# I Do these manifolds have a boundary?

1. May 3, 2017

### davidge

Any open subset of $\mathbb{R}^{n}$;
The n-Sphere, $\mathbb{S}^n$;
The Klein Bottle.

I guess they don't have a boundary, as a neighborhood of any point of them is homeomorphic to $\mathbb{R}^n$.
I'd like to know whether my guess is correct and whether the reason I'm giving for them not to have a boundary is valid.

(I actually found on web that the Klein Bottle does not have a boundary.)

Last edited: May 3, 2017
2. May 3, 2017

### lavinia

None of these manifolds have a boundary and your reason is correct.

3. May 3, 2017

Thnx

4. May 3, 2017

### meteo student

Can you explain that a bit more ? I understand what homeomorphism is from Wikipedia - https://en.wikipedia.org/wiki/Homeomorphism but why would being homeomorphic result in not having a boundary ?

5. May 3, 2017

### lavinia

What definition of the boundary of a manifold are you thinking of?

6. May 3, 2017

### davidge

I think he is looking for an answer of why manifolds without boundary have to be homeomorphic to $\mathbb{R}^n$.

7. May 3, 2017

### martinbn

They don't have to be, for example the sphere.

8. May 3, 2017

### davidge

I meant the points on the manifold have to be a neighborhood homeomorphic to $\mathbb{R}^n$. Is this not so?

9. May 3, 2017

### lavinia

yes but what do you mean by a boundary? There is a difference between a manifold and a manifold with boundary.

For instance a sphere has no boundary but a closed ball does.

10. May 3, 2017

### martinbn

May be, depends on what you mean by "the points". Do you mean all points i.e. the whole manifold? Then, no, it doesn't have to be homeomorphic to $\mathbb R^n$.

11. May 3, 2017

### davidge

Well, I mean something that is in agreement with our notion of a end point or a barrier, for familiar geometrical objects, like the Earth, a sphere, a disk. At the same time, this definition has to be correct for any kind of manifold we want to apply it. So, given
a manifold $M$ and a point $x \in M$,
an open subset $U$ of $M$ containing $x$;
Suppose we can find an homeomorphism $\Psi: U \rightarrow V \subseteq \mathbb{R}^n$. If this can happen for every $x \in M$ we say $M$ have no boundaries.

For the case when it's possible to define only a half region around $x$ where there are a homeomorphism as above, then we say $x$ is in the boundary of $M$. Thus $M$ have a boundary. All points on the boundary will get mapped to $y \in \mathbb{R}^n = (0,y_1,...,y_n)$ in the local mapping.

We see that by this way we can recover our notions of a boundary for the objects that I mentioned above (The sphere, etc...)
I've constructed a diagram showing these concepts for the case of the closed 1-ball, namely the disk. I'm not sure if that diagram is actually correct in everything, however.

I wrote some notes in my language, not in English. So I'm going to translate it

1 - Here we see the disk from far away. The disk consists of all points that lie within the red circle.

2 - Now we zoom into the border of the disk and we see something like this

3 - Here is the result of the map of several points of the border of the disk in $\mathbb{H}^2$

Of course objects have physical limitations, so the above diagram is only a representation.

12. May 3, 2017

### lavinia

For a manifold each point has to have a neighborhood that is homeomorphic to an open subset of $R^{n}$ as the OP said in the original post. For a manifold with boundary each point has to have a neighborhood that is homeomophic to an open subset of $R^{n+}$.

It is important to understand that a point can not both be an interior point and a boundary point according to this definition.

Last edited: May 3, 2017
13. May 3, 2017

### davidge

"[..] In technical language, a manifold with boundary is a space containing both interior points and boundary points. [..]", from https://en.wikipedia.org/wiki/Manifold#Manifold_with_boundary

14. May 3, 2017

### Staff: Mentor

That's not the correct way to state it. The correct way to state it is the way lavinia stated it in post #12. "Only a half region where there is a homeomorphism" doesn't really make sense.

15. May 3, 2017

### davidge

"Each point has to have a neighborhood that is homeomophic to an open subset of $\mathbb{R}^{n \ +}$ (which is the same as $\mathbb{H}^2$) as she stated, isn't the same thing as "there's some half-like-region around each point $x \in M$ that is homeomorphic to $\mathbb{H}^2$"?

16. May 3, 2017

### Staff: Mentor

No, because "half-region" implies that the neighborhood that is homeomorphic to $\mathbb{H}^2$ is "half" of something. It isn't. It's just a neighborhood that is homeomorphic to $\mathbb{H}^2$.

17. May 3, 2017

### davidge

Ok. Just to know, in my diagram above, is it correct to picture the region around the point as that circular red disk?

18. May 3, 2017

### Staff: Mentor

As the semicircular red region. Plus, the region includes the straight line boundary (since the point $x$ lies on it), but not the circular boundary (since $\mathbb{H}^2$ doesn't include that, it's open there).

19. May 3, 2017

### davidge

Ok, thanks

20. May 3, 2017

### meteo student

I think this is the subtlety I was looking for. The difference between a sphere and a closed ball.
https://en.wikipedia.org/wiki/Ball_(mathematics)
That helps !