Proposition 1.16 in Lee's Introduction to Smooth Manifolds

Question

I'm having some trouble with one of the statements in the proof for Prop. 1.16 in Lee's Introduction to Smooth Manifolds. A countable collection $\mathcal{B}$ of coordinate balls covering a manifold $M$ is given, and Lee asserts:

"For any pair of coordinate balls $B,B' \in \mathcal{B}$, the intersection $B \cap B'$ has at most countably many components, each of which is path-connected."

I'm trying to figure out why this is true. My thoughts for the "countably many components" part is as follows: If there were uncountably many components, then we could find a basis element in each of them, and since the components are disjoint, these basis elements must be distinct. This would then violate second countability. The hole here is that we can only find these basis elements if the components are open in $M$, and I don't know if this is true. Similarly I don't see why the components are path connected.

The fact that $M$, being a manifold, is locally path connected means that the components of $M$ are open and path connected, but I don't know if this is true of the components of $B \cap B'$. If open subspaces inherit local path connectedness then I think all this would be resolved, but is that the case?

Answer 1

It all follows from standard general topological facts. I need that the coordinate balls are open, separable and locally (path-)connected:

The coordinate balls are locally connected, and this means (by a standard theorem) that all components of open subsets are open, so this applies to $B \cap B'$ (which is open in $B$). And open subsets of locally path-connected subsets are still locally path-connected (and so this holds of for all components of $B \cap B'$ in particular).

Also, another standard theorem is that a connected, locally path-connected space is path-connected, so all these (open) components of $B \cap B'$ are path-connected as well.

That there are only at most countably many of such components follows from the fact that in a separable space there can be at most countably many disjoint open sets (each has to contain a distinct member from a counatble dense set) (in general, a separable space is ccc) and $B$ (and $B \cap B'$) is certainly separable.

Answer 2

the question on countability of components of $B \cap B'$ you have solved by observation that the balls form a countable basis, let show why each component is path connected: Let define on points of $B \cap B'$ an equiv relation: $x \sim y$ iff therer exist a finite chain of balls $B_1 , B_2 ,..., B_n$ such that $x \in B_1, y \in B_n, B_i \cap B_{i+1} $ not empty. Obviously by construction you can connect all point from same equivalence class by a path , prove that it's an equivalence relation which separates $B \cap B'$ in disjoint open pieces. But $B \cap B'$ is connected, thus there not exist two or more equivalence classes und thus all points are connectable by a path.

Answer 3

It is trivial that local path-connectedness is inherited by open subspaces. Suppose $M$ is locally path-connected, $U\subseteq M$ is open, $x\in U$, and $V\subseteq U$ is a neighborhood of $x$ in $U$. Since $U$ is open in $M$, $V$ is also a neighborhood of $x$ in $M$, so since $M$ is locally path-connected there is a path-connected open neighborhood $W\subset V$ of $x$ in $M$. Then $W$ is also a path-connected open neighborhood of $x$ in $U$.