Interior and boundary points of $n$-manifold with boundary I'm reading Lee, 'Introduction to Topological Manifolds', 2011. After he introduces $n$-manifold with a boundary 

An $n$-manifold with a boundary is a second countable Hausdorff space in which any point has a neighborhood which is homeomorphic either to an open subset of $\mathbb R^n$ or to an open subset of $\mathbb H^n = \{x\in \mathbb R^n:x_n\geq 0\}$ endowed with a Euclidean topology.

The interior chart is then $(U,\phi)$ where $U$ is open in $M$ and $\phi:U\to V\subseteq \mathbb R^n$ is a homeomorphism (here $V$ is open in $\mathbb R^n$) and boundary chart as $(U',\phi')$ where $U'$ is open in $M$ and $\phi':U'\to V'\subseteq \mathbb H^n$ is a homeomorphism (here $V$ is open in $\mathbb H^n$ and $V'\cap \partial \mathbb H^n\neq\emptyset$).
Then he writes:

If M is an $n$-manifold with boundary, a point $p$ in $M$ is called an interior point
  of $M$ if it is in the domain of an interior chart; and it is called a boundary point of
  $M$ if it is in the domain of a boundary chart that takes $p$ to $\partial \mathbb H^n$. 

The natural question is - can be $p$ both interior and boundary point of $M$? In 2011 version of the book Lee writes:

Every point of M is either an interior
  point or a boundary point: if $p \in  M$ is in the domain of an interior chart, then it is
  an interior point; on the other hand, if it is in the domain of a boundary chart, then it
  is an interior point if its image lies in $\operatorname{Int}\mathbb H^n$, and a boundary point if the image lies
  in $\partial \mathbb H^n$.

I cannot understand his argumentation. Moreover, in the very same book on the next page he writes explicitly that it is not possible to prove that $p$ is either interior or boundary at that stage. (Theorem 2.59). I have two question: 


*

*Do I miss something, or he states two contradictory facts?

*I tried to prove that any point is either boundary interior or boundary and come to the fact that any subset $V'$ open in $\mathbb H^n$ such that $V'\cap \partial \mathbb H^n\neq\emptyset$ is not homeomorphic to any open subset of $\mathbb R^n$ and couldn't prove that fact. Can you help me with that proof?
 A: 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)
A: The invariance of the boundary can be seen easily from the Local Inversion Theorem  of Calculus. Indeed, it is enough to see that a diffeomorphism $h$ (a change of charts) of a halfspace $\mathbb H^n$ onto another $\mathbb H^n$ maps $\{x_n\ge0\}$ into $\{x_n\ge0\}$. But suppose, say, $h(0)=(0,\dots,0,a)$ with $a>0$. Then the inverse $h^{-1}$ restricts to a differentiable mapping $W=\{x_n>0\}\to\mathbb R^n$ with non zero jacobian det. Thus, by local inversion, $h^{-1}$ maps some open ball $B\subset W$ around $h(0)$ onto an open nbhd $V$ of $a$ in $\mathbb R^n$. Such a $V$ cannot be included in $\mathbb H^n$, and we are done.
Just for the record, this uses local inversion, but the result is in fact more truly elementary: it can be deduced by a little jiggling with the notion of derivative. Although this is not easy to find in the literature...
