I think your confusion is based on interpreting "either ... or" as meaning "one or the other but not both" (exclusive or). Although this interpretation is not uncommon in everyday English, that's not how I was using the phrase in the passage you quoted. I meant "or" in the usual mathematical-logic sense of "one or the other or both" (inclusive or).
So my claim on page 43 is that "every point of $M$ is either an interior point or a boundary point," but I'm not claiming (at this stage) that a point cannot be both. The fact that it cannot be both is what Theorem 2.59 (Invariance of the Boundary) guarantees, but the tools for proving it in full generality are not developed until Chapter 13 (Homology).
The issue of exclusive vs. inclusive or is a common source of confusion. In ordinary English, "either...or" is probably most often interpreted as exclusive or, but not universally. The meaning depends on context: in "you can have either soup or salad with your entree," it's clearly exclusive, but in "you must have either a bachelor's degree or three years' experience," it's just as clearly inclusive. The same problem arises even more with "or" alone (if you delete "either" from those two sentences, it doesn't change the meaning of either one).
While there may be mathematicians who use "either A or B" when they really mean "either A or B but not both," I think most careful mathematical writers insert some phrase such as "but not both" when they really mean exclusive or, and otherwise interpret "or" and "either...or" as inclusive. That's certainly the way I write. I'm sorry it confused you -- I'll try to be more sparing with my use of "either" from now on.
John M. (Jack) Lee
(the author)
The portion where you show that $M$ is locally Euclidean is not correct and here's why:
You have to show that for every point $p \in M$ there exists a neighborhood $U \subseteq M$ of $p$ and a homeomorphism from $U$ to an open subset of either $\mathbb R^2$ or $\mathbb H^2$.
Now here's what you wrote:
- For point $p \in M$ take set $U$ open in $\mathbb H^2$ with $p \in U$
So your neighborhood $U$ is in $\mathbb H^2$ and not in $M$! While it is indeed true that $M \subseteq \mathbb H^2$ we cannot simply say that a neighborhood in $\mathbb H^2$ is a neighborhood in $M$. If $p$ is the point $p = (1, 1)$ then any neighborhood of $p$ in $\mathbb H^2$ will contain points that are not in $M$.
Edit:
To fix the proof remember that if $U \subseteq M$ contains part of the boundry of $M$ then the chart for $U$ will have to map that boundry to the boundry of $\mathbb H^2$. So for example if $p = (1, 1)$ then let $U = (.4, 1] \times (.4, 1]$ and for a chart try $\phi\colon U \to \mathbb H^2$ defined by $\phi(x, y) = (1 - x, 1 - y)$. I leave it to you to check that the image of $\phi$ is open in $\mathbb H^2$ and that $\phi$ is a homeomorphism onto this image. You'll also need to come up with charts covering the rest of $M$. I suggest you take
- $U_2 = [0, .6) \times (.4, 1]$
- $U_3 = (.4, 1] \times [0, .6)$
- $U_4 = [0, .6) \times [0, .6)$
as the neighborhoods for those charts.
Edit #2:
Whoops! My edit above has a mistake as pointed out by goobie. Also pointed out by goobie is the fix: Instead of the chart $\phi$ that I suggested take $\phi\colon U \to \mathbb H^2$ defined by $\phi(z) = z^2$ (here I'm using complex numbers to denote points in the plane). Then you'll just need to do some translation and rotation to handle the other corners.
Best Answer
The subset $[0,1[ \cup \{2\}$ of the real line is a manifold with boundary having two connected components of different dimensions. The component $[0,1[$ is a 1-dimensional manifold with boundary, and the single point {2} is a 0-dimensional manifold.