The boundary of a closed Mobius band

Question

The Mobius band is defined by the quotient: $r:I\times I\rightarrow M$, with the equivalence relation $(0,x)\sim (1,1-x)$ for all $x\in I$.

The boundary of Mobius band $M$ is defined as the set of points that have an open neighbourhood which is homeomorphic to the closed half space.

I know its boundary is $r(B)$ where $B=\{(x,y)\in I\times I: y\in\{0,1\}, x\in I\}$. We need to find an open neighbourhood for each point in $r(B)$ s.t. homeomorphic to the closed half space $H=\{(x,y)\in\mathbb{R}^2:y\geq 0\}$.

It is easy to prove $r((0,1)\times (0,1))$ is not in the boundary $\partial M$, and $r((0,1)\times\{0,1\})\subset\partial M$.

But how to prove $r(\{0,1\}\times (0,1))$ is not in the boundary, and $r(0,0)$ $r(1,0)$ are in the boundary?

Answer 1

But how to prove $r(\{0,1\}\times (0,1))$ is not in the boundary

Consider $[x,y]\in r(\{0,1\}\times(0,1))$. Note that $[x,y]=\{(0,t),(1,1-t)\}$ for some $t\in(0,1)$, as an equivalence class. Pick $0<r<\min(t,1-t)$ and consider two half open disc:

$$D_1=\big\{v\in I^2\ \big|\ \lVert v-(0,t)\rVert < r\big\}$$ $$D_2=\big\{v\in I^2\ \big|\ \lVert v-(1,1-t)\rVert < r\big\}$$

These are two disjoint and open subsets of $I^2$. They are not standard disks however we can define a continuous function

$$h:D_1\cup D_2\to \mathbb{R}^2$$ $$h(x,y)=\begin{cases} (x,y) &\text{if }x\leq 1/2 \\ (x-1, y-1) &\text{otherwise} \end{cases}$$

and note that the image of $h$ is precisely the open ball $B$ in $\mathbb{R}^2$ around $(0,t)$ of radius $r$. Of course $h$ is not a homeomorphism, because it is not injective, however it induces an injective map $H:(D_1\cup D_2)/\sim\to B$ on the quotient space, and thus $H$ is a homeomorphism. Note that $(D_1\cup D_2)/\sim$ is an open neighbourhood of $[x,y]$.

In particular $[x,y]$ has an open neighbourhood homeomorphic to $\mathbb{R}^2$ and thus it cannot be a boundary point.

and $r(0,0)$ $r(1,0)$ are in the boundary?

For that you can utilize the fact that if $v\in M$ is such that any open neighbourhood $U$ of $v$ satisfies $U\cap\partial M\neq \emptyset$ then $v\in\partial M$. This is a simple consequence of the fact that $\partial M$ is a closed subset of $M$. Which follows (as earlier) from the observation that $x\in M\backslash\partial M$ if and only if $x$ has an open neighbourhood homeomorphic to $\mathbb{R}^n$.

Answer 2

You gave a description of the Möbius band as $$ M = (I\times I)/{\sim}_1 \quad\text{where}\quad (0,y)\sim_1(1,1-y) \text{ for $t\in I$.} $$ To allow for less messy operations on points of $M$, you can also describe this as $$ M' = (\mathbb R\times I)/{\sim}_2 \quad\text{where}\quad (x,y)\sim_2(x+1,1-y) \text{ for $x\in\mathbb R$ and $y\in I$.} $$ Check that the inclusion $I\times I\hookrightarrow \mathbb R\times I$ induces a homeomorphism $M\to M'$.

The benefit of this description is that it doesn't treat the points $(x,y)$ with $x\in\{0,1\}$ any different from those with $x\in (0,1)$. In particular, we get an automorphism $M'\to M'$ given by $[(x,y)]_2\mapsto[(x+\tfrac 1 2, y)]_2$.

Since you have shown that $r_1((0,1)\times (0,1))$ does not contain boundary points of $M$, the above automorphism yields that $r_2(\mathbb R\times (0,1))$ does not contain any boundary points of $M'$ and hence $r_1(\{0,1\}\times(0,1))$ doesn't either.

You can think of the automorphism $M'\to M'$ in terms of $M$ as gluing $I\times I$ along $\{0,1\}\times I$ as you did, then cutting it open along $\{\tfrac 1 2\}\times I$ to obtain a new description as a quotient of $I\times I$, where points with $x\in\{0,1\}$ in the old quotient are now points with $x=\frac 1 2$ in the new one.