Number of components of intersection of coordinate balls.

Question

I encountered the following statement as part of the proof of Theorem 7.21 in Lee's Topological Manifolds textbook.

Statement: Let $M$ be a manifold, and let $\boldsymbol{U}$ be a countable cover of $M$ by coordinate balls. For each $U,U'\in\boldsymbol{U}$ the intersection $U\cap U'$ has at most countably many components...

The wording suggests that the number of components of $U\cap U'$ could greater than 1 or even countably infinite but why isn't $U\cap U'$ the only connected component of itself? My reasoning is that if $U$ and $U'$ are coordinate balls then they are homeomorphic to open balls in $\mathbb{R}^n$. The intersection $U\cap U'$ is then homeomorphic to its image which is the intersection of two open balls in $\mathbb{R}^n$ and hence connected.

Answer 1

Consider $M=S^1\subset\mathbb{R}^2$ and $U=M\setminus\{(0,1)\}$ and $U'=M\setminus\{(0,-1)\}$. Then $U$ and $U'$ are homeomorphic to $\mathbb{R}$, but $U\cap U'$ is disconnected: its two components are the right and left semicircles of $M$ (draw a picture if you don't see this!).

In higher dimensions, you can have even more complicated things happen. For instance, you can have two open sets $U,U'\subseteq\mathbb{R}^2$ which are both homeomorphic to open balls but such that $U\cap U'$ has infinitely many connected components. For instance, let $$U=(0,1)\times(0,2)\setminus \{1/2,1/3,1/4,\dots\}\times[1,2)$$ and $$U'=(0,1)\times(1,2).$$ Then $U$ and $U'$ are both homeomorphic to $\mathbb{R}^2$ (this takes some work to prove for $U$), but $U\cap U'$ has infinitely many connected components, namely $(1/2,1)\times(1,2)$, $(1/3,1/2)\times(1,2)$, $(1/4,1/3)\times(1,2)$, and so on. Again, I recommend drawing a picture.

What's going on is that $U$ and $U'$ are both individually homeomorphic to balls, but you can't find a single map defined on $U\cup U'$ that simultaneously is a homeomorphism from $U$ to a ball and a homeomorphism from $U'$ to a ball. As a result, you can't simultaneously think of $U$ and $U'$ as balls in $\mathbb{R}^n$ for the purposes of computing their intersection. Even if you can embed $U\cup U'$ in $\mathbb{R}^n$, the images of $U$ and $U'$ under such an embedding may not be balls (they are just sets homeomorphic to balls), and they may not look much like balls (for instance, they don't have to be convex). So their intersection could be quite complicated.