I'm reading a book about geometry, and after thinking and viewing the Möbius strip, I want to know whether the book is right or not.
The book says with a little description (that I can't write here despite of my little english...) that the Möbius strip is given by
$f(u) + v g(u), \{v, -1, 1\}, \{u, 0, 2 \pi \}$
where
$f(u) = \{2 sin (u), 2 cos(u), 0\}$
$g(u) = \{0, sin(u/2), cos(u/2)\}$
(You can see that $g(u)$ is always parallel to plane ${X=0}$).
You can see the shape in this file I programmed in Mathematica
but
when I try to view in my head the Möbius strip with the description of a piece of paper or any other material... I view any similar to this
$f(u) + v g(u), \{v, -1, 1\}, \{u, 0, 2 \pi \}$
where
$f(u) = \{2 sin(u), 2 cos(u), 0\}$
$g(u) = \{sin(u/2)cos(u/2), sin(u/2)sin(u/2), cos(u/2)\}$
You can see the shape in this file I programmed in Mathematica
To view the cdf files, you need the CDF Player http://www.wolfram.com/cdf-player/
or in a easy way vieweing 
I´m interested on this question considering the strip as a figure.
Thank you very much.
When discussing topological spaces we are usually only interested in them up to homeomorphism (bijections that are continuous in both directions). "Mild" deformations can be ignored. The two subsets of $\mathbb R^3$ that you describe are homeomorphic, which is best seen by the fact that they both are homeomorphic (via the $(u,v)\mapsto f(u)+vg(u)$) to the topological space $[-1,1]\times[0,2\pi]$ where the points $(x,0)$ and $(-x,2\pi)$ are identified.