Is there a manifold which requires infinitely many charts to cover it? [duplicate]

In the definition of a parametrised surface $S$, for every point in the surface, $p \in W \subseteq S$, where $W$ is open, there exists a coordinate chart or patch , $F :U\to \mathbb{R}^n$ that maps to $p$ from an open subset $U \in \mathbb{R}^n$

Is that right? If anyone knows of a more general definition, I'm willing to learn. It sounds a lot like a manifold, which I'm not entirely familiar with.

In this definition, the number of surface patches in the atlas is not stipulated. Given a parametrisable surface, is a finite number of charts sufficient to describe the surface? Can we find a surface that requires infinitely many patches to fully chart? If so, in which $\mathbb{R}^n$ does the first such surface occur? In which dimensions is it always possible to find a finite number of patches for any given surface?

EDIT: Added requirement that such a surface (manifold) be connected. One made from infinitely many disconnected subsets would have to be charted infinitely.

If $M$ is an $n$-dimensional manifold, then it is in particular a normal space of covering dimension $n$, and Ostrand's theorem tells us that that if we start with a locally finite open covering $\mathcal U$ of $M$ consisting of coordinate charts (such a thing exists), there is a refinement $\mathcal V$ of $U$ such that $\mathcal V$ is the union of $n+1$ subfamilies $\mathcal V_1$, $\dots$, $\mathcal V_{n+1}$ such that each $\mathcal V_i$ is a disjoint family.

If we allow for non-connected coordinate charts, then by looking at the union of each $\mathcal V_i$ we obtain an atlas consisting of $n+1$ charts. (If we insist on connected charts, we should be able to connect the components with thing open tubes...)

This gives Ryan's bound for $n=3$, for example.

Later. One knows that the clique number of a graph is a lower bound for the chromatic number: Ostrand's theorem says that up to refinement, equality can be achieved if the graph comes from a covering of a normal space. This is a very nice result!

I think for a compact connected surface (a 2-dimensional manifolds) you only need three charts to cover the manifold. The rough idea is like this: think of a minimal CW-decomposition of the surface. Your first chart is a regular neighbourhood of the maximal forest in the 1-skeleton. The 2nd chart will be the union of the regular neighbourhoods of the remaining 1-cells, plus little arcs that connect them all up. Your remaining chart will be the union of the interiors of the 2-cells, with little arcs connecting them together.

Similarly, I think all compact 3-manifolds can be covered by 4 or 5 charts.

You can get lower bounds on the number of charts needed for any manifold by cup product arguments -- see: http://en.wikipedia.org/wiki/Lusternik%E2%80%93Schnirelmann_category

In face, any triangulated $n$-manifold (not necessarily compact) can be covered by $n$ coordinate charts! However, this is a bit of a cheat — the elements of your atlas will be disconnected (in particular, this does not conflict with the bounds that Ryan Budney discussed coming from the Lusternik-Schnirelmann category, which come from connected covers).

I don't know a good reference for this, but here's the idea. Choose a triangulation of your manifold. For each vertex $v$ of your triangulation, choose a neighborhood $U^0_v$ such that $U^0_v$ is homeomorphic to a ball and such that $U^0_v \cap U^0_w = \emptyset$ for distinct vertices $v$ and $w$. Let $U^0$ be the union of all the $U^0_v$. It is clear that $U^0$ is homeomorphic to the disjoint union of countably many open balls in Euclidean space.

Next, for each edge $e$ of your triangulation, choose a neighborhood $U^1_e$ of $e \setminus U^0$ such that $U^1_e$ is homeomorphic to a ball and such that $U^1_e \cap U^1_f = \emptyset$ for distinct edges $e$ and $f$ (draw yourself a picture to convince yourself this is possible!). Let $U^1$ be the union of all the $U^1_e$. Again, it is clear that $U^1$ is homeomorphic to the disjoint union of countable many open balls in Euclidean space.

The pattern is now clear — repeat this procedure on the the $2$-simplices, the $3$-simplices, etc. The result is that $M = U^0 \cup U^1 \cup \cdots \cup U^n$.