Tube lemma
This article is about the statement of a simple but indispensable lemma in topology
Contents
Statement
Let be a compact space and any topological space. Consider endowed with the product topology. Suppose and is an open subset of containing the entire slice . Then, we can find an open subset of such that:
, and
In other words, any open subset containing a slice contains an open cylinder that contains the slice.
Proof
Given: A compact space , a topological space . , and is an open subset of containing the slice .
To prove: There exists an open subset of such that is contained in .
Proof:
- A collection of open subsets inside whose union contains : For each , we have , so by the definition of openness in the product topology, there exists a basis open subset containing . In particular, we get a collection of open subsets contained in , whose union contains .
- This collection yields a point-indexed open cover for : Note that since is a basis open set containing , is an open subset of containing , so the , form an open cover of .
- (Given data used: is compact): This cover has a finite subcover: Indeed, since is compact, we can choose a finite collection of points such that is the union of the s.
- If is the intersection of the corresponding s, then is open in , , and : First, is open since it is an intersection of finitely many open subsets of . Second, each contains , so . Third, if , then there exists such that . By definition, , so . Thus, .
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)