Separation One Coordinate at a Time
In a product space, two distinct points differ in at least one coordinate. The product topology lets us use an open set in that coordinate to separate the points, while leaving all other coordinates unrestricted. This gives exact criteria for when a product is \(T_0\) or \(T_1\). The converse uses coordinate projections: if the product has the separation property, then each factor does too.
Let \(\{(X_i,\mathcal{T}_i):i\in I\}\) be a family of topological spaces, and let \(X=\prod_{i\in I}X_i\) have the product topology. A point \(x\in X\) is a tuple with coordinate \(x_i\in X_i\). A basic open set is a product \(\prod_{i\in I}U_i\), where each \(U_i\in\mathcal{T}_i\) and \(U_i=X_i\) for all but finitely many indices. Equivalently, we can build basic open sets by restricting finitely many coordinates and leaving the rest unrestricted.
For \(i\in I\), the coordinate projection is \(\pi_i:X\to X_i\), defined by \(\pi_i(x)=x_i\). If \(U\) is open in \(X_i\), then \(\pi_i^{-1}(U)\) restricts only coordinate \(i\), so it is open in the product topology. Such a set is called a coordinate cylinder. If \(X\ne\varnothing\), each projection is also onto: start with any tuple in \(X\) and replace its \(i\)-th coordinate by any chosen point of \(X_i\).
The Product Criteria
Proof. First suppose every factor is \(T_0\), and let \(x,y\in X\) be distinct. There is an index \(i\) with \(x_i\ne y_i\). Since \(X_i\) is \(T_0\), there is an open set \(U\in\mathcal{T}_i\) containing one of \(x_i,y_i\) but not the other. The coordinate cylinder \(\pi_i^{-1}(U)\) is open in \(X\) and contains exactly one of \(x,y\). Thus \(X\) is \(T_0\). If every factor is \(T_1\), the same choice of \(i\) gives open sets \(U,V\in\mathcal{T}_i\) with \(x_i\in U\), \(y_i\notin U\), \(y_i\in V\), and \(x_i\notin V\). Their cylinders \(\pi_i^{-1}(U)\) and \(\pi_i^{-1}(V)\) separate \(x\) from \(y\) in both directions. Hence \(X\) is \(T_1\).
For the converse, fix \(i\in I\) and distinct points \(u,v\in X_i\). Since \(X\ne\varnothing\), choose a tuple \(z\in X\). Construct \(x,y\in X\) by setting \(x_i=u\), \(y_i=v\), and \(x_j=y_j=z_j\) for every \(j\ne i\). These tuples differ only at coordinate \(i\). If \(X\) is \(T_0\), an open set in \(X\) contains one of \(x,y\) but not the other. Choose a basic open set \(\prod_j W_j\) containing that tuple and contained in the separating open set. At coordinate \(i\), \(W_i\) must contain the corresponding one of \(u,v\) and exclude the other: if \(W_i\) contained both, the basic open set would contain both tuples, since they agree at every other coordinate. Thus \(X_i\) is \(T_0\).
If \(X\) is \(T_1\), choose an open set containing \(x\) but not \(y\), and a second open set containing \(y\) but not \(x\). Inside each, choose a basic open set containing the specified tuple. The \(i\)-th coordinate restrictions of these basic sets are open in \(X_i\), contain \(u\) and \(v\), respectively, and exclude the other point. Therefore \(X_i\) is \(T_1\). This works for each \(i\), proving both converses. \(\square\)
The nonempty-product hypothesis in the theorem serves the converse: it supplies a tuple whose coordinates can be held fixed while one coordinate is changed. For an arbitrary family of nonempty spaces, the axiom of choice guarantees that the product is nonempty; this is where choice is relevant for an unrestricted product. If the index set is empty, the product is a singleton and the condition on all factors is vacuous; that product is both \(T_0\) and \(T_1\).
Worked Products
Worked Example: A Product of Two Sierpinski Spaces
Let \(S=\{a,b\}\) have topology \(\{\varnothing,\{a\},S\}\). Determine whether \(S\times S\), with the product topology, is \(T_0\) and whether it is \(T_1\).
The factor \(S\) is \(T_0\): for the distinct points \(a,b\), the open set \(\{a\}\) contains \(a\) but not \(b\). The product criterion therefore gives that \(S\times S\) is \(T_0\). Concretely, any distinct pairs differ in at least one coordinate. If the differing coordinate is \(a\) in one pair and \(b\) in the other, the cylinder restricting that coordinate to \(\{a\}\) contains the first pair and excludes the second.
The factor is not \(T_1\). In the product, consider \((b,b)\) and \((a,b)\). The only open set in \(S\) containing \(b\) is \(S\). Consequently, the only basic open set containing \((b,b)\) is \(S\times S\), and every open neighborhood of \((b,b)\) is the whole product. There is no open set containing \((b,b)\) but excluding \((a,b)\), so the product is not \(T_1\).
Check: \(\{a\}\times S\) is open because it restricts the first coordinate to the open set \(\{a\}\) and leaves the second unrestricted. It contains \((a,b)\) and excludes \((b,b)\), confirming the \(T_0\) separation for this pair. The failed reverse separation confirms the \(T_1\) failure.
Worked Example: A Product That Is \(T_1\)
Let \(A\) be an infinite set with the finite-complement topology, whose open sets are \(\varnothing\) and the subsets with finite complement. Let \(D=\{0,1\}\) have the discrete topology, in which every subset is open. Show that \(A\times D\) is \(T_1\).
First, \(A\) is \(T_1\): for distinct \(p,q\in A\), the set \(A\setminus\{q\}\) is open, contains \(p\), and excludes \(q\); reversing \(p,q\) gives the other separation. The discrete space \(D\) is \(T_1\), since each singleton is open. By the product criterion, \(A\times D\) is \(T_1\).
We can also verify the closed-singleton test directly. For \(p\in A\) and \(d\in D\), the complement of the singleton \(\{(p,d)\}\) is \[ (A\setminus\{p\})\times D\ \cup\ A\times(D\setminus\{d\}). \] The first set consists exactly of pairs whose first coordinate is not \(p\); the second adds the remaining pair with first coordinate \(p\). Thus their union is precisely the complement. Both sets are open coordinate cylinders, so the complement is open and \(\{(p,d)\}\) is closed. This holds for every \((p,d)\), as required for a \(T_1\) space.
Worked Example: One Indiscrete Factor Prevents \(T_0\)
Let \(J=\{u,v\}\) have the indiscrete topology \(\{\varnothing,J\}\), and let \(D=\{0,1\}\) be discrete. In \(J\times D\), compare \(x=(u,0)\) and \(y=(v,0)\).
A basic open set containing \(x\) must use \(J\) in the first coordinate, since \(J\) has no proper nonempty open set. In the second coordinate it can use \(\{0\}\) or \(D\). Either choice also contains \(y\). The same reasoning applies to every basic open set containing \(y\). Since every open set is a union of basic open sets, every open set containing either point contains both. Thus no open set distinguishes \(x\) from \(y\), and the product is not \(T_0\).
This is exactly what the criterion predicts: the factor \(J\) is not \(T_0\), so the product cannot be \(T_0\). The factor \(D\) being \(T_1\) does not compensate for the failure in \(J\).
A Common Mistake: Restricting Too Many Coordinates
A frequent error is to try to separate product points using a set that fixes every coordinate to a singleton. For example, in the Sierpinski space above, \(\{a\}\times\{b\}\) is not open: \(\{b\}\) is not open in \(S\). The argument that this set separates points therefore does not establish anything about the product topology.
The correction is to restrict only a coordinate where the points differ, using an open set in that factor, and to leave the other coordinates unrestricted. In the same example, \(\{a\}\times S\) is open and separates \((a,b)\) from \((b,b)\). For an arbitrary product, the corresponding set is the coordinate cylinder \(\pi_i^{-1}(U)\). It is open by the definition of the product topology, regardless of whether any singletons in the other factors are open.
The criteria also distinguish the roles of \(T_0\) and \(T_1\). For \(T_0\), one open set must distinguish each pair in at least one direction. For \(T_1\), open sets must distinguish the pair in both directions. The product inherits exactly the coordinatewise version of whichever requirement is imposed; the two conditions should not be conflated.
Check Your Understanding
Use basic open sets and coordinate projections to justify your answers.
- In a product, two distinct tuples differ at some coordinate. Explain how a \(T_0\) factor at that coordinate gives a separating open cylinder.
- Why does proving that a product is \(T_1\) require an open set in each direction for a pair of distinct points?
- Give a basic open neighborhood of \((a,b)\) in \(S\times S\) that excludes \((b,b)\), where \(S\) is the Sierpinski space used above.
- Where is nonemptiness of the product used in proving that product \(T_0\) or \(T_1\) implies the same property for each factor?
- In \(J\times D\) from the third example, why does every basic open set containing \((u,0)\) also contain \((v,0)\)?