For any point x of the (say open) Moebius band M in R^3, there is an open 2-disk neighborhood D of x in M, and D lies in a 3-ball B in R^3 that is "cylindrical" in that it is homeomorphic to D x (-1,1), and such that D x {0} corresponds to B \int M. The two components -- or more accurately the *fact* of the two components -- D x (-1,0) and D x (0,1) of B - Dx{0} is why the Moebius band (or any surface embedded in R^3 ) is LOCALLY two-sided. But if you try to do this GLOBALLY, any embedded normal bundle of M in R^3 (aka a "tubular neighborhood"), *minus* the 0-section (the part corresponding to M itself), has only ONE component. This fact is why it's called one-sided (rigorously). --Dan P.S. Plus, if you make one, any 5-year-old can plainly see it has only one side. ARRIVISTE A STRIVER, I rwg wrote: << We like to say that a Moebius strip embedded in 3-space has one side, whereas a unit disk there has two. How can we say that? What are the "sides" of a 2-dimensional set of points? It is its own boundary.

...

_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele