04-07-2024

Problem: Let $x,y \in \mathbb{R} $. Define a function \[ \delta : \mathbb{R} \times \mathbb{R} \to \mathbb{R},\ \delta (x,y) = \left\vert \tan ^{-1} x - \tan ^{-1} y \right\vert. \] Then show that $\delta $ is a metric on $\mathbb{R} $.
Solution: We need to show that the function \[ \delta (x,y) = \left\vert \tan ^{-1} x - \tan ^{-1} y \right\vert \] is a metric on $\mathbb{R} $. We will verify each property of a metric.
  • $\delta (x,y) \geq 0$ for any $x,y \in \mathbb{R} $.
    This is clear from the definition.
  • $\delta (x,y) = 0$ if and only if $x = y$.
    Let \begin{align*} \delta (\times y) = 0 & \implies \left\vert \tan ^{-1} x - \tan ^{-1} y \right\vert = 0 \\ & \implies \tan ^{-1} x - \tan ^{-1} y = 0 \\ & \implies \tan ^{-1} x = \tan ^{-1} y \\ & \implies x = y, \end{align*} where the last implication is true because $\tan ^{-1} $ is an injective function.
  • $\delta (x,y) = \delta (y,x)$.
    This is also clear as \begin{align*} \delta (x,y) & = \left\vert \tan ^{-1} x - \tan ^{-1} y \right\vert \\ & = \left\vert \tan ^{-1} y - \tan ^{-1} x \right\vert \\ & = \delta (y,x). \end{align*}
  • $\delta (x,z) \leq \delta (x,y) + \delta (y,z)$ for any $x,y,z$.
    Consider \begin{align*} \delta (x,z) & = \left\vert \tan ^{-1} x - \tan ^{-1} y \right\vert\\ & = \left\vert \tan ^{-1} x \textcolor{red}{- \tan ^{-1} y + \tan ^{-1} y} - \tan ^{-1} z \right\vert \\ & \leq \left\vert \tan ^{-1} x - \tan ^{-1} y \right\vert + \left\vert \tan ^{-1} y - \tan ^{-1} z \right\vert \\ & = \delta (x,y) + \delta (y,z). \end{align*}
Thus, we proved that $\delta $ is a metric on $\mathbb{R} $.
Note that in the above problem, if we replace $\tan ^{-1} $ by any injective function, then the same result will be true. That is, if $f: X\to X $ is an injective function and $d$ is a metric on $X$, then $\delta (x,y) = d(f(x), f(y))$ is again a metric on $X$.