Do these manifolds have a boundary?

[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.)

 

[lavinia]

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.)

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

 

[davidge]

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

Thnx

 

[meteo student]

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

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 ?

 

[lavinia]

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 ?

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

 

[davidge]

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

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

 

[martinbn]

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

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

 

[davidge]

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

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

 

[lavinia]

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

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.

 

[martinbn]

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

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 $\mathbb R^n$.

 

[davidge]

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 $\mathbb R^n$.

Please see below.

yes but what do you mean by a boundary?

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.

 

[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.

 

[davidge]

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

"[..] 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

 

[PeterDonis]

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 $M$.

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.

 

[davidge]

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.

"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 $\mathbb{H}^2$"?

 

[PeterDonis]

"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 $\mathbb{H}^2$"

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 $\mathbb{H}^2$.

 

[davidge]

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 $\mathbb{H}^2$.

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

 

[PeterDonis]

is it correct to picture the region around the point as that circular red disk?

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).

 

[davidge]

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).

Ok, thanks

 

[meteo student]

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.

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 !

 

[lavinia]

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 !

The sphere has no boundary but it is the boundary of the closed ball. One might ask more generally when a manifold that has no boundary is itself the boundary of another - one higher dimensional manifold. For instance a torus has no boundary but it is the boundary of a solid torus - a doughnut filled with strawberry jelly. The Klein bottle is also the boundary of a solid Klein bottle. However, the projective plane is a surface without boundary yet it is not the boundary of any manifold. There is no way to solidify it.

The question of when a compact manifold without a boundary is the boundary of a one higher dimensional manifold can be asked for any dimension.

 

[meteo student]

So in my case the earth(land + ocean + atmosphere) has no boundary am I correct ?

 

[lavinia]

So in my case the earth(solid Earth + ocean + atmosphere) has no boundary am I correct ?

The sold Earth does have a boundary. It is the surface of the Earth - approximately a sphere.

 

[meteo student]

The sold Earth does have a boundary. It is the surface of the Earth - approximately a sphere.

What if you include the atmosphere ? I mean the Earth as a sphere or an ellipsoid.

 

[WWGD]

What if you include the atmosphere ? I mean the Earth as a sphere or an ellipsoid.

Same issue: if you include the outer layer, it is the same as the Earth with its boundary.

 

[meteo student]

Same issue: if you include the outer layer, it is the same as the Earth with its boundary.

So the earth(and by that I mean the atmosphere and the land and the ocean) is mathematically from the viewpoint of manifolds an open surface am I correct ?

 

[WWGD]

So the earth(and by that I mean the atmosphere and the land and the ocean) is mathematically from the viewpoint of manifolds an open surface am I correct ?

As long as the outer layer itself , yes. It is a sphere without its peel in both cases.