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:

1. 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 $X\times \{a\}$.
2. 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 $X$.
3. (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.
4. 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)