Finite Discrete Factors and Product Compactness
A product of finite sets can be enormous: even a sequence of binary choices has infinitely many coordinates. Compactness nevertheless persists when each coordinate has the discrete topology, provided we use the product topology. The key is that a basic open set restricts only finitely many coordinates. This tutorial applies Tychonoff’s theorem to such products and uses compactness to organize coordinate constraints, including constraints indexed by an arbitrary set.
Let \(I\) be an index set, and for each \(i\in I\) let \(F_i\) be a finite set with the discrete topology, meaning every subset of \(F_i\) is open. The product \(\prod_{i\in I}F_i\) is given the product topology. A point \(x\) of this product is a function with \(x(i)\in F_i\) for each \(i\). A basic open set restricts \(x(i)\) to a specified open subset of \(F_i\) for each \(i\) in some finite set of coordinates, and places no restriction on the other coordinates.
The finiteness of \(J\) is essential: basic open sets in the product topology impose conditions on only finitely many coordinates. The factors are compact because they are finite. Indeed, for any open cover of a finite set, choose one covering set for each point; these finitely many choices give a finite subcover. The empty finite set is compact as well, since its empty subcover covers it.
The Compactness Theorem
Proof. Each \(F_i\) is compact by the finite-cover argument above. Tychonoff’s theorem states that a product of compact spaces, with the product topology, is compact; its ultrafilter proof was given in the earlier material on Tychonoff’s theorem. Applying it to the family \((F_i)_{i\in I}\) proves the claim. This use of Tychonoff’s theorem for arbitrary index sets relies on the axiom of choice, as in its general form. If \(I=\varnothing\), the product is a singleton and is compact. If some \(F_i=\varnothing\), the product is empty and is compact. Thus these cases are included. \(\square\)
The theorem gives compactness, not a guarantee that the product has a point: an empty factor makes the product empty. In the examples below, we verify nonemptiness by explicitly giving a point whenever it is needed.
Worked Example: The Binary Sequence Space
Consider \(X=\{0,1\}^{\mathbb{N}}\), where each copy of \(\{0,1\}\) has the discrete topology and \(X\) has the product topology. Its points are binary sequences \(x=(x_1,x_2,\ldots)\). By the theorem, \(X\) is compact.
For instance, the conditions \(x_1=0\) and \(x_3=1\) define the cylinder $$ C=\{x\in X:x_1=0,\ x_3=1\}. $$ It is nonempty: the sequence \((0,0,1,0,0,\ldots)\) belongs to it. It is a basic open set because it restricts only the two coordinates \(1\) and \(3\). Its complement is the union of the two open cylinders \(\{x:x_1=1\}\) and \(\{x:x_3=0\}\). Therefore \(C\) is also closed.
More generally, fixing any values at a finite collection of coordinates gives a closed and open cylinder: it is open by the basis description, and its complement is a finite union of open cylinders, one for each way at least one of the specified coordinate values can fail. The compactness theorem applies to the whole sequence space, not because its points are finite sequences, but because all its factors are compact.
Check: the displayed sequence has first coordinate \(0\) and third coordinate \(1\), so it lies in \(C\). The sequence \((1,0,1,0,0,\ldots)\) fails the first condition and lies outside \(C\), as required.
Closed Constraints and Finite Checks
A useful consequence of compactness is that closed constraints which are consistent in every finite collection are consistent all together. A family of sets has the finite intersection property if every finite subfamily has nonempty intersection. Here is the compactness argument in the form needed for products.
Proof. Suppose instead that \(\bigcap_{\alpha\in A}C_\alpha=\varnothing\). Then the open sets \(X\setminus C_\alpha\) cover \(X\): each point fails to belong to at least one \(C_\alpha\), and therefore belongs to its complement. Compactness gives a finite subcover \(X\setminus C_{\alpha_1},\ldots,X\setminus C_{\alpha_m}\). Taking complements, this says \(C_{\alpha_1}\cap\cdots\cap C_{\alpha_m}=\varnothing\), contrary to the finite intersection property. If \(A\) is empty, its intersection is \(X\), which is nonempty unless \(X\) is empty; the proposition’s hypothesis of the finite intersection property alone does not rule out \(X=\varnothing\). In that case the claimed conclusion can fail, so the proposition is understood with \(X\ne\varnothing\), or with the additional condition that the intersection of the empty subfamily, namely \(X\), is nonempty. \(\square\)
For products of nonempty finite sets, the empty-product issue in that proposition does not arise: the product has a point whenever one has chosen a value in each factor. In the examples that follow, we give explicit points, so the nonemptiness needed for the conclusion is directly checked.
Worked Example: Constraints at Arbitrary Binary Coordinates
Let \(I\) be any set, possibly uncountable, and consider \(X=\{0,1\}^{I}\) with the product topology. Choose disjoint subsets \(J,K\subseteq I\), and impose the conditions \(x(i)=1\) for \(i\in J\) and \(x(i)=0\) for \(i\in K\). Define $$ C=\{x\in X:x(i)=1\text{ for }i\in J,\ x(i)=0\text{ for }i\in K\}. $$
For each \(i\in I\), the coordinate condition \(x(i)=a\), where \(a\in\{0,1\}\), defines a closed set: its complement is the other coordinate condition, which is open. Thus \(C\), an intersection of closed coordinate sets, is closed. Since \(X\) is compact, \(C\) is compact as a closed subset of a compact space. To verify nonemptiness without making any selections, define \(x(i)=1\) when \(i\in J\), and \(x(i)=0\) when \(i\notin J\). Disjointness gives \(x(i)=0\) for every \(i\in K\), so \(x\in C\).
If instead \(J\cap K\ne\varnothing\), choose \(i\in J\cap K\). The conditions require both \(x(i)=1\) and \(x(i)=0\), which no binary coordinate can satisfy; hence \(C=\varnothing\). The disjointness condition is exactly what makes the displayed constraints compatible.
Check: when \(J\cap K=\varnothing\), the point just defined satisfies the \(J\)-conditions by its first defining case and the \(K\)-conditions because every \(i\in K\) lies outside \(J\). When \(i\in J\cap K\), the two required values differ, confirming the contradiction.
The same argument can be expressed through finite checks. If a family of closed coordinate constraints in a nonempty compact product has the finite intersection property, then all the constraints have a common solution. In particular, for binary coordinates, any collection of requirements of the form \(x(i)=a\) that is consistent in each finite subcollection is globally consistent: a conflict would already appear at a coordinate where two different values were required.
Worked Example: A Product with Different Finite Factors
For \(n\in\mathbb{N}\), let \(F_n=\{0,1,\ldots,n\}\) with the discrete topology and let \(X=\prod_{n\in\mathbb{N}}F_n\) have the product topology. Every factor is finite, so \(X\) is compact. Consider the constraints \(x_1=0\) and \(x_2=2\). They define a closed cylinder: each coordinate condition is closed, and their intersection is closed.
The point \(x=(0,2,0,0,0,\ldots)\) belongs to \(X\): its first entry \(0\) lies in \(F_1=\{0,1\}\), its second entry \(2\) lies in \(F_2=\{0,1,2\}\), and every later entry \(0\) lies in \(F_n\). It satisfies both constraints. The resulting closed subset is compact because it is closed in the compact space \(X\).
Check: \(x_1=0\) and \(x_2=2\) by the displayed coordinates. For \(n\ge3\), \(0\in\{0,1,\ldots,n\}=F_n\), so every remaining coordinate is valid.
A Common Mistake: Restricting Infinitely Many Coordinates
A frequent error is to treat a set that specifies every coordinate as a basic open set. In \(X=\{0,1\}^{\mathbb{N}}\), that would incorrectly make each singleton \(\{x\}\) open. The error is that a basic open set in the product topology can restrict only finitely many coordinates.
To check this directly, let \(x\in X\) and let \(U\) be any basic open neighborhood of \(x\). The restrictions defining \(U\) involve a finite set \(J\subseteq\mathbb{N}\). Since \(\mathbb{N}\) is infinite, choose \(n\in\mathbb{N}\setminus J\). Define \(y\) by \(y(n)=1-x(n)\) and \(y(k)=x(k)\) for \(k\ne n\). Then \(y\ne x\), but \(y\) still satisfies every restriction defining \(U\), so \(y\in U\). Thus no basic open neighborhood of \(x\) is contained in \(\{x\}\), and \(\{x\}\) is not open.
This also corrects a tempting but invalid compactness argument: the singleton sets cover the binary sequence space, but they do not form an open cover, so they cannot be used to contradict compactness. Product topology records finite-coordinate information locally; compactness then provides a global conclusion for appropriate closed constraints.
| Question | Answer | Reason |
|---|---|---|
| Is each finite discrete factor compact? | Yes | Every open cover of a finite set has a finite subcover. |
| Which topology is used on the product? | The product topology | Its basic open sets restrict finitely many coordinates. |
| Is an arbitrary product of these factors compact? | Yes | Apply Tychonoff’s theorem. |
| Does compactness alone guarantee a point in the product? | No | An empty factor makes the product empty. |
| Can finitely consistent closed constraints be solved together? | Yes, in a nonempty compact space | The closed-set finite intersection argument applies. |
Check Your Understanding
Use the product topology throughout, and state explicitly when a product or constraint set is nonempty.
- Describe a basic open neighborhood of a point in \(\{0,1\}^{I}\) when \(I\) is uncountable. How many coordinates may it restrict?
- Explain why \(\prod_{n\in\mathbb{N}}\{0,1,\ldots,n\}\) is compact.
- For binary coordinates, what happens if two constraints require different values at the same index?
- Why does compactness not imply that a product with an empty factor has a point?
- In \(\{0,1\}^{\mathbb{N}}\), explain why a basic neighborhood of a sequence contains another sequence.