tube lemma
Tube lemma - Let and be topological spaces such that is compact. If is an open set of containing a ”slice” , then contains some ”tube” , where is a neighborhood of in .
Proof : is a union of basis elements , with and open sets in and respect. Since is compact (it is homeomorphic to ), only a finite number of such basis elements cover .
We may assume that each of the basis elements actually intersects , since otherwise we could discard it from the finite collection and still have a covering of .
Define . The set is open and contains because each intersects by the previous remark.
We now claim that . Let be a point in . The point is in some and so . We also know that .
Therefore as desired.
References
- 1 J. Munkres, Topology (2nd edition), Prentice Hall, 1999.
Title | tube lemma |
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Canonical name | TubeLemma |
Date of creation | 2013-03-22 17:25:39 |
Last modified on | 2013-03-22 17:25:39 |
Owner | asteroid (17536) |
Last modified by | asteroid (17536) |
Numerical id | 7 |
Author | asteroid (17536) |
Entry type | Theorem |
Classification | msc 54D30 |