Help showing the distance function is uniformly continuous.

Question

Let $X$ be a metric space and $q \in X$. I want to show that the distance function $d(q,p)$ is a uniformly continuous function of $p$.

I know how to show that $d$ is continuous, but I am stuck on how to show UC.

Given $\epsilon >0$ let $\delta =?$. Then if $d(x,y) <\delta$, then $|d(q,x)-d(q,y)|<\epsilon$.

I cannot figure out how to choose $\delta$.

Please help :). Thank you.

Answer 1

$d(q,x) \leq d(q,y) + d(y,x)$ and $d(q,y) \leq d(q,x)+d(x,y)$ so $|d(q,x)-d(q,y)| \leq |d(x,y)| <\epsilon$ if $d(x,y)<\epsilon$.

Answer 2

$d(q,x)\leq d(q,y)+d(y,x)$ so that $d(q,x)-d(q,y)\leq d(y,x)=d(x,y)$

By symmetry: $d(q,y)-d(q,x)\leq d(x,y)$

This together allows the conclusion that $|d(q,x)-d(q,y)|\leq d(x,y)$