# A connected manifold can be covered by finite open subsets

In Vick's $$\textit{An Introduction to Algebraic Topology}$$,

6.5 theorem states that “if M is a connected, noncompact n-manifold without boundary, then $$H_n(M) = 0$$".

Consequently, I have a question: If M is a connected n-manifold without boundary, can M be covered by finite open subsets $$\{U_1, \dots , U_k\}$$ such that every $$U_i$$ is homeomorphic to $$\mathbb{R}^n$$?

Even though I have tried many examples, but I didn't construct a counterexample. I don't know whether it is right. Here are some related results: (1)https://mathoverflow.net/questions/58988/an-example-of-a-complex-manifold-without-a-finite-open-cover

• You can always cover a (second-countable) manifold with finitely many coordinate charts, but the way I know to do this does not guarantee that these charts are connected. Commented Dec 23, 2020 at 11:50
• @Tyrone Thank you, are there some references about the way to cover the manifold? Commented Dec 23, 2020 at 12:03
• Connections, Curvature and Cohomology Vol I by Greub, Halperin and Vanstone. See $\S1.2$ (the final corollary appears on pg. 20 in the copy I have). The statement there is for topological manifolds, but the smooth case is no different. Commented Dec 23, 2020 at 12:08
• Commented Dec 23, 2020 at 18:19

Good question. The answer is positive.

First of all, as in my answer here, there exist $$n+1$$ families $${\mathcal F}_1,..., {\mathcal F}_{n+1}$$ of subsets in $$M$$ such that:

1. Each $${\mathcal F}_i$$ is a union of closed (and tame) pairwise disjoint $$n$$-balls $$B_{ij}, j\in J_i$$, where each $$J_i$$ is (at most) countable.

2. $$\bigcup_{i=1}^{n+1} \bigcup_{j\in J_i} int(B_{ij})=M.$$

3. Each compact $$K$$ in $$M$$ intersects only finitely many closed balls in the above collections.

I will show that each $$V_i= \bigcup_{j\in J_i} int(B_{ij})$$ is contained in an (open) subset $$U_i$$ of $$M$$ homeomorphic to $$R^n$$. This will do the job. To construct $$U_i$$, I will treat each $$B_{ij}$$ as a $$0$$-handle in $$M$$. I will identify each $$J_i$$ with an interval in $${\mathbb N}$$ (finite or infinite). Next, for each pair of consecutive indices $$j, j+1\in J_i$$, connect $$B_{ij}, B_{ij+1}$$ by a 1-handle $$H_{ij}$$ in $$M$$ so that distinct 1-handles are pairwise disjoint and intersect only the balls $$B_{ij}, B_{ij+1}$$ in $${\mathcal F}_i$$ and only along disks in their boundaries. A 1-handle is a thickened (tame, simple) arc $$a_{ij}$$ in $$M$$ connecting boundary spheres of $$B_{ij}, B_{ij+1}$$. Making these arcs $$a_{ij}$$ pairwise disjoint is easier if $$M$$ has dimension $$\ge 3$$ (first take any locally finite collection of tame arcs and then away any accidental intersection). Constructing these arcs in the case of surfaces is not hard but tedious, I can explain how to do so if you like. (Connectivity of $$M$$ is used to ensure the existence of an arc connecting $$B_{ij}, B_{ij+1}$$.)

Informally, the union of the 0-handles $$B_{ij}$$ and $$1$$-handles $$H_{ij}$$ is a "chain of closed balls" (finite or infinite). Arguing inductively, one sees that each finite chain $$W_i:=B_{i1}\cup H_{i1}\cup B_{i2} \cup ... \cup H_{i,k-1}\cup B_{ik}$$ is homeomorphic to the closed $$n$$-ball. Similarly, each infinite chain $$W_i:=B_{i1}\cup H_{i1}\cup B_{i2} \cup ... \cup H_{i,k-1}\cup B_{ik} \cup ...$$ is homeomorphic to the closed $$n$$-dimensional half-space. The interior $$U_i$$ of each chain is homeomorphic to $${\mathbb R}^{n+1}$$.

Thus, $$M= U_1\cup ...\cup U_{n+1}.$$

A similar argument works in PL and smooth categories.