tube lemma
Tube lemma - Let X and Y be topological spaces such that Y is compact. If N is an open set of X×Y containing a ”slice” x0×Y, then N contains some ”tube” W×Y, where W is a neighborhood of x0 in X.
Proof : N is a union of basis elements U×V, with U and V open sets in X and Y respect. Since x0×Y is compact (it is homeomorphic to Y), only a finite number U1×V1,…,Un×Vn of such basis elements cover x0×Y.
We may assume that each of the basis elements Ui×Vi actually intersects x0×Y, since otherwise we could discard it from the finite collection and still have a covering of x0×Y.
Define W:=. 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 |