# Tube lemma

## 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:

1. 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 .
2. 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 .
3. (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.
4. 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)