The Standard Bounded Metric

The Standard Bounded Metric

Recall from the Metric Spaces page that if $M$ is a nonempty set then a function $d : M \times M \to [0, \infty)$ is called a metric if for all $x, y, z \in M$ we have that the following three properties hold:

  • $d(x, y) = d(y, x)$.
  • $d(x, y) = 0$ if and only if $x = y$.
  • $d(x, y) \leq d(x, z) + d(z, y)$.

Furthermore, the set $M$ with the metric $d$, denoted $(M, d)$ is called a metric space.

We will now look at another type of metric known as a standard bounded metric - which can be obtained from another metric.

Definition: Let $(M, d)$ be a metric space. The Standard Bounded Metric of $d$ is the metric $\bar{d} : M \times M \to [0, \infty)$ defined for all $x, y \in M$ by $\bar{d}(x, y) = \min \{ 1, d(x, y) \}$.

For example, consider the metric space $(\mathbb{R}, d)$ where $d : \mathbb{R} \times \mathbb{R} \to [0, \infty)$ is the standard Euclidean metric defined for all $x, y \in \mathbb{R}$ as $d(x, y) = \mid x - y \mid$. Then the standard bounded metric of $d$ is given for all $x, y \in \mathbb{R}$ by:

\begin{align} \quad \bar{d}(x, y) = \min \{ 1, d(x, y) \} \end{align}

Let's verify that the standard bounded metric $\bar{d}$ satisfies all three of the conditions above.

For the first condition, since $d(x, y) = d(y, x)$ for all $x, y \in \mathbb{R}$ we have that:

\begin{align} \quad \bar{d}(x, y) = \min \{1, d(x, y) \} = \min \{ 1, d(y, x) \} = \bar{d} (y, x) \end{align}

For the second condition, suppose that $\bar{d}(x, y) = 0$. Then $d(x, y) = 0$ which happens if and only if $x = y$. Now suppose that $x = y$. Then $d(x, y) = 0$ and $\bar{d}(x, y) = \min \{ 1, 0 \} = 0$.

For the third condition, we have that for all $x, y, z \in \mathbb{R}$ that $d(x, y) \leq d(x, z) + d(z, y)$, i.e., $\mid x - y \mid \leq \mid x - z \mid + \mid z - y \mid$. So:

\begin{align} \quad \bar{d}(x, y) = \min \{ 1, \mid x - y \mid \} \leq \min \{ 1, \mid x - z \mid + \mid z - y \mid \} \leq \min \{ 1, \mid x - z \mid \} + \min \{ 1, \mid z - y \mid \} = \bar{d} (x, z) + \bar{d} (z, y) \end{align}

Therefore $\bar{d}$ is a metric.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License