Hi, can someone please explain how the following is true:
"For any $c>0$ the relations
$$(x,y)\sim (x, \,y+ k c)\quad(k\in{\mathbb Z}), \qquad (x,y)\sim \bigl(x+\ell,\,(-1)^\ell y\bigr) \quad(\ell\in{\mathbb Z})$$
define a Klein bottle $K_c$ of "length" $1$ and "width" $c$ as a quotient of the $(x,y)$-plane , and with a rectangle $[0,1]\times[0,c]$ as fundamental domain."

I cannot see how this is the same as the definition I learned which is $(x,0)\sim (x,1)$ and $(0,y)\sim (c,1-y)$. Thank you!

Take a sequence $a_n\to 1$, such that $0 < a_n < 1$ for every $n$. There exists $x_n$ such that $d(x_n,M) \ge a_n$ for all $n$. Since $$d(x_n,M) = \inf_{m\in M} \|x_n - m\|$$
$d(x_n,M) \ge a_n$ implies $\|x_n - m\| \ge a_n$ for all $n$. $S$ is compact, so $x_n$ has a convergent subsequence $x_{n_k} \to x \in S$. $$\|x_{n_k} - m\| \ge a_{n_k} \quad \forall m\in M, \forall k\in \mathbb N$$
Taking $k\to\infty$, $$\|x - m\| \ge 1 \quad \forall m\in M$$
Taking $\inf$ over all $m\in M$, $d(x,M) \ge 1$.

You may restrict the equivalence relation to the fundamental domain, so the first space you describe is $[0,1]\times[0,c]$ with the identifications $(x,0)\sim(x,c)$ and $(0,y)\sim(1,c-y)$, whereas the second space you describe is $[0,c]\times[0,1]$ with the identifications $(x,0)\sim(x,1)$ and $(0,y)\sim(c,1-y)$. I suggest drawing pictures.
Both are rectangles with one side of length $1$ and one side of length $c$, where one pair of edges gets identified with the same orientation and the other pair of edges gets identified with the opposite orientation. However, which pair is the short one …