Some doubts about the first part of the proof of Lemma 1.10 in Introduction to smooth manifolds by John Lee, 2nd edition. I have some doubts about the first part of the proof of Lemma 1.10 in Introduction to smooth manifolds by John Lee, 2nd edition.
Lemma 1.10. Every topological manifold has a countable basis of precompact coordinate balls.
Proof. Let $M$ be a topological $n$-manifold. First we consider the special case in which $M$ can be covered by a single chart. Suppose $\varphi: M \to \hat U \subseteq \mathbb{R}^n$ is a global coordinate map, and let $\mathscr{B}$ be the collection of all open balls $B_r(x)\subseteq  \mathbb{R}^n$ such that $r$ is rational, $x$ has rational coordinates, and $B_{r'}(x)\subseteq  \hat U$ for some $r' > r$. Each such ball is precompact in $\hat U$, and it is easy to check that $\mathscr{B}$ is a countable basis for the
topology of $\hat U$. Because $\varphi$ is a homeomorphism, it follows that the collection of sets of the form $ \varphi^{-1}(B)$ for$ B \in \mathscr{B}$ is a countable basis for the topology of $M$; consisting of precompact coordinate balls, with the restrictions of $\varphi$ as coordinate maps.
My questions are
1) Why do we need to consider another ball $"B_{r'}(x)\subseteq  \hat U$ for some $r' > r$"?Is the $x$ in $B_r(x) $ the same as the $x$ in   $B_{r'}(x)$ ?   It makes more sense to me that $r' < r$, so that there is enough room for the closure of $B_{r}$ in $U$ to stay inside $\hat U$, because if I close a set in $\hat U$, the closure cannot get out of $\hat U$, can it? Can someone clear this up?
2)"... Each such ball is precompact in $\hat U$". My proof of this is: The closure of an open ball in $\hat U$ is obviously closed in $\hat U$  and since an open ball in $\mathbb{R}^n$ is bounded and in  $\mathbb{R}^n$ :closed + bounded equals compact, we have that the closure of $B_{r}(x)$ in $\hat U$  is compact, therefore $B_{r}(x)$ is precompact.  Is this correct?
3) How do I check that $\mathscr{B}$ is a countable basis for the topology of $\hat U$? My try: The set $\mathscr{B}$  is a countable basis for $\hat U$ if it is made of open subsets  of $\hat U$ and  any open set of $\hat U$ is a countable union of elements of $\mathscr{B}$. So because of the first requisite open subsets of $\hat U$ should have the form : $\hat U \cap U$ with U some open set of $\mathbb{R^n}$, right? And how do I prove the second requisite?
thank you!
 A: In the order:
1 and 2: If $r<r'$, then $B_r(x)\subset B_{r'}(x)$ (draw a picture).
In turns out that in fact we also have $\bar B_r(x) \subset B_{r'}(x)$.
Since a closed ball in a finite dimensional vector space is compact, if $r'$ is chosen such that $B_{r'}(x) \subset \hat U$, then $B_r(x)$ is precompact in $\hat U$: its closure in $\hat U$ is precisely $\bar B_r(x)$, which is compact.
As mentioned my @Malkoun in the comment section, putting $B_{r'}(x)$ between $B_r(x)$ and $\hat U$ prevents you from being too close to the boundary of $\hat U$.
If $A\subset \hat U$, the closure of $A$ in $\hat U$ is not the usual closure $\bar A$ you know in $\Bbb R^n$, but $\bar A \cap \hat U$.
Staying away from the boundary ensures that these two notions of closure are the same.
3: The balls (all the balls) in $\hat U$ form a basis of the topology (that is, any open subset of $\hat U$ is a union of balls).
To show that the rational balls (whose centre has rational coordinates as well as the radius) form a basis of the topology of $\hat U$, it suffices to show that any ball $B_R(z)\subset \hat U$ is the union of such rational balls.
As a consequence of the density of $\Bbb Q$ in $\Bbb R$, it holds that
$$B_R(z) = \bigcup_{(x,r) \text{ rationals such that } B_r(x)\subset \hat B_R(z)}B_r(x).$$
(Approach the coordinates of $z$ by some close rationals, etc.)
To conclude, just invoke the fact that the map defined on the set of rational balls in $\hat U$ by $B_r(x) \mapsto (x,r)\in \Bbb Q^{n+1}$ is injective, and thus there are only countably many rational balls in your basis of the topology.
