Separation in a Product Is Tested One Coordinate at a Time
A product has many coordinates, but its basic open sets restrict only finitely many of them. This makes point separation a coordinatewise question: if two product points differ, can one coordinate distinguish their entries? The answer gives a precise classification for products of discrete, cofinite, and Sierpinski spaces. One qualification matters: for an arbitrary family of nonempty spaces, its product need not be nonempty unless one assumes the axiom of choice. We will state the classification with the product itself assumed nonempty, so no choice of coordinates is needed in the proof.
Recall the separation conditions. A space is T0 if for any distinct points, some open set contains one but not the other. It is T1 if for any distinct points \(x,y\), there is an open set containing \(x\) but not \(y\) (and therefore, applying the condition to the reversed pair, one containing \(y\) but not \(x\)). It is Hausdorff if any two distinct points have disjoint open neighborhoods.
For a basic open set containing a product point \(x\), each restricted coordinate gives an open neighborhood of \(x(i)\). Conversely, choosing open neighborhoods in finitely many coordinates and leaving all other coordinates unrestricted gives a basic open neighborhood. These two observations are the tools used below.
The Coordinatewise Criterion
Proof. First suppose every factor is T0. Let \(x,y\in P\) be distinct. They differ at some index \(i\), so \(x(i)\ne y(i)\). Since \(X_i\) is T0, an open set \(U_i\) contains one of these entries but not the other. The cylinder \(\{z\in P:z(i)\in U_i\}\) is open in the product and contains the corresponding product point but not the other. Thus \(P\) is T0.
Now suppose every factor is T1. For distinct \(x,y\in P\), choose an index \(i\) with \(x(i)\ne y(i)\). There is an open set \(U_i\) in \(X_i\) containing \(x(i)\) but not \(y(i)\). Its cylinder contains \(x\) but not \(y\). This proves that \(P\) is T1.
If every factor is Hausdorff, take distinct \(x,y\in P\) and an index \(i\) where they differ. Choose disjoint open neighborhoods \(U_i\) of \(x(i)\) and \(V_i\) of \(y(i)\) in \(X_i\). Their cylinders are disjoint: a point in both would have its \(i\)-th coordinate in \(U_i\cap V_i=\varnothing\). Hence \(P\) is Hausdorff.
For the converses, use the assumption \(P\ne\varnothing\): fix one point \(p\in P\). This single point supplies entries \(p(j)\) in all coordinates; we do not make a separate choice from every factor.
If some factor \(X_i\) is not T0, there are distinct \(u,v\in X_i\) such that no open set distinguishes them in either direction. Form \(x,y\in P\) by taking \(x(i)=u\), \(y(i)=v\), and \(x(j)=y(j)=p(j)\) for \(j\ne i\). They are distinct. Any basic neighborhood of \(x\) that restricts coordinate \(i\) uses an open set containing \(u\), which must also contain \(v\); its other restrictions contain the shared entries \(p(j)\). Thus it contains \(y\). The same argument with \(u,v\) reversed shows that every basic neighborhood of \(y\) contains \(x\). Since every open neighborhood contains a basic neighborhood, no open set distinguishes \(x\) and \(y\). So \(P\) is not T0.
If \(X_i\) is not T1, there are distinct \(u,v\in X_i\) such that no open set containing \(u\) excludes \(v\). Construct \(x,y\) as before, with entries \(u,v\) at coordinate \(i\) and the shared entries from \(p\) elsewhere. Every basic neighborhood of \(x\) contains \(y\): at coordinate \(i\), any restriction containing \(u\) also contains \(v\). Therefore no open neighborhood of \(x\) excludes \(y\), and \(P\) is not T1.
Finally, if \(X_i\) is not Hausdorff, choose distinct \(u,v\in X_i\) with no disjoint open neighborhoods. Construct \(x,y\) using these entries at \(i\) and \(p(j)\) at all other coordinates. Suppose there were disjoint open neighborhoods of \(x\) and \(y\). Choose basic neighborhoods inside them. Their coordinate restrictions at \(i\) are open neighborhoods of \(u\) and \(v\), so they intersect. At every other restricted coordinate, both neighborhoods contain the shared entry \(p(j)\). Consequently the two basic neighborhoods intersect, contradicting disjointness. Thus \(P\) is not Hausdorff. All three converses are proved. \(\square\)
The nonempty-product hypothesis cannot simply be dropped from the reverse implications. The empty space satisfies T0, T1, and Hausdorff vacuously, since it has no pair of distinct points. For example, a product with an empty factor is empty, even if another factor is not T0. Also, if the index set is empty, the product is a singleton and has all three properties. These cases explain why the theorem explicitly assumes a nonempty product.
What the Common Factors Contribute
A discrete topology on a set \(D\) is \(\mathcal{P}(D)\), the collection of all subsets. Every singleton is open, so distinct points have disjoint singleton neighborhoods. Thus every discrete space is Hausdorff (including the empty space, vacuously).
For a set \(X\), the cofinite topology consists of the empty set and all subsets whose complements in \(X\) are finite. Every cofinite space is T1: for \(x\ne y\), the set \(X\setminus\{y\}\) is open, contains \(x\), and excludes \(y\). If \(X\) is finite, every subset has finite complement, so the topology is discrete and the space is Hausdorff. If \(X\) is infinite, it is not Hausdorff. Indeed, two nonempty open sets each have finite complement, and their intersection has finite complement as well. An infinite set cannot be exhausted by that finite complement, so the intersection is nonempty. No two nonempty open neighborhoods can be disjoint.
The Sierpinski space is \(S=\{a,b\}\) with topology \(\{\varnothing,\{a\},S\}\). It is T0 because \(\{a\}\) contains \(a\) but not \(b\). It is not T1: the only open set containing \(b\) is \(S\), so no open set contains \(b\) and excludes \(a\). It is not Hausdorff, since every open neighborhood of \(b\) is \(S\), which intersects every neighborhood of \(a\).
Worked Example: A Product of Two Finite Discrete Spaces
Let \(X=\{r,s,t\}\) have the cofinite topology and let \(Y=\{0,1\}\) have the discrete topology. Both sets are finite, so every subset of either set is open: the cofinite topology on \(X\) is discrete. The product \(X\times Y\) is therefore Hausdorff by the criterion, since both factors are Hausdorff.
Here is a direct separation calculation. Consider \((r,0)\) and \((s,1)\). The open sets \(\{r\}\times\{0\}\) and \(\{s\}\times\{1\}\) contain the respective points. Their intersection is empty because \(\{r\}\cap\{s\}=\varnothing\) (and also \(\{0\}\cap\{1\}=\varnothing\)). More generally, if two distinct points in \(X\times Y\) differ in their first coordinate, use distinct singleton neighborhoods in \(X\); if they have the same first coordinate, they differ in the second, so use distinct singleton neighborhoods in \(Y\). In each case the corresponding cylinders are disjoint.
Check: the product has six points. The two cylinders used above are singletons, since \(\{r\}\times\{0\}=\{(r,0)\}\) and \(\{s\}\times\{1\}=\{(s,1)\}\). This verifies the separation for the chosen pair and agrees with the coordinatewise argument.
Worked Example: An Infinite Cofinite Factor
Let \(X=\mathbb{R}\) have the cofinite topology and \(D=\{0,1\}\) have the discrete topology. The product \(X\times D\) is T1 because both factors are T1. It is not Hausdorff because the factor \(X\) is infinite and cofinite, hence not Hausdorff.
To see the failure explicitly, consider \(x=(0,0)\) and \(y=(1,0)\). Any open neighborhood of \(x\) contains a basic neighborhood \(U\times V\), where \(U\) is cofinite and contains \(0\), and \(V\) is open in \(D\) and contains \(0\). Any open neighborhood of \(y\) contains a basic neighborhood \(W\times Z\), where \(W\) is cofinite and contains \(1\), and \(Z\) contains \(0\). Since \(D\) is discrete, \(V\cap Z\) contains \(0\). Also \(U\cap W\ne\varnothing\): its complement is contained in the union of the two finite sets \(\mathbb{R}\setminus U\) and \(\mathbb{R}\setminus W\), so it is still cofinite in the infinite set \(\mathbb{R}\). Hence \((U\cap W)\times(V\cap Z)\) is nonempty, and the two neighborhoods cannot be disjoint.
Check: the ordered pair is distinct because \(0\ne1\) in the first coordinate. The argument applies to any neighborhoods of this pair, since each contains a basic neighborhood of the displayed form. Thus this is a concrete failure of Hausdorffness, not just a failure of a sufficient test.
Worked Example: A Countable Product of Sierpinski Spaces
Give \(P=\prod_{n\in\mathbb{N}}S\) the product topology, where every factor is the Sierpinski space. This product is nonempty: the point \(p\) with \(p(n)=a\) for every \(n\) belongs to it. Each factor is T0, so the coordinatewise criterion shows that \(P\) is T0. Since every factor is not T1, the product is not T1 and is not Hausdorff.
For a direct T0 test, take any distinct \(x,y\in P\). There is an index \(n\) where their entries differ. Since the entries are \(a,b\), the open set \(\{a\}\) in coordinate \(n\) contains one entry and not the other. The cylinder restricting only coordinate \(n\) distinguishes \(x,y\).
For a direct failure of T1, let \(x(n)=b\) for every \(n\), and let \(y(1)=a\) while \(y(n)=b\) for \(n\ne1\). Every basic neighborhood of \(x\) restricts finitely many coordinates, and every open set in \(S\) containing \(b\) must be \(S\). Thus the restrictions at those coordinates also contain the corresponding entries of \(y\); all unrestricted coordinates impose no condition. Every neighborhood of \(x\) therefore contains \(y\), so it cannot be an open set containing \(x\) and excluding \(y\).
Check: \(x\ne y\) because \(x(1)=b\) and \(y(1)=a\). The cylinder \(\{z:z(1)=a\}\) contains \(y\) and excludes \(x\), which is consistent with T0. It does not provide the missing open neighborhood of \(x\) excluding \(y\).
A Common Mistake: Confusing T1 with Hausdorffness
A common incorrect conclusion is: “If every factor is T1, then the product is Hausdorff.” The coordinatewise theorem gives the correct conclusion: the product is T1. Hausdorffness requires every factor to be Hausdorff as well. The product \(\mathbb{R}_{\mathrm{cof}}\times\{0,1\}\) above is the counterexample: both factors are T1, so the product is T1, but the infinite cofinite factor prevents the product from being Hausdorff.
Another tempting shortcut is to reason that one can always pick a point in every other factor when comparing two coordinates. For arbitrary families, that silently assumes a nonempty product. The proof avoided this assumption by explicitly requiring \(P\ne\varnothing\) and taking all shared coordinates from one point \(p\in P\). If one instead assumes that every factor is nonempty, the assertion that the product is nonempty for an arbitrary index set uses the axiom of choice. Keeping the product-nonempty hypothesis visible makes the coordinate argument valid without that extra assumption.
The classification can be used as a quick checklist:
| Factor type | T0? | T1? | Hausdorff? |
|---|---|---|---|
| Discrete space | Yes | Yes | Yes |
| Finite cofinite space | Yes | Yes | Yes |
| Infinite cofinite space | Yes | Yes | No |
| Sierpinski space | Yes | No | No |
For a nonempty product of these spaces, check the corresponding column in every factor. The product has a given separation property exactly when every coordinate factor has that property.
Check Your Understanding
In each question, use the product topology and justify the classification from coordinate neighborhoods.
- Classify \(\mathbb{N}\times\mathbb{R}_{\mathrm{cof}}\) as T0, T1, and Hausdorff, where \(\mathbb{N}\) is discrete and \(\mathbb{R}_{\mathrm{cof}}\) is the infinite cofinite space.
- Classify \(S\times\{0,1\}\), where \(S\) is Sierpinski and \(\{0,1\}\) is discrete. Identify a pair witnessing any failure.
- Explain why a product of a non-Hausdorff factor and a nonempty family of other factors cannot be Hausdorff when the whole product is nonempty.
- What are the T0, T1, and Hausdorff properties of the empty product, which is a singleton?
- Why does the coordinatewise converse proof use a point of the product rather than merely asserting that each factor has a point?