Let $(X,d)$ be a metric space. Let us fix two points $\alpha,\beta\in X$. My question how can I bound the following expression: $$ \sup_{x\in X}|d(\alpha,x)-d(\beta,x)| $$
If you consider the plane $\mathbb{R}^2$ with the Euclidean distance, it is easy to see that if you move outwards from $\alpha$ and $\beta$, this value can be bounded by $d(\alpha,\beta)$ times some constant. I search some of this bound but with a general metric space.
Thanks in advance.
The inequality $$\tag{1} \lvert d(\alpha, x)-d(\beta, x)\rvert\le d(\alpha, \beta)$$ holds for every metric space. It is actually a consequence of the triangle inequality for $d$ and the symmetry $d(a, b)=d(b, a)$.
Proof: by triangle inequality, $$ d(\alpha, x)\le d(\alpha, \beta)+d(\beta, x)$$ therefore $d(\alpha, x)-d(\beta, x)\le d(\alpha, \beta)$. Redoing this with $\alpha$ and $\beta$ reversed, $$ d(\beta, x)\le d(\beta, \alpha)+d(\alpha, x).$$ Since $d(\beta, \alpha)=d(\alpha, \beta)$, we conclude that $d(\beta, x)-d(\alpha, x)\le d(\alpha, \beta)$, and the two inequalities coalesce into (1).