Upper bound on the number of charts needed to cover a topological manifold If $M^n$ is a compact topological manifold (not necessarily with additional structure), is there an upper bound on the number of charts needed to cover $M$ ? Does this bound depend on the dimension of $M$ ?
Thanks in advance...
Cheers
 A: This is an old question, but maybe somebody still cares. 
First, let us define what a "chart" means in the context of topological manifolds $M^n$:
Definition. A chart on $M$ is an open subset $U\subset R^n$ and a topological embedding $f: U\to M^n$. We say that $M$ is covered by charts $(U_i, f_i), i\in J$, if
$$
M=\bigcup_{i\in J} f_i(U_i).
$$
Note that in this definition I do not require the open sets $U$ to be connected, which is important. I also require $M$ to be Hausdorff and 2nd countable. 
Then:
Theorem. Every topological $n$-dimensional manifold $M$ (compact or not) admits a cover by $n+1$ charts. 
Proof. Note that $M$ has topological dimension $n$ and is a normal topological space (since $M$ embeds in some $R^N$ and, hence, is metrizable). Let ${\mathcal W}$ be an open cover of $M$ by subsets homeomorphic to open balls in $R^n$. Thus, by Ostrand's theorem on colored dimension, there exists a set $\{{\mathcal V}_i: i=0,...,n\}$ such that:


*

*Elements of each ${\mathcal V}_i$ are certain pairwise disjoint open subsets of $M$. 

*The union
$$
{\mathcal V}=\bigcup_{i=0}^{n} {\mathcal V}_i
$$
is an open cover of $M$. 

*Each element of ${\mathcal V}$ is contained in an element of ${\mathcal W}$. 
By 2nd countability property of $M$, each  ${\mathcal V}_i$ is at most countable and each of its elements is homeomorphic to an open subset of $R^n$ (part 3). Therefore, for each $i$, the (disjoint!) union
$$
T_i=\bigcup_{V\in {\mathcal V}_i} V 
$$
admits a topological embedding $g_i: T_i\to U_i=g_i(T_i)\subset R^n$. Its inverse $f_i$ is a chart on $M$. Since ${\mathcal V}$ is a cover of $M$, it follows that we obtained a cover of $M$ by $n+1$ charts $(U_i, f_i), i=0,...,n$. qed   
