothing\). We have separated every pair of distinct points in \(X_i\), so \(X_i\) is Hausdorff. Since \(i\) was arbitrary, every factor is Hausdorff. For the final assertion, the axiom of choice ensures that a product of nonempty sets is nonempty; the two implications now follow from this proof and the preceding theorem. \(\square\)

The converse uses a tuple in the product, not just the fact that each factor is nonempty in isolation. For an arbitrary family, the axiom of choice is what guarantees a simultaneous choice of one coordinate value from every factor. If the product is empty, its Hausdorff condition alone cannot imply that its factors are Hausdorff.

A Common Pitfall: Restricting Infinitely Many Coordinates

In the product topology, a basic open set restricts only finitely many coordinates. A set that restricts every coordinate may be open in the box topology, but that does not make it open in the product topology.

For a concrete check, let \(X=\prod_{n\in\mathbb{N}}\mathbb{R}\) with the product topology, and let \(x_n=0\) and \(y_n=1\) for every \(n\). The sets \(U=\prod_{n\in\mathbb{N}}(-1/3,1/3)\) and \(V=\prod_{n\in\mathbb{N}}(2/3,4/3)\) contain \(x\) and \(y\), and \(U\cap V=\varnothing\). The mistaken step is to call \(U\) and \(V\) product-open: each restricts infinitely many coordinates.

For example, \(U\) is not open. If it were open, it would contain a basic open neighborhood \(B\) of \(x\). That basic neighborhood restricts some finite set \(F\subseteq\mathbb{N}\) only. Choose \(m\in\mathbb{N}\setminus F\). The tuple that agrees with \(x\) at every coordinate except \(m\), where its value is \(1\), belongs to \(B\), because coordinate \(m\) is unrestricted there. It does not belong to \(U\), since \(1\notin(-1/3,1/3)\). This contradicts \(B\subseteq U\).

The correct separation uses just coordinate \(1\): \(\pi_1^{-1}((-1/3,1/3))\) and \(\pi_1^{-1}((2/3,4/3))\) are product-open, contain \(x\) and \(y\), and are disjoint. The successful argument needs no infinite-coordinate neighborhood.

What the Product Theorem Gives

The theorem provides a direct test for a product: if every factor is Hausdorff, then any two distinct tuples can be separated by looking at one coordinate where they differ. Conversely, when the product is nonempty, Hausdorffness of the product forces Hausdorffness of each factor. The nonempty-product condition in the converse matters: an empty space is Hausdorff vacuously, regardless of the separation properties of sets that might otherwise have been its factors.