One Neighborhood That Works Across a Compact Factor
Suppose an open set in a product contains every point of a slice \(\{x\}\times Y\). At each point of that slice, openness supplies a product neighborhood, but the neighborhood in \(X\) might depend on the point of \(Y\). The tube lemma says that compactness of \(Y\) lets us reduce these possibly many neighborhoods to finitely many, and then choose one neighborhood in \(X\) that works for all of \(Y\).
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. The product topology on \(X\times Y\) has basis consisting of the sets \(U\times V\), where \(U\in\mathcal{T}_X\) and \(V\in\mathcal{T}_Y\). Thus, if \(W\) is open in \(X\times Y\) and \((x,y)\in W\), there are open sets \(U\subseteq X\) and \(V\subseteq Y\) such that \((x,y)\in U\times V\subseteq W\). A space is compact if every open cover of it has a finite subcover.
The conclusion is uniform in \(y\): one open set \(U\) works for every point of \(Y\). It does not say that \(W\) itself is a product, or that \(U\) is unique. If \(Y=\varnothing\), the hypothesis is automatic and the conclusion holds for any open neighborhood \(U\) of \(x\); the proof below handles the nonempty case.
Proof by a Finite Subcover
Proof. If \(Y=\varnothing\), choose \(U=X\), which is open and contains \(x\); then \(U\times Y=\varnothing\subseteq W\). Now suppose \(Y\ne\varnothing\). For each \(y\in Y\), the point \((x,y)\) belongs to \(W\). Since \(W\) is open in the product topology, the basis property gives open sets \(U_y\subseteq X\) and \(V_y\subseteq Y\) such that \((x,y)\in U_y\times V_y\subseteq W\). In particular, \(x\in U_y\) and \(y\in V_y\). Therefore the open sets \(V_y\), for \(y\in Y\), cover \(Y\). Compactness gives finitely many points \(y_1,\ldots,y_n\in Y\) such that \(Y=V_{y_1}\cup\cdots\cup V_{y_n}\). Define \(U=U_{y_1}\cap\cdots\cap U_{y_n}\). A finite intersection of open sets is open, and every \(U_{y_i}\) contains \(x\), so \(U\) is an open neighborhood of \(x\). To check the required inclusion, take any \((u,y)\in U\times Y\). Since the \(V_{y_i}\) cover \(Y\), there is an index \(i\) with \(y\in V_{y_i}\). Also \(u\in U\subseteq U_{y_i}\). Hence \((u,y)\in U_{y_i}\times V_{y_i}\subseteq W\). Thus \(U\times Y\subseteq W\). \(\square\)
The finite intersection is the key step: it ensures that the chosen \(U\) meets every local requirement at once. Without compactness, the cover \(\{V_y:y\in Y\}\) need not have a finite subcover, and an infinite intersection of the corresponding \(U_y\)'s need not be open or even a neighborhood of \(x\).
Worked Examples
Worked Example: A Finite Discrete Factor
Let \(X=\mathbb{R}\) with its standard topology and let \(Y=\{a,b\}\) have the discrete topology, in which every subset is open. Consider \(W=((-2,2)\times\{a\})\cup((-1,3)\times\{b\})\). Show that the tube lemma applies at \(x=0\), and find a suitable \(U\).
First, \(Y\) is compact: from any open cover, choose one member containing \(a\) and one member containing \(b\); these at most two sets cover \(Y\). Each displayed piece of \(W\) is a product of open sets, so \(W\) is open in \(X\times Y\). The slice is \(\{0\}\times Y=\{(0,a),(0,b)\}\). Because \(-2<0<2\), we have \((0,a)\in(-2,2)\times\{a\}\). Because \(-1<0<3\), we have \((0,b)\in(-1,3)\times\{b\}\). Thus the whole slice lies in \(W\).
Take \(U=(-2,2)\cap(-1,3)=(-1,2)\). This is an open neighborhood of \(0\). For every \(u\in U\), we have \(-2<u<2\), so \((u,a)\in W\), and \(-1<u<3\), so \((u,b)\in W\). Therefore \(U\times Y\subseteq W\). In fact, for this particular \(W\), the set of real numbers \(u\) for which both \((u,a)\) and \((u,b)\) lie in \(W\) is exactly \((-2,2)\cap(-1,3)=(-1,2)\), confirming the computation.
Worked Example: A Compact Interval as the Factor
Let \(X=\mathbb{R}\) with its standard topology and \(Y=[0,1]\) with the subspace topology inherited from \(\mathbb{R}\). The interval \(Y\) is compact by the Heine–Borel theorem. Define \(V_1=[0,3/4)\), \(V_2=(1/4,1]\), \(U_1=(-2,1)\), and \(U_2=(-1,2)\), where \(V_1,V_2\) carry the subspace topology. Set \(W=(U_1\times V_1)\cup(U_2\times V_2)\). Find a neighborhood \(U\) of \(0\) such that \(U\times Y\subseteq W\).
The sets \(V_1=Y\cap(-1,3/4)\) and \(V_2=Y\cap(1/4,2)\) are open in \(Y\). The interval sets \(U_1\) and \(U_2\) are open in \(X\), so both product sets in the definition of \(W\) are open in \(X\times Y\), and hence \(W\) is open. The sets \(V_1\) and \(V_2\) cover \(Y\): if \(0\le y<3/4\), then \(y\in V_1\); if \(3/4\le y\le1\), then \(y>1/4\) and \(y\in V_2\). Since \(0\in U_1\cap U_2\), each point \((0,y)\) with \(y\in V_i\) belongs to \(U_i\times V_i\). Thus \(\{0\}\times Y\subseteq W\).
The tube-lemma construction gives \(U=U_1\cap U_2=(-1,1)\). To verify the inclusion directly, take \(u\in(-1,1)\) and \(y\in[0,1]\). If \(y<3/4\), then \(y\in V_1\) and \(-2<u<1\), so \((u,y)\in U_1\times V_1\subseteq W\). If \(y\ge3/4\), then \(y>1/4\), so \(y\in V_2\), and \(-1<u<2\), so \((u,y)\in U_2\times V_2\subseteq W\). These cases include every \(y\in[0,1]\), and therefore \(U\times Y\subseteq W\).
Check: \(0\in(-1,1)\), and the two cases just verified show that for every \(y\in[0,1]\), \((0,y)\in W\). The proposed neighborhood contains the entire slice as required.
Worked Example: What Fails for a Noncompact Factor
Let \(X=\mathbb{R}\) with its standard topology and \(Y=\mathbb{N}=\{1,2,\ldots\}\) with the discrete topology. Define \(W=\bigcup_{n\in\mathbb{N}}((-1/n,1/n)\times\{n\})\). This \(W\) is open because each rectangle in the union is a product of open sets. For every \(n\in\mathbb{N}\), \(0\in(-1/n,1/n)\), so \((0,n)\in W\), and consequently \(\{0\}\times Y\subseteq W\).
However, no open neighborhood \(U\) of \(0\) satisfies \(U\times Y\subseteq W\). To prove this, let \(U\) be any such neighborhood. There is an \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq U\). Choose \(n\in\mathbb{N}\) with \(1/n<\varepsilon\). Then \(1/n\in U\), so \((1/n,n)\in U\times Y\). But among the rectangles defining \(W\), only the rectangle with second coordinate \(n\) can contain a point whose second coordinate is \(n\), and that rectangle requires the first coordinate to satisfy \(-1/n<x<1/n\). At \(x=1/n\), the strict inequality fails. Hence \((1/n,n)\notin W\), proving that \(U\times Y\) is not a subset of \(W\).
The factor \(Y\) is indeed noncompact: the open cover \(\{\{n\}:n\in\mathbb{N}\}\) has no finite subcover, since a finite union of singletons omits every natural number not among the finitely many selected. This example shows why compactness is a genuine hypothesis, not a dispensable condition.
A Closed-Projection Consequence
For \(F\subseteq X\times Y\), its projection onto \(X\) is the set \(\pi_1(F)=\{x\in X:\text{ there exists }y\in Y\text{ with }(x,y)\in F\}\). The tube lemma gives a useful closedness criterion when \(Y\) is compact.
Proof. We show that \(X\setminus\pi_1(F)\) is open. Fix \(x\in X\setminus\pi_1(F)\). By the definition of projection, no \(y\in Y\) has \((x,y)\in F\). Thus \(\{x\}\times Y\subseteq(X\times Y)\setminus F\). Since \(F\) is closed, its complement is open. Apply the tube lemma to \(W=(X\times Y)\setminus F\) to obtain an open neighborhood \(U\) of \(x\) with \(U\times Y\subseteq W\). No point of \(U\) can belong to \(\pi_1(F)\): if \(u\in U\cap\pi_1(F)\), there is \(y\in Y\) with \((u,y)\in F\), contradicting \(U\times Y\subseteq(X\times Y)\setminus F\). Hence \(U\subseteq X\setminus\pi_1(F)\). Every point of this complement has an open neighborhood contained in the complement, so the complement is open and \(\pi_1(F)\) is closed. If \(Y=\varnothing\), then \(\pi_1(F)=\varnothing\), which is closed, so that case also holds. \(\square\)
For a concrete check, take \(X=\mathbb{R}\), \(Y=[0,1]\), and \(F=(-\infty,0]\times[0,1]\). This \(F\) is closed in the product: its complement is \((0,\infty)\times[0,1]\), which is open in \(\mathbb{R}\times[0,1]\). Its projection is \(\pi_1(F)=(-\infty,0]\), a closed subset of \(\mathbb{R}\). The proposition explains the role of the compact factor in this kind of argument: it converts the local neighborhoods around each point of a whole slice into one neighborhood that avoids the closed set.
A Common Pitfall
Common mistake. In the finite discrete example, if the second rectangle is instead \((1,3)\times\{b\}\), one might write \(U=(-2,2)\cap(1,3)=(1,2)\) and claim that this is a neighborhood of \(0\). The arithmetic intersection is correct, but the conclusion is not: \(0\notin(1,2)\). More importantly, the hypotheses of the tube lemma fail, because \((0,b)\notin(1,3)\times\{b\}\), so \(\{0\}\times\{a,b\}\) is not contained in the proposed \(W\). The correction is to use an interval containing \(0\), such as \((-1,3)\), for the second rectangle. Then \(U=(-2,2)\cap(-1,3)=(-1,2)\) is an open neighborhood of \(0\), and the whole slice is contained in \(W\). Before applying the lemma, check both that \(W\) is open and that every point of the specified slice really belongs to \(W\).
Check Your Understanding
Use the product-basis definition and the finite-subcover step to justify your answers.
- In the tube lemma, which space must be compact, and why does the proof require only finitely many neighborhoods in \(X\)?
- Let \(Y=\{a,b,c\}\) have the discrete topology. Explain why it is compact and how the construction of \(U\) works if the slice is covered by three product neighborhoods.
- In the noncompact example, why does choosing \(n\) with \(1/n<\varepsilon\) ensure that \((1/n,n)\) is in \(U\times\mathbb{N}\) but not in \(W\)?
- State the closed-projection proposition and identify where closedness of \(F\) is used.
- Does the tube lemma require \(X\) to be compact? Identify the step in its proof that uses compactness and the space to which it applies.