Topological Proof Workshop II

From Compact Slices to Compact Products

In “Topological Proof Workshop I,” continuity was used to pull open neighborhoods back to a domain, while density was tested against arbitrary nonempty open sets. This workshop uses a related open-cover strategy: compactness lets us replace a family of neighborhoods by finitely many, and that finite choice gives control across an entire slice of a product.

The key step is the tube lemma. If an open set contains every point of a slice \(\{x_0\}\times K\), with \(K\) compact, then it contains a product neighborhood \(V\times K\) of that whole slice. The neighborhood \(V\) works simultaneously for every point of \(K\). We will use this uniformity to prove that a product of two compact spaces is compact.

The Tube Lemma

Recall that a basis for the product topology consists of products of open sets in the factors. Thus, near each point \((x_0,y)\) of a slice, an open set in the product contains a basic open rectangle. Compactness of the second factor allows us to select finitely many such rectangles; intersecting their first-coordinate neighborhoods then gives one neighborhood that works along the whole compact set.

Proof. If \(K=\varnothing\), take \(V=X\); then \(V\times K=\varnothing\subseteq U\). Suppose \(K\ne\varnothing\). For each \(y\in K\), the point \((x_0,y)\) belongs to \(U\). By the basis property for the product topology, there are open sets \(A_y\subseteq X\) and \(B_y\subseteq Y\) such that \(x_0\in A_y\), \(y\in B_y\), and \(A_y\times B_y\subseteq U\).

The family \(\{B_y:y\in K\}\) covers \(K\). Since \(K\) is compact, there are \(y_1,\ldots,y_m\in K\) such that \(K\subseteq\bigcup_{j=1}^{m}B_{y_j}\). Define \(V=\bigcap_{j=1}^{m}A_{y_j}\). This is an open neighborhood of \(x_0\), because it is a finite intersection of open sets, each containing \(x_0\). Take any \((x,z)\in V\times K\). Since the sets \(B_{y_j}\) cover \(K\), there is an index \(j\) such that \(z\in B_{y_j}\). Also \(x\in V\subseteq A_{y_j}\), so \((x,z)\in A_{y_j}\times B_{y_j}\subseteq U\). Therefore \(V\times K\subseteq U\), as required. \(\square\)

The compactness assumption is used precisely to extract finitely many of the sets \(B_y\). A finite intersection of the corresponding neighborhoods \(A_y\) is still open and still contains \(x_0\). Without that finite selection, intersecting all the \(A_y\) need not produce an open neighborhood.

Compactness of a Product of Two Spaces

We now apply the Tube Lemma to an arbitrary open cover. For each fixed point in the first factor, compactness of the second factor gives finitely many cover members that cover the slice above that point. Their union is an open set containing the slice, so the Tube Lemma expands that coverage to a whole neighborhood in the first factor times the second factor. Compactness of the first factor then reduces the number of such neighborhoods to finitely many.

Proof. If either \(X\) or \(Y\) is empty, then \(X\times Y=\varnothing\), which is compact. Suppose both are nonempty, and let \(\mathcal{U}\) be an open cover of \(X\times Y\).

Fix \(x\in X\). For each \(U\in\mathcal{U}\), define \(U_x=\{y\in Y:(x,y)\in U\}\). The map \(i_x:Y\to X\times Y\), given by \(i_x(y)=(x,y)\), is continuous: its coordinate maps are the constant map with value \(x\) and the identity map on \(Y\), and the product-topology criterion for continuity applies. Therefore each \(U_x=i_x^{-1}(U)\) is open in \(Y\). The sets \(U_x\), as \(U\) ranges over \(\mathcal{U}\), cover \(Y\), because \(\mathcal{U}\) covers \(X\times Y\). By compactness of \(Y\), finitely many members \(U_1,\ldots,U_r\) of \(\mathcal{U}\) have sections that cover \(Y\). Their union \(W_x=U_1\cup\cdots\cup U_r\) is open in \(X\times Y\) and contains \(\{x\}\times Y\).

Apply the Tube Lemma with compact set \(Y\) to \(W_x\). There is an open neighborhood \(V_x\) of \(x\) such that \(V_x\times Y\subseteq W_x\). As \(x\) varies over \(X\), the sets \(V_x\) cover \(X\). Compactness of \(X\) provides \(x_1,\ldots,x_s\in X\) such that \(X\subseteq\bigcup_{k=1}^{s}V_{x_k}\).

For each \(k\), the set \(V_{x_k}\times Y\) is covered by the finite collection of members of \(\mathcal{U}\) chosen to form \(W_{x_k}\). These finitely many collections, one for each \(k=1,\ldots,s\), together contain only finitely many members of \(\mathcal{U}\). They cover \(X\times Y\): given \((x,y)\in X\times Y\), choose \(k\) with \(x\in V_{x_k}\); then \((x,y)\in V_{x_k}\times Y\subseteq W_{x_k}\). Thus the original open cover has a finite subcover, and \(X\times Y\) is compact. \(\square\)

Finite Products and Further Applications

The two-factor theorem extends to any finite number of compact factors by induction. A product of zero factors is a one-point space, which is compact. If a product of \(n\) compact spaces is compact, then its product with one more compact space is compact by the theorem just proved. Therefore every finite product of compact spaces, with the product topology, is compact.

The finite-product theorem is useful whenever a problem involves several compact parameters. It lets us control all coordinates at once through a single compactness conclusion. The proof also identifies exactly why the result is finite: it uses compactness of one factor to choose finitely many cover members over each slice, and compactness of the other factor to choose finitely many slices. For an arbitrary infinite product, this argument does not provide the required finite selection; a general product theorem requires additional machinery.

A common pitfall is to say that a product is compact merely because each coordinate space is compact, without specifying the topology or the number of factors. Here the topology is the product topology, and the proof establishes the result for two factors and, by induction, for finitely many factors. Another pitfall is to assume that a neighborhood working at each point of a compact slice automatically works uniformly on the whole slice. The Tube Lemma supplies precisely the finite-cover argument needed to justify that step.