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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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    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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    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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    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