Tube lemma

This article is about the statement of a simple but indispensable lemma in topology

Statement

Let $X$ be a compact space and $A$ any topological space. Consider $X\times A$ endowed with the product topology. Suppose $a\in A$ and $U$ is an open subset of $X\times A$ containing the entire slice $X\times \{a\}$ . Then, we can find an open subset $V$ of $A$ such that:

$a\in V$ , and $X\times V\subseteq U$

In other words, any open subset containing a slice contains an open cylinder that contains the slice.

Proof

Given: A compact space $X$ , a topological space $A$ . $a\in A$ , and $U$ is an open subset of $X\times A$ containing the slice $X\times \{a\}$ .

To prove: There exists an open subset $V$ of $A$ such that $a\in V$ $X\times V$ is contained in $U$ .

Proof:

A collection of open subsets inside $U$ whose union contains $X\times \{a\}$ : For each $x\in X$ , we have $(x,a)\in U$ , so by the definition of openness in the product topology, there exists a basis open subset $M_{x}\times N_{x}\subseteq U$ containing $(x,a)$ . In particular, we get a collection $M_{x}\times N_{x},x\in X$ of open subsets contained in $U$ , whose union contains $X\times \{a\}$ .
This collection yields a point-indexed open cover for $X$ : Note that since $M_{x}\times N_{x}$ is a basis open set containing $(x,a)$ , $M_{x}$ is an open subset of $X$ containing $X$ , so the $M_{x},x\in X$ , form an open cover of $X$ .
(Given data used: $X$ is compact): This cover has a finite subcover: Indeed, since $X$ is compact, we can choose a finite collection of points $\{x_{1},x_{2},\dots ,x_{n}\}\subseteq X$ such that $X$ is the union of the $M_{x_{i}}$ s.
If $V$ is the intersection of the corresponding $N_{x_{i}}$ s, then $V$ is open in $A$ , $a\in V$ , and $X\times V\subseteq U$ : First, $V$ is open since it is an intersection of finitely many open subsets of $Y$ . Second, each $N_{x_{i}}$ contains $a$ , so $a\in V$ . Third, if $(x,v)\in X\times V$ , then there exists $x_{i}$ such that $x\in M_{x_{i}}$ . By definition, $v\in N_{x_{i}}$ , so $(x,v)\in M_{x_{i}}\times N_{x_{i}}\subseteq V$ . Thus, $(x,v)\in U$ .

References

Textbook references

Topology (2nd edition) by James R. Munkres, ^{More info}, Page 168, Lemma 26.8, Chapter 3, Section 26 (the proof is given before the theorem, as Step 1 of the proof of Theorem 26.7 on page 167)