Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A subset $X$ of $\mathbb{R}^3$ is called compact if it is closed (i.e., the set of points in $\mathbb{R}^3$ that are not in $X$ is open) and bounded (i.e., $X$ is contained in some open ball). If $S$ is a surface, then the surface is compact iff $S$ is compact.

Why is the Möbius strip not a compact surface? Can you give me another example of a limited non-compact surface with boundary?

share|improve this question
2  
If the Möbius strip is not compact, it must be because the edge is not included in the set we're considering. With the boundary, any reasonable embedding of the Möbius strip in $\mathbb R^3$ should be compact. –  Henning Makholm Oct 27 '11 at 18:21
    
@HenningMakholm is correct, the Möbius strip is compact, unless you are not including the boundary, in which case, the problem with it is that it isn't closed. –  Thomas Andrews Oct 27 '11 at 18:25
6  
Perhaps it doesn't allow something that includes a boundary to be considered a "surface" in the first place? –  Henning Makholm Oct 27 '11 at 18:43
1  
@Lmn, I'm not a good person to ask about that. I never learned differential geometry formally, so I generally just extrapolate the theory from Wikipedia as I go along. The definition you quote clearly prevents the surface from including a boundary. –  Henning Makholm Oct 28 '11 at 15:41
1  
@Lmn6: Note that that is not possible if $S$ is the Mobius strip with boundary and $p$ is a point on the boundary; there is no open set $W\subseteq\mathbb{R}^3$ containing $p$ such that $S\cap W$ is homeomorphic to an open set $U\subseteq\mathbb{R}^2$. –  Zev Chonoles Oct 28 '11 at 15:41

2 Answers 2

up vote 2 down vote accepted

One can think of a Möbius strip independently of any particular embedding inside $\mathbb{R}^3$, IE just as a two dimensional manifold. One definition is the set $\mathbb{R} \times [0,1]$ with $(x,0)$ glued to $(-x,1)$ for each $x \in \mathbb{R}$. A bunch of copies of $\mathbb{R}$, one for each element of $[0,1]$, glued to together at the ends with a reversing twist, if you will. Let $M$ be the Möbius strip defined this way. The first nice property of this definition is that every point is the same in the sense that, 'locally' everything looks like $\mathbb{R}^2$.

In the construction above you can use an open interval $(-1,1)$ instead of $\mathbb{R}$. But if you use a closed interval $[-1,1]$ as an alternate definition you don't really have a Möbius strip, but a related object, a 'manifold with boundary'. These aren't quite so nice: nearby every point it may look like $\mathbb{R}^2$, or for boundary points it may look like the closed upper half plane. Manifolds with boundary are not as easy to work with.

The Möbius strip is a good first example of a (non-trivial) vector bundle. Another example is the cylinder, which is what you get from same construction above, but without the twist. While the cylinder is a vector bundle, it is a trivial one: it's essentially just the Cartesian product of a circle with a vector space (the real number line).

For a better info on vector bundles see: http://en.wikipedia.org/wiki/Vector_bundle.

share|improve this answer
2  
How do the smooth functions $M\to\mathbb R$ fail to form a vector space? –  Henning Makholm Oct 27 '11 at 19:01
    
Also, in a vector bundle, the vector spaces don't overlap at all - they partition the total space. –  Jason DeVito Oct 27 '11 at 19:28
    
Yes, of course HM and JD, thanks. –  Skolem Oct 28 '11 at 3:10
2  
Even with the edit, I still have to mildly object. The word "trivial" for vector bundle almost never means "just a vector space, taken as a whole" because the base space almost never has the structure of a vector space in a compatible fashion. If you want "trivial" to mean this, then an easier example of a "nontrivial" bundle is the trivial vector bundle over 2 points. The Mobius band is an example of a nontrivial bundle in the sense that it's not bundle isomorphic to a product $\mathbb{R}\times S^1$ even though locally it is. –  Jason DeVito Oct 28 '11 at 14:52
2  
My objection is unchanged. The answer still claims that the smooth functions from $M$ to $\mathbb R$ do not form a vector space, which is just wrong. The pointwise sum of two such functions is itself a smooth function; the scaling of a smooth function by any real number is also a smooth function. That makes them a vector space. –  Henning Makholm Oct 28 '11 at 18:44

The Möbius strip without boundary, i.e. without its edge, is not compact. Similarly, the cylinder $$\mathbb{S}^1\times(0,1)=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2=1,\;\;0<z<1\}$$ is a bounded non-compact surface. This is because, without their boundaries, these subsets of $\mathbb{R}^3$ are not closed. With their boundaries, both of these surfaces are closed, and hence compact, e.g. $$\mathbb{S}^1\times[0,1]=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2=1,\;\;0\leq z\leq 1\}$$ is compact.

By the way, it is a theorem (the Heine-Borel theorem) that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded; where compact has this more general meaning.

share|improve this answer
    
Can I ask you what is the exact definition of surface you are using? Thanks –  user14174 Oct 28 '11 at 15:29
    
Since you are asking about $\mathbb{R}^3$, I mean a 2-dimensional embedded submanifold of $\mathbb{R}^3$. One can also define "surface" without reference to an ambient space, by simply declaring that a surface is a 2-dimensional manifold. See the wiki article for more explanation. –  Zev Chonoles Oct 28 '11 at 15:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.