If $M$ is a $n$-dimensional manifold which is Hausdorff and satisfies second countable axiom, for each $p \in M$, there is an open neighborhood ${U_{r(p)}}(p)$ which, through a differential mapping ${\varphi _p}$ , is diffeomorphic to ${B_{r(p)}} = \{ x \in {R^n}|\left| x \right| < r(p)\} $ and ${\varphi _p}(p) = 0$, then can we find a at most countably many {U_i} = {U_{r'({p_i})}}({p_i}) \subset {U_{r({p_i})}}({p_i}) (that is, to shrink the open ball), such that {${U_i}$} is a locally finite open covering of $M$?
The shrinkage of open covering
1 Answers
Yes. See Michael Spivak's "A Comprehensive Introduction to Differential Geometry" Volume I, Chapter 2, Theorems 13 and 14. There a proof is given for a general cover, you'll have to find the transition to the case of balls (in case you really need it) yourself.
(the countability is a consequence of the second countability axiom in your assumptions)
Edit: as Georges Elencwajg has pointed out a more detailed answer seems to be desired, which shows that the refining cover can be taken by using only (countably many of the) balls which are obtained by shrinking the balls we started with.
Denote the original cover by $\cal U$
Take a refining locally finite cover $\cal O$ as guaranteed by the theorems cited. Since we get it as a refinement of the original cover we may assume the closure of each $O\in \cal O$ is compact. For each $p \in M$ choose $O(p) \in \cal O$ such that $p\in O(p)$. Also chose $U_r(p)\in \cal U$ and a radius $s$ such that $p\in U_s(p) \subset U_r(p)\cap O(p)$. Denote by ${\cal V} $ the cover consisting of these $U_s(p)$
Clearly {\cal O}^' := \{O(p) : p\in M\} is a subcover of $\cal O$.
It is also clear that
$E(p) := \overline{O(p)} \subset \cup_{q\in M} U_s(q)$
where the $U_s(q)$ are taken from $\cal V$. As $E(p)$ is compact, $E(p) \subset \cup\{ U_{k} : k \in J(O(p))\}$ where the $U_k$ are still taken from $\cal V$, but with a finite index set $J(O(p))$. We may also assume that for $k\in J((O(p))$, $U_k \cap O(p)\neq \emptyset$
Note the index set is chosen to depend on $O(p)$, not on $p$ - we may well have $O(p) = O(q)$ for $p\neq q$. Now let $ {\cal W} = \{U_k \in {\cal V}: k\in J(O(p))\quad \mbox{for some } p\in M\}$
Claim: this is a countable locally finite open subcover of $\cal V$. It is obviously a subcover, it remains to show it's locally finite and countable. Since {\cal O}^' is countable and each $O(p)\in \cal O$ contributes at most finitely many $U_k$, $\cal W$ is countable.
Now choose $p\in M$ and let ${\cal W}(p) := \{U\in {\cal W}: p \in U\}$. For each $U\in {\cal W}(p)$, $U\subset O(q)$ for some $q$ by construction, so $p\in O(q)\subset \overline{O(q)}$. Since {\cal O}^' is locally finite and $\overline{O(q)}$ is compact, there are only finitely many $O(z)$ with $O(z)\cap \overline{O(q)} \neq \emptyset$, hence also at most finitely many $O(z)$ such that $O(z)\cap U \neq \emptyset$. By construction of $\cal W$, $U$ is from one of the finite open covers of some of these $O(z)$, hence there are only finitely many $U\in {\cal W}(p)$
-
0@Adterram Out of curiosity: since you seem to think the proof is correct, and no one else provided an answer, would you mind considering to tag the answer accordingly? – 2011-12-20