Given a metric space \(X\) and a bounded subset \(S\subseteq X\). Let \(r>0\). For \(x\in S_r^c\), \(p\in S\) and every \(y\in S\), we have
\[ C^{-1} d(x,p) \le d(x,y)\le C d(x,p), \]where
\[ C=(1+\tfrac{\diam S}{r}). \]
Remarks
- If \(r=\alpha \diam S\) for a number \(\alpha\ge 1\), then \[ C=1+\alpha. \]
Proof
Let \(y\in S\). Then
\begin{align*} d(x,y) &\le d(x,p)+d(y,p)\\ &\le d(x,p)+ \diam S\\ &\le \Bigl(1+\frac{ \diam S}{r}\Bigr) d(x,p), \end{align*}where we used the triangle inequality and the estimate \(d(x,p)\ge r\). This proves the upper bound of \(d(x,y)\).
By the same argument, we also find
\[ d(x,p)\le \Bigl(1+\frac{\diam S}{r}\Bigr) d(x,y), \]which implies the lower bound.