From Local Rectangles to One Product Neighborhood
In the product topology, every open set containing a point contains a basic open rectangle around that point. But a compact slice contains many points, and the rectangles supplied at those points may have different neighborhoods in the second factor. Compactness is what lets us replace those varying choices with one common neighborhood.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces, and give \(X\times Y\) the product topology. Its basic open sets are rectangles \(U\times V\), where \(U\in\mathcal{T}_X\) and \(V\in\mathcal{T}_Y\). A subset \(K\subseteq X\) is compact if every cover of \(K\) by open subsets of \(X\) has a finite subcover. We will use compactness to choose finitely many rectangles and then coordinate their sides.
Proof. If \(K=\varnothing\), take \(U=\varnothing\) and \(V=Y\). The required inclusions hold, including \(U\times V=\varnothing\subseteq W\). Now suppose \(K\ne\varnothing\). For each \(x\in K\), the point \((x,y_0)\) belongs to the open set \(W\). By the basis for the product topology, there are open sets \(U_x\subseteq X\) and \(V_x\subseteq Y\) with \(x\in U_x\), \(y_0\in V_x\), and \(U_x\times V_x\subseteq W\). The family of all \(U_x\), for \(x\in K\), covers \(K\). Compactness gives points \(x_1,\ldots,x_n\in K\) such that \(K\subseteq U_{x_1}\cup\cdots\cup U_{x_n}\). Set \(U=\bigcup_{i=1}^n U_{x_i}\) and \(V=\bigcap_{i=1}^n V_{x_i}\). The finite union \(U\) and finite intersection \(V\) are open; moreover \(K\subseteq U\) and \(y_0\in V\). If \((u,v)\in U\times V\), then \(u\in U_{x_i}\) for some \(i\), while \(v\in V\subseteq V_{x_i}\). Thus \((u,v)\in U_{x_i}\times V_{x_i}\subseteq W\). Therefore \(U\times V\subseteq W\). \(\square\)
The important choice is the intersection in the second coordinate. Every point of \(U\) may come from a different rectangle, so the common \(V\) must fit inside every selected rectangle’s second-coordinate neighborhood.
Worked Example: A Rectangle Around a Compact Interval
In \(\mathbb{R}\times\mathbb{R}\), with the product of the standard topologies, let \(K=[-1,1]\) and \(W=\{(x,y): |x|+|y|<2\}\). Find a product neighborhood \(U\times V\) of \(K\times\{0\}\) contained in \(W\).
First, \(W\) is open. If \((a,b)\in W\), let \(\delta=2-|a|-|b|>0\). When \(|x-a|<\delta/4\) and \(|y-b|<\delta/4\), the triangle inequality gives \(|x|+|y|\le |a|+|b|+|x-a|+|y-b|<2-\delta+\delta/2<2\). So a basic open rectangle around each point of \(W\) lies in \(W\). Also, for every \(x\in[-1,1]\), \(|x|+|0|\le 1<2\), so \(K\times\{0\}\subseteq W\).
Choose \(U=(-5/4,5/4)\) and \(V=(-1/2,1/2)\). These are open, \(K\subseteq U\), and \(0\in V\). For every \((x,y)\in U\times V\), the strict endpoint conditions give \(|x|<5/4\) and \(|y|<1/2\). Hence \(|x|+|y|<5/4+1/2=7/4<2\), so \(U\times V\subseteq W\).
Check: the rectangle contains the entire slice because \([-1,1]\subset(-5/4,5/4)\) and \(0\in(-1/2,1/2)\); its points satisfy the defining strict inequality for \(W\), as computed above.
The Compact-Factor Version
A closely related form has a compact set in the second coordinate and a single point in the first. It is often called the tube lemma as well. Here the entire compact set is covered by rectangles, and the neighborhoods in the first coordinate are intersected.
Proof. If \(Y=\varnothing\), take \(U=X\); then \(U\times Y=\varnothing\subseteq W\). Suppose \(Y\ne\varnothing\). For each \(y\in Y\), choose 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 \(Y\). By compactness, finitely many, say \(V_{y_1},\ldots,V_{y_m}\), cover \(Y\). Let \(U=\bigcap_{j=1}^m U_{y_j}\). This is open and contains \(x_0\). For any \(y\in Y\), choose \(j\) with \(y\in V_{y_j}\). For any \(x\in U\), we have \(x\in U_{y_j}\), so \((x,y)\in U_{y_j}\times V_{y_j}\subseteq W\). This holds for every \(x\in U\) and every \(y\in Y\), proving \(U\times Y\subseteq W\). \(\square\)
Worked Example: Coordinating Two Finite Choices
Take \(X=Y=\mathbb{R}\) with their standard topologies, \(K=\{-1,1\}\), and \(W=\bigl((-2,0)\times(-1,1)\bigr)\cup\bigl((0,2)\times(-2,2)\bigr)\). The set \(W\) is open because it is a union of open rectangles. It contains \(K\times\{0\}\): the point \((-1,0)\) belongs to the first rectangle, and \((1,0)\) belongs to the second.
The finite set \(K\) is compact: from any open cover, choose one covering set for \(-1\) and one covering set for \(1\). The two first-coordinate neighborhoods \(U_1=(-2,0)\) and \(U_2=(0,2)\) cover \(K\). The common second-coordinate neighborhood must lie in both \((-1,1)\) and \((-2,2)\), so take \(V=(-1,1)\). Taking \(U=U_1\cup U_2=(-2,0)\cup(0,2)\), we obtain \(K\subseteq U\), \(0\in V\), and \(U\times V\subseteq W\): if \(x\in U_1\), then \(y\in V\subseteq(-1,1)\); if \(x\in U_2\), then \(y\in V\subseteq(-2,2)\). In either case \((x,y)\in W\).
Check: \(-1\in(-2,0)\) and \(1\in(0,2)\), with both inclusions strict; and \((-1,1)\subset(-2,2)\). Thus the chosen \(U\) and \(V\) have the required containment.
A Product-Space Application: Closed Projections
For a subset \(F\subseteq X\times Y\), its projection onto \(X\) is \(\pi_X(F)=\{x\in X:\text{ there is }y\in Y\text{ with }(x,y)\in F\}\). Projection need not preserve closedness in general. The compact-factor tube lemma supplies the missing control when the second factor is compact.
Proof. Let \(x_0\in X\setminus\pi_X(F)\). By the definition of projection, there is no \(y\in Y\) for which \((x_0,y)\in F\). Hence \(\{x_0\}\times Y\subseteq W\), where \(W=(X\times Y)\setminus F\). Since \(F\) is closed, \(W\) is open. The compact-factor tube lemma gives an open neighborhood \(U\) of \(x_0\) with \(U\times Y\subseteq W\). Therefore no \(x\in U\) belongs to \(\pi_X(F)\): if \(x\in\pi_X(F)\), some \(y\in Y\) has \((x,y)\in F\), contradicting \(U\times Y\subseteq W\). We have found an open neighborhood of every point in \(X\setminus\pi_X(F)\) that stays inside that complement. Thus the complement is open, and \(\pi_X(F)\) is closed. If \(Y=\varnothing\), then \(\pi_X(F)=\varnothing\), which is closed, so that case also satisfies the claim. \(\square\)
Worked Example: Projecting a Closed Set
Let \(X=\mathbb{R}\), \(Y=[0,1]\) with the subspace topology inherited from \(\mathbb{R}\), and \(F=\{(x,y)\in\mathbb{R}\times[0,1]:xy=1\}\). First verify that \(F\) is closed. The function \(g:\mathbb{R}\times[0,1]\to\mathbb{R}\), \(g(x,y)=xy\), is continuous: at \((a,b)\), for \(|x-a|<1\), \[ |xy-ab|\le (|a|+1)|y-b|+|b||x-a|. \] Given \(\varepsilon>0\), requiring \(|y-b|<\varepsilon/(2(|a|+1))\) and \(|x-a|<\min(1,\varepsilon/(2(|b|+1)))\) makes the right side less than \(\varepsilon\). Consequently \(F=g^{-1}(\{1\})\) is closed, because \(\{1\}\) is closed and a continuous preimage of a closed set is closed.
To compute the projection, \(xy=1\) forces \(y\ne0\) and \(x=1/y\). Since \(0<y\le1\), this gives \(x\ge1\). Conversely, if \(x\ge1\), then \(y=1/x\) satisfies \(0<y\le1\) and \(xy=1\). Therefore \(\pi_X(F)=[1,\infty)\), which is closed in \(\mathbb{R}\). The theorem also explains why points outside the projection have neighborhoods that miss it. For \(x_0<1\), the open interval \((-\infty,1)\) contains \(x_0\) and is disjoint from \([1,\infty)\).
Check: \(1\cdot1=1\), so \(1\) is in the projection; if \(x<1\), no \(y\in[0,1]\) can satisfy \(xy=1\), since \(y=0\) gives \(xy=0\) and \(y>0\) gives \(x=1/y\ge1\).
Common Pitfall: Union Instead of Intersection
In the compact-slice proof, the second-coordinate neighborhoods are intersected, not united. A union may include points that are not allowed by the rectangle chosen for a particular first-coordinate point.
For the finite example above, an incorrect choice would be \(V_{\mathrm{wrong}}=(-1,1)\cup(-2,2)=(-2,2)\). Then \((-1,3/2)\in U\times V_{\mathrm{wrong}}\), because \(-1\in U\) and \(3/2\in(-2,2)\). But \((-1,3/2)\notin W\): it is not in the first rectangle because \(3/2\notin(-1,1)\), and it is not in the second because \(-1\notin(0,2)\). The correction is to use \(V=(-1,1)\cap(-2,2)=(-1,1)\); then every point of \(U\times V\) belongs to the rectangle corresponding to its first coordinate.
The tube argument is useful whenever the goal is to make a local choice uniform across a compact set. It is not a claim that compactness is unnecessary: without it, infinitely many local neighborhoods may have no common neighborhood that works across the entire slice.
Check Your Understanding
Use the definitions of compactness and the product topology to justify each answer.
- In the compact-slice tube lemma, why is the second-coordinate neighborhood an intersection of finitely many sets?
- State the compact-factor tube lemma when \(Y\) is empty, and explain why its conclusion still holds.
- Give an example of an open subset of \(\mathbb{R}\times\mathbb{R}\) that contains \(\mathbb{R}\times\{0\}\) but contains no set of the form \(\mathbb{R}\times V\) for an open neighborhood \(V\) of \(0\).
- Which hypothesis in the closed-projection theorem ensures that one neighborhood works for every point of the second factor?
- In the finite example, explain why \((-1,3/2)\) belongs to \(U\times V_{\mathrm{wrong}}\) but not to \(W\).