Supremum Metric

Last updated Nov 1, 2022

Let $(M_{i}, d_{i}){i \in I}$ be Metric Spaces with Index Set $I$ and let $M = \prod\limits{i \in I}^{} M_{i}$. Then the function $d: M \times M \to \overline{\mathbb{R}{\geq 0}}$ defined $$d(x, y) := \sup\limits{i \in I} d_{i}(x_{i}, y_{i})$$ (where $d(x,y) = \infty$ if $d_{i}(x_{i},y_{i})$ is unbounded) is an Extended Distance Function. It is known as the Supremum Metric.

We show that it satisfies the Axioms of a Distance Function:

1. Let $x, y \in M$. Suppose $d(x, y) = 0$. Then, by definition of Supremum, $d_{i}(x_{i}, y_{i}) \leq 0$ for all $i \in I$. But already $d_{i}(x_{i}, y_{i}) \geq 0$ for all $i \in I$, so we have $d_{i}(x_{i}, y_{i}) = 0$. By definition of Distance Function, this means $x_{i} = y_{i}$ for all $i \in I$. Then $x = y$. $\checkmark$
2. Let $x, y \in M$. Observe $$d(x,y) = \sup\limits_{i \in I} d_{i}(x_{i}, y_{i}) = \sup\limits_{i \in I} d_{i}(y_{i}, x_{i}) = d(y,x)$$ $\checkmark$
3. Let $x,y,z \in M$. Then $$\begin{align*} d(x,y) + d(y,z) &= \sup\limits_{i \in I} d_{i}(x_{i}, y_{i}) + \sup\limits_{i \in I} d_{i}(y_{i}, z_{i}) \\ &\geq\sup\limits_{i \in I} (d_{i}(x_{i}, y_{i}) + d_{i}(y_{i}, z_{i}))\\ &\geq \sup\limits_{i \in I} d_{i}(x_{i}, z_{i})\\ &= d(x, z) \end{align*}$$ $\checkmark$

1. The metric induced by a Supremum Norm is a Supremum Metric.

• Cartesian Product
• Sum of Supremum bounds Supremum of Sum