Mobius strip-an intuition

Question

I do not follow how in the snippet below in the example 7.3 the space we are taking the quotient of is the whole of $[0,1]\times\mathbb{R}$? I mean, the diameter of the Mobius strip is bounded, but $\mathbb{R}$ is not.

Answer 1

This is $[0,1]\times (0,1)$ and if you identify $(0,t)$ with $(1,1-t)$ you will obtain the open strip Mobius because points on the dashed lines are not in the square.

Answer 2

Both the Möbius strip and the vector bundle are defined as topological spaces, not as metric spaces, so the notion of "bounded" doesn't apply to them. The real line is, as you said, unbounded, provided you use the usual metric $|x-y|$; the same topological space admits other metrics, for example $$ \frac{|x-y|}{1+|x-y|}, $$ that are not bounded.