Showing that every path can be well-divided?

Question

Let $\gamma: [0,1] \rightarrow S^1$ be a path. We'll say that $\gamma$ is well-divided if there are $a_1,...a_n$ such that:

1. $a_1=0$, $a_n=1$
2. $\forall_{1\leq i < n}: a_i<a_{i+1}$
3. $\forall_{i\neq j} : \gamma(a_i)\neq \gamma(a_j) $
4. $\forall_{1\leq i < n}:||\gamma(a_i) - \gamma(a_{i+1}) ||<1$

I would like to show that every non-constant path is well-divided. I'm not really sure on where to begin in here..and why isn't it correct on a constant path? is it just because of condition #3? without it, it just seems trivial, but correct.

Answer 1

(1) Divide segment [0,1] into finite number of subsegments $I_1,...,I_m$ such that diameter of $\gamma(I_i)$ less than 1.

To find such finite splitting one can simply covers compact [0,1] by open counter images of small (diam < 1) open parts of $S^1$, then transform this covering to covering of open intervals and finally take a finite subcovering consisting of intervals $J_q$ for each of them $\gamma(J_q) < 1$.

(2) For each subsegment $I_i = [u_{i-1}, u_i]$ do the following.

Let $M = \gamma(I_i)$ and $γ_0=γ(u_{i-1})$, $γ_1=γ(u_i)$ and $A \subset [0,1]$ is a finite set which contains of all previously selected points (from all segments $I_j$, j < i, look further).

Take $k-2$ arbitrary distinct points $p_2,...,p_{k-1}$ from $M \setminus (\{γ_0,γ_1\} \cup \gamma(A))$ such that $\|p_i−p_j\|<1$. For each point $p_i$ take one its counter image and order all k-2 counter images. So we get points $γ_0=a^i_1 < a^i_2 ... a^i_{k-1} < a^i_k=γ_1$ $\in$ $I_i$

Merge all such points lists we get required set $a_1,...,a_n$:

$$ \{a_1,...,a_n\} = {\bigcup}_{i=1}^m \{a^i_1,...,a^i_{k_i}\} $$

Answer 2

Fix six points $p_i$ in $S^1$ each a distance $1$ apart. Let $U_i$ be the union for points $x\in\gamma^{-1}(p_i)$ of the connected component of $I$ around $x$ mapping to points of distance strictly less than $1$ from $p_i$. Every point of $S^1$ is within distance $1$ of some $p_i$, so the $U_i$ cover $I$. Each $U_i$ is a union of open intervals, so we have an open covering of $I$ with a finite subcover by intervals $I_k$. Each of these intervals was the collection of all points around some $x_k$ mapping to within $1$ of $\gamma(x_k)$, so that the boundary points of the $I_k$ all map to an $p_j$ distinct from $\gamma(x_k)$, with the possible exception of $0$ and $1$.

Choose all these boundary points for your $a_{2i+1}$. Each $\gamma(a_{2i-1})$ is either exactly $1$ or $0$ away from $\gamma(a_{2i+1})$; if $1$ away, there's $a_{2i}$ in between within $1$ of each of them and with $\gamma a_{2i}\neq \gamma_{2i-1},\gamma_{2i+1}$. If $\gamma(a_{2i-1})=p_j=\gamma(a_{2i+1})$ we need to check $\gamma$ is nonconstant on the intervening interval: but the points in $(a_{2i-1},a_{2i+1})$ are strictly within $1$ of some $x_k,$ which can't map to $p_j$ by the definition of the $a$s. So $\gamma$ is nonconstant between our two points, and we can find an appropriate $a_{2i}$.

Note that the reasoning that $\gamma(a_i)\neq \gamma(x_k)$ relies on at least some $a_i$ actually making it all the way to $1$ away from $p_k$-so if for instance $\gamma$ was constant every interval in our original union would be $I$ and we'd get $a_0=0,a_1=1$ with $\gamma$ mapping both to the same place. So condition 3 is needed.