The Polygonal Inequality for Metric Spaces

Recall from the Metric Spaces page that if $M$ is a set and $d : M \times M \to [0, \infty)$ is a function that satisfies the following properties:

• For all $x, y \in M$, $d(x, y) = d(y, x)$.

• For all $x, y \in M$, $d(x, y) = 0$ if and only if $x = y$.

• For all $x, y, z \in M$, $d(x, y) \leq d(x, z) + d(z, y)$.

Then $d$ is called a metric on $M$, and the pair $(M, d)$ is called a metric space.

The third property of a metric $d$ above is called the triangle inequality and can be extended to a more general polygonal inequality which we prove in the following theorem.

Theorem 1 (The Polygonal Inequality for Metric Spaces): Let $(M, d)$ be a metric space. Then for all $x_1, x_2, ..., x_n \in M$, $\displaystyle{d(x_1, x_m) \leq \sum_{k=1}^{n-1} d(x_k, x_{k+1})}$.

• Proof: Let $x_1, x_2, ..., x_m \in M$. By repeated use of the triangle inequality property of the metric $d$ we have that:

(1)

\begin{align} \quad d(x_1, x_n) & \leq d(x_1, x_2) + d(x_2, x_n) \\ \quad d(x_1, x_n) & \leq d(x_1, x_2) + d(x_2, x_3) + d(x_3, x_n) \\ \quad d(x_1, x_n) & \leq d(x_1, x_2) + d(x_2, x_3) + d(x_3, x_4) + d(x_4, x_n) \\ & \vdots \\ \quad d(x_1, x_n) & \leq d(x_1, x_2) + d(x_2, x_3) + ... + d(x_{n-1}, x_n) \\ \quad d(x_1, x_n) & \leq \sum_{k=1}^{n-1} d(x_k, x_{k+1}) \quad \blacksquare \end{align}

Geometrically, the polygonal identity can be depicted as follows: