Metric for connected path space. I'm trying to prove the next function is a metric for the space of connected paths $T_{x,y}(X)$ where $x,y\in X\subset\mathbb{R}^{n}:$
$$d(x,y)=\inf\{L(\sigma):\sigma\in T_{x,y}(X)\},$$ where $L(\sigma)=\sup\{\sum_{k=1}^{m}||\sigma(t_{k-1})-\sigma(t_{k})||:0=t_{0}<t_{1}<\cdots<t_{m}=1\}$ and $||(\cdot)||$ is the usual euclidean norm. 
My problem is with triangular inequality.I think the other requierements I got them. The only achieve that I have is a result of triangular inequality of euclidean norm: $||x-y||\leq L(\sigma_{x,z})+L(\sigma_{z,y}).$ I expected to bound $L(\sigma_{x,y})$ with the previous; but it's useless. 
Would you mind give me a hand?
Thanks for advance.
 A: If we assume for a moment that the infimum were always attained, that is, we had
$$d(x,y) = \min \{ L(\sigma) : \sigma \in T_{x,y}(X)\},$$
then we could state the triangle inequality as "the shortest path from $x$ to $y$ is not longer than the shortest path from $x$ to $y$ that passes through $z$", and it would be immediate from picking paths of minimal length. Since in general the infimum is not attained, however, we need to approximate, but the formulation can serve as a guide.
For any three points $x,y,z\in X$, we consider
$$T_{x,y;\,z}(X) := \bigl\{ \sigma \in T_{x,y}(X) : \bigl(\exists s\in (0,1)\bigr)\bigl(\sigma(s) = z\bigr)\bigr\},$$
the set of paths from $x$ to $y$ passing through $z$. On the one hand, from $T_{x,y;\,z}(X) \subset T_{x,y}(X)$, it follows immediately that
$$d(x,y) = \inf \{ L(\sigma) : \sigma \in T_{x,y}(X)\} \leqslant \inf \{ L(\sigma) : \sigma \in T_{x,y;\,z}(X)\}.\tag{1}$$
On the other hand, it is not difficult to show that
$$l(x,y;z) := \inf \{ L(\sigma) : \sigma \in T_{x,y;\,z}(X)\} = d(x,z) + d(z,y).$$
For if $\sigma \in T_{x,y;\,z}(X)$ and $\sigma(s) = z$ for some $s\in (0,1)$, then we can split the path $\sigma$ in two,
\begin{align}
\sigma_1(t) &= \sigma(s\cdot t) &&\text{for } t \in [0,1],\\
\sigma_2(t) &= \sigma\bigl(s + (1-s)\cdot t\bigr) && \text{for } t \in [0,1],
\end{align}
with $\sigma_1 \in T_{x,z}(X)$ and $\sigma_2 \in T_{z,y}(X)$. It is easily seen that $L(\sigma) = L(\sigma_1) + L(\sigma_2)$, and thus
$$l(x,y;z) \geqslant d(x,z) + d(z,y).\tag{2}$$
Conversely, for any $\sigma_1 \in T_{x,z}(X)$ and $\sigma_2 \in T_{z,y}(X)$, we can form the composition of paths
$$(\sigma_1 \ast \sigma_2)(t) = \begin{cases} \sigma_1(2t) &, t \in \bigl[0,\tfrac{1}{2}\bigr]\\ \sigma_2(2t-1) &, t \in \bigl[\tfrac{1}{2},1\bigr].\end{cases}$$
Thus we obtain a map $\ast \colon T_{x,z}(X) \times T_{z,y}(X) \to T_{x,y;\,z}(X)$, and we have $L(\sigma_1 \ast \sigma_2) = L(\sigma_1) + L(\sigma_2)$. By definition of $\inf$, for every $\varepsilon > 0$ we can find a path $\sigma_1 \in T_{x,z}(X)$ with $L(\sigma_1) < d(x,z) + \varepsilon/2$, and a path $\sigma_2 \in T_{z,y}(X)$ with $L(\sigma_2) < d(z,y) + \varepsilon/2$. Composing these two paths shows
$$l(x,y;z) < d(x,z) + d(z,y) + \varepsilon,\tag{3}$$
since $L(\sigma_1 \ast \sigma_2) = L(\sigma_1) + L(\sigma_2) < d(x,z) + \varepsilon/2 + d(z,y) + \varepsilon/2$. Since $(3)$ holds for all $\varepsilon > 0$, we obtain $l(x,y;z) \leqslant d(x,z) + d(z,y)$, and together with $(1)$ the equality
$$l(x,y;z) = d(x,z) + d(z,y),$$
which, inserted into $(1)$, is the triangle inequality.
