5

I'm studying the proof of the Theorem 2.4.16 (page 42) of this textbook (A Course in Metric Geometry by D. Burago, Y. Burago and S. Ivanov); I quote the statement:

Theorem 2.4.16. Let $ (X,d) $ a complete metric space.

  1. If for every $x,y \in X$ there exist a midpoint, then $d$ is strictly intrinsic.
  2. If for every $x,y \in X$ and every positive $\epsilon$ there exists an $\epsilon$-midpoint, then $d$ is intrinsic.

I remind that a point $z \in X$ is called a midpoint between point $x,y$ if $d(x,z) = d(z,y) = \frac{1}{2} d(x,y) $.

The first lines of the proof give us the idea:

To prove that a metric is intrinsic we have to show that for any two points $x$ and $y$ there are paths connecting $x$ and $y$ whose lengths approssimate $d(x,y)$ with any given precision. In the case of strictly intrinsic metric, there must be a path whose length is equal to $d(x,y)$. [...] We will construct a path $\gamma : [0,1] \to X$ between $x$ and $y$ such that $\gamma (0) = x $, $\gamma (1) = y$ and $L(\gamma) = d(x,y)$.

The first question (connected with the second one) is: who is $L(\gamma)$ here? Because the idea of this chapter is to show that a length space is a kind of primitive notion: we can define a function $L$ called length and then observe that there is a metric induced by this length ( - this metric is called intrinsic). But if we already have a metric, we can also define a length (from this metric).

Moreover I don't understand the end of the proof: practically the path is built on dyadic rational numbers and extended to entire interval $[0,1]$ because the metric space is complete. Then the book says:

According to our construction, for every two dyadic rationals $t_i$, $t_j$ $$d(\gamma(t_i),\gamma(t_j)) = |t_i - t_j | \cdot d(x,y) $$ [...] This implies that $L(\gamma) = d(x,y)$.

Why? I don't get it.

Thank you in advance!

gangrene
  • 1,517

1 Answers1

1

Note that your map is $d(x,y)-$Lipschitz, and when you extend it to the whole space,it's still being $d(x,y)-$Lipschitz. Thus, for every $t,t'\in[0,1]$, it's $$d(\gamma(t),\gamma(t,'))=|t-t'|d(x,y).$$ From the above identity, $$L(\gamma)=\sup\sum_{i=0}^{n-1} d(\gamma(y_i),\gamma(y_{i+1}))=\sup\sum_{i=0}^{n-1}|y_i-y_{i+1}|d(x,y)=d(x,y),$$ where the supremum is taken over all partitions $0=y_0<y_1<\cdots<y_n=1$ of $[0,1]$.

user279325
  • 389