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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 $M$?

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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 $U\in {\cal W}(p)$

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Dear Thomas, it is completely clear that you can find a denumerable locally finite open cover refining any given open cover of $M$: just take a refining locally finite open cover (possible by paracompactness of $M$) and extract from it a countable cover (possible by second countability of $M$, as you judiciously remark).The difficulty is that the OP wants the final denumerable cover to be composed only of balls concentric to his fixed collection of open balls . –  Georges Elencwajg Dec 18 '11 at 14:37
    
@GeorgesElencwajg: thanks for pointing that out. I extended the answer to cover that requirement. –  user20266 Dec 18 '11 at 18:07
    
It is a nice proof. Besides, I think the last step can try to prove that for every $O \in{\cal O}^'$, the cardinal of all $U\in {\cal W}$ such that $U \subset O$ is finite. –  Hezudao Dec 19 '11 at 2:24
    
@Adterram This is implied, once the result is known ($\cal W$ being locally finite), since the closeure of $O$ is compact. –  user20266 Dec 19 '11 at 7:22
    
@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? –  user20266 Dec 20 '11 at 17:50

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