Showing arclength is supremum of all polygonal approximations Show that the arclength of a rectifiable parametrized path is the supremum of all polygonal approximations of the arclength.
A parametrized path $\gamma:[a,b] \rightarrow \mathbb{R}^n$ is rectifiable if there is a number $l$ such that for each $\epsilon>0$, there is a $\delta>0$ such that if $P=\{x_0,...,x_m\}$ of $[a,b]$ with $||\mathcal{P}||<\delta$ then $$|\sum\limits_{i=1}^{m}||\gamma(x_i)-\gamma(x_{i-1})||-l| <\epsilon$$.
I had no trouble showing that $l-r$ is not an upper bound for any $r>0$ since this result is basically built into the definition. However I am having a difficulty showing $l$ is an upper bound of the set of all polygonal approximations of arclength. I tried to show that given any partition $P$ and refinement $P'$ of $P$, that the polygonal approximation with respect to the partition $P'$ is greater than $P$. However, I could not figure out how to use that to show that $l$ is an upper bound for all polygonal approximations. How would I proceed to show that?
I also tried to show that if $P$, $P'$ are any two partitions of $[a,b]$ with $||P'||<||P||$, then the polygonal approximation for the partition with respect to $P'$ should be greater than for $P$. Then since $l=\lim\limits_{||P|| \rightarrow 0}\sum\limits_{i=1}^m||\gamma(x_i)-\gamma(x_{i-1})||$ it should be that $l$ is an upper bound, but I do not know. I also saw a solution that said the answer is obvious by geometry, which intuitively seems right, but I did not know how to show this using geometry.
 A: Note that any refinement of a polygonal approximation must have at least as long a path length. This follows from the triangle inequality. For if we have a polygonal approximation $P = \{x_0, ..., x_m\}$, and we replace $x_i, x_{i + 1}$ with $x_i, y, x_{i + 1}$, then we replace the length $||\gamma(x_i) - \gamma(x_{i + 1})||$ with the length $||\gamma(x_i) - \gamma(y)|| + ||\gamma(y) - \gamma(x_{i + 1})||$, and we know that $||\gamma(x_i) - \gamma(x_{i + 1})|| = ||\gamma(x_i) - \gamma(y) + \gamma(y) - \gamma(x_{i + 1})|| \leq ||\gamma(x_i) - \gamma(y)|| + ||\gamma(y) - \gamma(x_{i + 1})||$ by the triangle inequality.
Now consider some polygonal approximation $P = \{x_0, ..., x_m\}$. We wish to show that $\sum\limits_{i = 0}^{m - 1} ||\gamma(x_i) - \gamma(x_{i + 1})|| \leq l$. Suppose the opposite: that is, that $\sum\limits_{i = 0}^{m - 1} ||\gamma(x_i) - \gamma(x_{i + 1})|| > l$. Then let $\epsilon = \sum\limits_{i = 0}^{m - 1} ||\gamma(x_i) - \gamma(x_{i + 1})|| - l$. Take $\delta$ such that for all $||Q|| = ||\{y_0, ..., y_n\}|| < \delta$, $||\sum\limits_{i = 0}^{n - 1} ||\gamma(y_i) - \gamma(y_{i + 1})|| - l|| < \epsilon$. Then take a refinement $Q = \{y_0, ..., y_m\}$ of $P$ such that $||Q|| < \delta$. Then we see that $\sum\limits_{i = 0}^{n - 1} ||\gamma(y_i) - \gamma(y_{i + 1})|| \geq \sum\limits_{i = 0}^{m - 1} ||\gamma(x_i) - \gamma(x_{i + 1})|| > l$. Then $||\sum\limits_{i = 0}^{n - 1} ||\gamma(y_i) - \gamma(y_{i + 1})|| - l|| = \sum\limits_{i = 0}^{n - 1} ||\gamma(y_i) - \gamma(y_{i + 1})|| - l \geq \sum\limits_{i = 0}^{m - 1} ||\gamma(x_i) - \gamma(x_{i + 1})|| - l = \epsilon$. Contradiction.
Thus, we see that $\sum\limits_{i = 0}^{m - 1} ||\gamma(x_i) - \gamma(x_{i + 1})|| \leq l$. So $l$ is an upper bound, as required.
