Why Compactness of a Product Needs an Argument
The product topology on \(X\times Y\) is built from open rectangles \(U\times V\), where \(U\) is open in \(X\) and \(V\) is open in \(Y\). Even if both factors are compact, an open cover of the product can consist of sets that are not rectangles. Compactness of each factor therefore cannot be applied to that cover directly. The Tube Lemma supplies the bridge: it turns a neighborhood of a whole fiber into a neighborhood of its base point times the entire other factor.
A topological space is compact if every cover by open sets has a finite subcover. We use the product topology throughout: its basis consists of the open rectangles \(U\times V\). If either factor is empty, then \(X\times Y\) is empty and is compact, since the empty subfamily covers it. We may therefore assume both factors are nonempty when proving the main result.
Proof. If \(Y=\varnothing\), take \(U=X\). Otherwise, for each \(y\in Y\), the point \((x_0,y)\) lies in the open set \(W\). By the basis definition of the product topology, there are open sets \(U_y\subseteq X\) and \(V_y\subseteq Y\) such that \(x_0\in U_y\), \(y\in V_y\), and \(U_y\times V_y\subseteq W\). The sets \(V_y\) cover \(Y\). Compactness gives finitely many \(y_1,\ldots,y_m\) for which \(V_{y_1},\ldots,V_{y_m}\) cover \(Y\). Set \(U=\bigcap_{j=1}^{m}U_{y_j}\). This is an open neighborhood of \(x_0\), since it is a finite intersection of open sets containing \(x_0\). If \(u\in U\) and \(z\in Y\), choose \(j\) with \(z\in V_{y_j}\). Then \(u\in U_{y_j}\) as well, so \((u,z)\in U_{y_j}\times V_{y_j}\subseteq W\). Thus \(U\times Y\subseteq W\). \(\square\)
The Product Theorem
Proof. The case in which either factor is empty was handled above. Assume now that both are nonempty, and let \(\mathcal{O}\) be an arbitrary open cover of \(X\times Y\). We will find a finite subfamily of \(\mathcal{O}\) that covers the product.
Fix \(x\in X\). The fiber \(\{x\}\times Y\) is compact: the map \(g_x:Y\to X\times Y\) given by \(g_x(y)=(x,y)\) is continuous, and its image is that fiber. To check continuity, the preimage of a basis rectangle \(U\times V\) is \(V\) if \(x\in U\), and is empty if \(x\notin U\); both are open in \(Y\). The continuous image of a compact space is compact.
Since \(\mathcal{O}\) covers the product, it covers \(\{x\}\times Y\). Compactness of that fiber gives finitely many members \(O_{x,1},\ldots,O_{x,k_x}\) of \(\mathcal{O}\) that cover it. Let \(W_x=O_{x,1}\cup\cdots\cup O_{x,k_x}\). This is open in \(X\times Y\) and contains the entire fiber \(\{x\}\times Y\). Apply the Tube Lemma to obtain an open neighborhood \(U_x\) of \(x\) with \(U_x\times Y\subseteq W_x\).
As \(x\) varies over \(X\), the sets \(U_x\) form an open cover of \(X\). Compactness of \(X\) gives \(x_1,\ldots,x_n\in X\) such that \(U_{x_1},\ldots,U_{x_n}\) cover \(X\). The finitely many sets \(O_{x_i,j}\), for \(1\le i\le n\) and \(1\le j\le k_{x_i}\), cover \(X\times Y\): given \((u,y)\in X\times Y\), choose \(i\) with \(u\in U_{x_i}\). Then \((u,y)\in U_{x_i}\times Y\subseteq W_{x_i}\), and \(W_{x_i}\) is covered by its selected members \(O_{x_i,1},\ldots,O_{x_i,k_{x_i}}\). We have found a finite subcover of \(\mathcal{O}\). Since \(\mathcal{O}\) was arbitrary, \(X\times Y\) is compact. \(\square\)
The two uses of compactness have distinct roles. Compactness of \(Y\) first reduces each fiber cover to finitely many open sets; the Tube Lemma thickens that finite cover to a neighborhood of the fiber’s base point. Compactness of \(X\) then reduces the resulting neighborhoods to finitely many. Without the tube step, a finite cover of each individual fiber would not yet give a finite cover of all fibers together.
Worked Applications
Worked Example: An Explicit Cover of an Interval Times a Two-Point Space
Let \(I=[0,1]\) with its usual subspace topology from \(\mathbb{R}\), and let \(D=\{a,b\}\) have the discrete topology, in which every subset is open. The interval \(I\) is compact by the open-cover compactness theorem for closed intervals, and \(D\) is compact because it is finite. Consider the following open sets in \(I\times D\), for each integer \(n\ge2\) and each \(i\in D\):
These sets are open in the product topology. Indeed, \(U_n=I\cap(-1,1-1/n)\) and \(V=I\cap(1/2,2)\) are open in \(I\), while \(\{i\}\) is open in \(D\). They cover the product: for any \(x\in I\) with \(x<1\), choose an integer \(n\ge2\) large enough that \(1/n<1-x\). Then \(x<1-1/n\), so \(x\in U_n\). The point \(1\) belongs to \(V\). This covers each point \((x,i)\) by a set associated with its second coordinate.
In fact, four sets suffice. Take \(U_3\times\{a\}\), \(V\times\{a\}\), \(U_3\times\{b\}\), and \(V\times\{b\}\), where \(U_3=[0,2/3)\). For each fixed \(i\), \(U_3\cup V=[0,2/3)\cup(1/2,1]=I\): the intervals overlap, \(0\in U_3\), and \(1\in V\). Hence the four sets cover \(I\times D\). This explicit subcover is consistent with the theorem; the theorem guarantees a finite subcover even when an open cover is not presented in such a conveniently structured form.
Worked Example: A Cofinite Space Times a Finite Discrete Space
Give \(X=\mathbb{R}\) the finite-complement topology: its open sets are \(\varnothing\) and the subsets whose complements in \(\mathbb{R}\) are finite. Let \(D=\{0,1\}\) have the discrete topology. The space \(X\) is compact: in any open cover, choose a nonempty member \(G\); its complement is finite, and each point of that finite complement can be covered by one additional member. Thus finitely many members cover \(X\). The finite space \(D\) is compact as well.
For \(n\in\mathbb{N}\) and \(i\in D\), define \(O_{n,i}=(\mathbb{R}\setminus\{n\})\times\{i\}\). These are open rectangles. They cover \(\mathbb{R}\times D\): for any \(x\in\mathbb{R}\) and \(i\in D\), \(x\) cannot equal both \(1\) and \(2\), so at least one of \(O_{1,i}\) or \(O_{2,i}\) contains \((x,i)\). In fact, the four sets \(O_{1,0},O_{2,0},O_{1,1},O_{2,1}\) cover the product. For each \(i\), the first two sets on that sheet have union \(\mathbb{R}\times\{i\}\), because \((\mathbb{R}\setminus\{1\})\cup(\mathbb{R}\setminus\{2\})=\mathbb{R}\).
This example uses a topology on \(\mathbb{R}\) different from the usual one. The cofinite topology is not Hausdorff: any two nonempty open sets have finite complements, so their intersection is nonempty. The product theorem did not require either factor to be Hausdorff.
Worked Example: Products of Indiscrete Spaces
Let \(X\) and \(Y\) be any nonempty sets, each with the indiscrete topology \(\{\varnothing,X\}\) and \(\{\varnothing,Y\}\), respectively. Each is compact: an open cover of a nonempty indiscrete space must include the whole space, since the only other open set is empty. The basis rectangles for the product topology are \(\varnothing\) and \(X\times Y\), so the product topology is also indiscrete. Any open cover of \(X\times Y\) must therefore include \(X\times Y\) itself, and one member already covers the product.
This direct check agrees with the theorem and again shows that no Hausdorff hypothesis is part of the product compactness result. Compactness is an open-cover property; it does not, by itself, assert that distinct points can be separated by neighborhoods.
A Common Pitfall: Projecting the Cover
A tempting but invalid shortcut is to project every open set in a cover onto \(X\), use compactness of \(X\) to select finitely many projections, and conclude that the corresponding original open sets cover \(X\times Y\). A projection records which first coordinates occur somewhere in the set; it does not record which second coordinates occur with them.
For a concrete check, let \(I=[0,1]\) with its usual topology in both factors, and set \(O=(I\times I)\setminus\{(0,0)\}\). This is open: the singleton \(\{(0,0)\}\) is closed in the metric space \(I\times I\). Both coordinate projections of \(O\) are all of \(I\). For each \(x\in I\), the point \((x,1)\) lies in \(O\), and for each \(y\in I\), the point \((1,y)\) lies in \(O\). Nevertheless, \(O\) misses \((0,0)\). Even if we add \(V=[0,1/2)\times[0,1/2)\), which is open in the product topology, then \(\{O,V\}\) covers \(I\times I\) but the single set \(O\) does not. Its projection already covers each factor, so choosing based only on projected coverage cannot certify coverage of the product. The Tube Lemma avoids this error by producing neighborhoods whose products with the entire other factor are contained in the chosen open sets.
Check Your Understanding
Use the open-cover definition and the Tube Lemma to organize your reasoning.
- In the product proof, where is compactness of \(Y\) used, and where is compactness of \(X\) used?
- Why is the fiber \(\{x\}\times Y\) compact when \(Y\) is compact? Give the map and identify its image.
- Let \(I=[0,1]\) and \(D=\{a,b\}\) be discrete. Verify directly that \(U_4=[0,3/4)\) and \(V=(1/2,1]\) cover \(I\).
- Why does knowing that the projections of a set cover both factors not show that the set covers their product?
- Which step of the proof would fail if \(Y\) were not compact, and which step would fail if \(X\) were not compact?