tube lemma
Tube lemma - Let $X$ and $Y$ be topological spaces^{} such that $Y$ is compact^{}. If $N$ is an open set of $X\times Y$ containing a ”slice” ${x}_{0}\times Y$, then $N$ contains some ”tube” $W\times Y$, where $W$ is a neighborhood^{} of ${x}_{0}$ in $X$.
Proof : $N$ is a union of basis elements $U\times V$, with $U$ and $V$ open sets in $X$ and $Y$ respect. Since ${x}_{0}\times Y$ is compact (it is homeomorphic to $Y$), only a finite number ${U}_{1}\times {V}_{1},\mathrm{\dots},{U}_{n}\times {V}_{n}$ of such basis elements cover ${x}_{0}\times Y$.
We may assume that each of the basis elements ${U}_{i}\times {V}_{i}$ actually intersects ${x}_{0}\times Y$, since otherwise we could discard it from the finite collection^{} and still have a covering of ${x}_{0}\times Y$.
Define $W:={U}_{1}\cap \mathrm{\dots}\cap {U}_{n}$. The set $W$ is open and contains ${x}_{0}$ because each ${U}_{i}\times {V}_{i}$ intersects ${x}_{0}\times Y$ by the previous remark.
We now claim that $W\times Y\subseteq N$. Let $(x,y)$ be a point in $W\times Y$. The point $({x}_{0},y)$ is in some ${U}_{i}\times {V}_{i}$ and so $y\in {V}_{i}$. We also know that $x\in W={U}_{1}\cap \mathrm{\dots}\cap {U}_{n}\subseteq {U}_{i}$.
Therefore $(x,y)\in {U}_{i}\times {V}_{i}\subseteq N$ as desired. $\mathrm{\square}$
References
- 1 J. Munkres, Topology (2nd edition), Prentice Hall, 1999.
Title | tube lemma |
---|---|
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 |