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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?

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    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.2011-10-27
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    @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.2011-10-27
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    @Henning: Pressley's book "Elementary Differential Geometry" (2ed) says (at page 111, Corollary 5.4.5) that "Every compact surface is orientable"... so the Möbius strip should be not compact. I really don't understand the definition of compact surface cited by the book.2011-10-27
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    Perhaps it doesn't allow something that includes a boundary to be considered a "surface" in the first place?2011-10-27
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    @Henning: yes, I started to believe that it is just as you say. Do you mind if I ask you to write down the definition of surface you get used to? Pressley's is this: "a subset $S$ of $\mathbb{R}^3$ is a surface if, for every point $p \in S$, there is an open set $U$ in $\mathbb{R}^2$ and an open set $W$ in $\mathbb{R}^3$ containing $p$ such that $S \cap W$ is homeomorphic to $U$."2011-10-28
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    @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.2011-10-28
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    @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 $U\subseteq\mathbb{R}^2$.2011-10-28
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    Ok, the last two comments (by Henning and Zev) are a lot interesting because now I think I probably understood what is the meaning of "compact surface" in Pressley's book. I believe it's http://en.wikipedia.org/wiki/Surface#Closed_surfaces2011-10-29

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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.

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    How do the smooth functions $M\to\mathbb R$ fail to form a vector space?2011-10-27
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    Also, in a vector bundle, the vector spaces don't overlap at all - they partition the total space.2011-10-27
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    Yes, of course HM and JD, thanks.2011-10-28
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    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.2011-10-28
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    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.2011-10-28
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    I think I finally fixed it... now that it's just archival. :)2011-12-29
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The Möbius strip without boundary, i.e. without its edge, is not compact. Similarly, the cylinder $$\mathbb{S}^1\times(0,1)=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2=1,\;\;0 is a bounded non-compact surface. This is because, without their boundaries, these subsets of $\mathbb{R}^3$ are not closed. With their boundaries, both of these surfaces are closed, and hence compact, e.g. $$\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.

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    Can I ask you what is the exact definition of surface you are using? Thanks2011-10-28
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    Since you are asking about $\mathbb{R}^3$, I mean a 2-dimensional [embedded submanifold](http://en.wikipedia.org/wiki/Submanifold) of $\mathbb{R}^3$. One can also define "surface" without reference to an [ambient space](http://en.wikipedia.org/wiki/Ambient_space), by simply declaring that a surface is a 2-dimensional [manifold](http://en.wikipedia.org/wiki/Manifold). See [the wiki article](http://en.wikipedia.org/wiki/Surface) for more explanation.2011-10-28