Why Induction Fits Finite Products
Compactness is defined by open covers, so proving that a product is compact means starting with an arbitrary open cover of the product and finding a finite subcover. For two factors, the tube lemma gives the key step: the product of two compact spaces is compact. For a finite number of factors, induction lets us reuse that result, provided we check that grouping coordinates respects the product topology.
A topological space \((X,\mathcal{T})\) is compact if every cover of \(X\) by open sets has a finite subcover. In other words, whenever \(X\subseteq\bigcup_{\alpha\in J}U_\alpha\) with each \(U_\alpha\) open, there is a finite subset \(J_0\subseteq J\) such that \(X\subseteq\bigcup_{\alpha\in J_0}U_\alpha\). The empty space is compact: the empty subfamily already covers it.
We use the two-factor compactness theorem: if \(Y\) and \(Z\) are compact topological spaces, then \(Y\times Z\), with the product topology, is compact. This theorem was proved in the earlier material on compactness of the product of two spaces. The finite-product proof below uses that result and also verifies the coordinate regrouping needed to apply it repeatedly.
Regrouping Coordinates
For \(n\geq 1\), the product \(\left(\prod_{i=1}^n X_i\right)\times X_{n+1}\) can be identified with \(\prod_{i=1}^{n+1}X_i\) by the map that sends \(((x_1,\ldots,x_n),x_{n+1})\) to \((x_1,\ldots,x_n,x_{n+1})\). This is a bijection. To use the two-factor theorem, we must check that this identification preserves the relevant topology, not merely the underlying points.
Proof. A basic open set in the grouped product has the form \(O\times U_{n+1}\), where \(O\) is open in \(\prod_{i=1}^n X_i\) and \(U_{n+1}\) is open in \(X_{n+1}\). By the definition of the finite product topology, \(O\) is a union of basic sets \(\prod_{i=1}^n U_i\). Therefore \(O\times U_{n+1}\) is a union of sets \(\prod_{i=1}^{n+1}U_i\), each open in the \((n+1)\)-factor product topology. Conversely, a basic open set \(\prod_{i=1}^{n+1}U_i\) corresponds under the identification to \(\left(\prod_{i=1}^n U_i\right)\times U_{n+1}\). The first factor \(\prod_{i=1}^n U_i\) is open in \(\prod_{i=1}^n X_i\), so this is open in the grouped product topology. Each topology therefore has exactly the same open sets under the identification, proving the claim. \(\square\)
The Finite-Product Theorem
Proof. We use induction on \(n\). When \(n=1\), the product is \(X_1\), which is compact by hypothesis. Now let \(n\geq 2\), and suppose the claim holds for products of \(n-1\) compact spaces. The induction hypothesis says that \(Y=\prod_{i=1}^{n-1}X_i\) is compact. The space \(X_n\) is compact by assumption. The two-factor compactness theorem therefore says that \(Y\times X_n\) is compact. By the coordinate-regrouping proposition, this space has the same topology as \(\prod_{i=1}^n X_i\). Hence \(\prod_{i=1}^n X_i\) is compact, completing the induction.
This argument also covers empty factors: if \(X_j=\varnothing\) for some \(j\), then the full product is empty and hence compact. If there are zero factors, we use the usual convention that the empty product is a one-point space; it is compact because every open cover has a one-set subcover (or the empty subcover if the point is not required to be covered, which cannot occur for a cover). Thus the finite-product statement includes both degenerate cases. \(\square\)
The induction step has two distinct ingredients. Compactness of the product of the first \(n-1\) factors comes from the induction hypothesis; compactness of the next grouped product comes from the two-factor theorem. The topology check ensures that the grouped product is genuinely the product topology on all \(n\) coordinates.
Worked Example: A Product of Finite Discrete Spaces
Let \(X_1=\{a,b\}\), \(X_2=\{0,1,2\}\), and \(X_3=\{u,v\}\), each with the discrete topology, in which every subset is open. Each factor is compact: for any open cover of a finite set, choose one covering set for each point; these choices form a finite subcover.
The product has \(2\cdot3\cdot2=12\) points. Listing them by the first and second coordinates gives \((a,0,u),(a,0,v),(a,1,u),(a,1,v),(a,2,u),(a,2,v)\), \((b,0,u),(b,0,v),(b,1,u),(b,1,v),(b,2,u),(b,2,v)\). In particular, each factor count contributes every choice of its coordinate: there are \(2\) choices for the first coordinate, \(3\) for the second, and \(2\) for the third.
To check compactness directly, take any open cover of this product. For each of its 12 points, choose one member of the cover containing that point. The chosen subfamily has at most 12 members and covers the whole product. The product is compact. This agrees with the induction theorem.
Check: The singleton \(\{(a,1,v)\}\) is open in the product topology because it is the basic open set \(\{a\}\times\{1\}\times\{v\}\). Thus the listed points really are separate points of the product, and the finite-point cover argument applies to all 12 of them.
Worked Example: An Interval with a Finite Factor
Take \(X_1=[0,1]\) with its subspace topology from the standard topology on \(\mathbb{R}\), and \(X_2=\{r,s\}\) with the discrete topology. The interval is compact by the open-cover theorem for closed intervals; the two-point space is compact because it is finite. The theorem gives compactness of \([0,1]\times\{r,s\}\).
Here a basic open set has the form \(U\times V\), where \(U\) is open in \([0,1]\) and \(V\) is one of \(\varnothing,\{r\},\{s\},\{r,s\}\). For example, \(U\times\{r\}\) is open for every open \(U\subseteq[0,1]\), and it contains exactly those product points whose second coordinate is \(r\) and whose first coordinate lies in \(U\). This description is the product topology being used; the theorem applies to the full product, not just to either coordinate slice.
For a direct check on one open cover, consider \(U_1=[0,1]\times\{r\}\) and \(U_2=[0,1]\times\{s\}\). Both are open, and their union is the whole product: a point has second coordinate either \(r\) or \(s\). The two sets themselves form a finite subcover. The general theorem gives the stronger conclusion that every open cover of this product has a finite subcover.
Check: The product has two disjoint slices, \([0,1]\times\{r\}\) and \([0,1]\times\{s\}\), and both are needed to cover all points in the particular cover just given. No claim that a general open cover has this form is being made; compactness concerns arbitrary open covers.
Worked Example: A Finite-Complement Factor
Let \(X_1=\mathbb{R}\) with the finite-complement topology: its open sets are \(\varnothing\) and the subsets whose complements in \(\mathbb{R}\) are finite. Let \(X_2=\{0,1\}\) and \(X_3=\{p,q,r\}\), both discrete. First verify compactness of \(X_1\). Given an open cover, if it has no nonempty member then it cannot cover \(\mathbb{R}\), so choose a nonempty member \(U\). Its complement \(F=\mathbb{R}\setminus U\) is finite. For each point of \(F\), choose a covering set containing that point. The set \(U\) together with these finitely many chosen sets is a finite subcover. The other two factors are finite and therefore compact.
It follows that \(\mathbb{R}\times\{0,1\}\times\{p,q,r\}\), with the product topology, is compact. A basic open set is \(U\times V\times W\), with \(U\) open in the finite-complement topology and \(V,W\) open in their discrete topologies. Consider the family, indexed by \(m\in\mathbb{N}\), \(G_m=(\mathbb{R}\setminus\{m\})\times\{0,1\}\times\{p,q,r\}\). Each \(G_m\) is open because \(\mathbb{R}\setminus\{m\}\) is cofinite and the other two factors are open in their discrete topologies.
This family covers the product. Indeed, for any real \(x\), at least one of \(x\ne1\) or \(x\ne2\) holds, so \(x\) belongs to \(\mathbb{R}\setminus\{1\}\) or \(\mathbb{R}\setminus\{2\}\); the other coordinates are unrestricted. More precisely, \(G_1\cup G_2\) is the entire product, since \((\mathbb{R}\setminus\{1\})\cup(\mathbb{R}\setminus\{2\})=\mathbb{R}\). Thus \(G_1,G_2\) are a finite subcover of this particular cover, in agreement with the theorem.
Check: The cover is not a single-set cover: \(G_m\) omits every point whose first coordinate is \(m\), for example \((m,0,p)\). Two sets suffice because no real number equals both \(1\) and \(2\).
Where the Induction Argument Stops
The theorem concerns finite products. At each induction step there is one new factor, and the product already formed is compact by the induction hypothesis. This does not by itself prove compactness of a product with infinitely many factors: there is no final induction step that adds all remaining coordinates at once.
A related pitfall is to replace the product topology by the box topology in an infinite product. For a family of spaces, the box topology has basis sets \(\prod_i U_i\) with every \(U_i\) open, while in the product topology only finitely many coordinates are restricted in a basic set. The distinction matters. For example, take countably many copies of \(\{0,1\}\), each discrete and compact, and give their infinite product the box topology. For any sequence \(x=(x_1,x_2,\ldots)\), the singleton \(\{x\}=\prod_i\{x_i\}\) is box-open. The product has infinitely many distinct points, such as the sequence of all zeros and the sequences that have a \(1\) in exactly one coordinate. Its cover by all singleton sets is open and has no finite subcover: a finite union of singletons cannot contain all those distinct points. Thus the finite-product theorem cannot be extended to infinite box products by repeating its conclusion.
The useful principle is precise: for a finite number of compact factors, group the coordinates into a compact product and one additional compact space, use the two-factor theorem, and justify the grouping with the product topology. The number of factors must be finite for this induction argument.
Check Your Understanding
For each question, identify which topology is being used and where compactness enters the argument.
- Explain the induction step that proves compactness of \(X_1\times X_2\times X_3\times X_4\) when all four factors are compact.
- Why must the coordinate regrouping proposition be checked before applying the two-factor theorem?
- Prove directly that a finite topological space is compact, without assuming its topology is discrete.
- In the finite-complement example, why does a nonempty open set have only finitely many points outside it, and how does that fact produce a finite subcover?
- For the infinite box product of two-point discrete spaces, explain why the singleton sets form an open cover with no finite subcover.