# Proof that the function $f(x)=d(x,y)$ is uniformly continuous?

Consider a metric space $(M, {\rm d})$ and $y$ fixed in $M$.

I want to prove that the function $f$ defined by $f(x)\colon={\rm d}(x,y)$ is uniformly continuous.

So I know that if this function is uniformly continuous, then for all $\epsilon>0$, there exists a $\delta>0$ such that if the ${\rm d}(x_1,x_2)<\delta$, then this implies the ${\rm d}(f(x_1),f(x_2))<\epsilon$.

So for all epsilon greater than zero, there exists a delta greater than zero such that ${\rm d}(x_1,x_2)<\delta \implies |{\rm d}(x_1,y)-{\rm d}(x_2,y)|<\epsilon$.

I can't figure out how to manipulate this equation to solve for delta!! I tried using the triangle inequality but it gets me nowhere. Please help!

Let $\epsilon > 0$, choose $\delta = \epsilon$. Take $x_1,x_2 \in M$ such that ${\rm d}(x_1,x_2) < \delta$. Then: $$|f(x_1)-f(x_2)| = |{\rm d}(x_1,y)-{\rm d}(x_2,y)| \leq {\rm d}(x_1,x_2) < \delta = \epsilon.$$

Obs.: by the triangle inequality, we have that $${\rm d}(x_1,y)\leq {\rm d}(x_1,x_2)+{\rm d}(x_2, y)\implies {\rm d}(x_1,y)-{\rm d}(x_2,y)\leq {\rm d}(x_1,x_2).$$ Swapping $x_1$ and $x_2$ you get: ${\rm d}(x_2,y)-{\rm d}(x_1,y)\leq {\rm d}(x_2,x_1) = {\rm d}(x_1,x_2).$ This way, the inequality $|{\rm d}(x_1,y)-{\rm d}(x_2,y)| \leq {\rm d}(x_1,x_2)$ follows from the definition of absolute value.

• why can you swap x1 and x2? – marry Jan 16 '15 at 0:18
• It is because what I've done earlier is valid for any points $x_1,x_2$. It was a quick way to saying that you can repeat the same argument. Like when books say "analogously, yadda yadda yadda". In details, what I omitted was: $${\rm d}(x_2,y)\leq {\rm d}(x_2,x_1)+{\rm d}(x_1, y)\implies {\rm d}(x_2,y)-{\rm d}(x_1,y)\leq {\rm d}(x_2,x_1).$$ Do compare with what I've done before. – Ivo Terek Jan 16 '15 at 0:20
• i see, that makes sense. Thanks!! how did you come up with this so quickly? – marry Jan 16 '15 at 0:21
• Practice dealing with these problems, I reckon. I think I'm a little biased, since I'm taking a summer course about metric spaces here (: – Ivo Terek Jan 16 '15 at 0:23
• pouco haha i just noticed your profile says universidade de sao paulo:) – marry Jan 16 '15 at 0:57

You have to show that for fixed $\;m\in X$= the metric space, the function $\;f(x):=d(m,x)\;$ is unif. continuous, so you have to show that

$$\forall\,\epsilon>0\;\exists\,\delta >0\;\;s.t.\;\;d(x,y)<\delta\implies |f(x)-f(y)|=|d(m,x)-d(m,y)|<\epsilon$$

But all you have to do then is use the triangle inequality and some symmetry:

$$d(m,x)\le d(m,y)+d(y,x)\implies |d(m,x)-d(m,y)|\le d(x,y)$$

and take now $\;\delta=\epsilon$ .

• It might confuse OP that you changed the fixed point to $m$, and used $y$ as variable, since he was using $y$ as a fixed point. I don't know. Just my opinion, though (: – Ivo Terek Jan 16 '15 at 0:16
• Indeed so, @IvoTerek . Yet I began answering this question before you edited it and didn't even see he did so. I think though that someone dealing with this must already be able to overcome that. Thank you. – Timbuc Jan 16 '15 at 0:20
• thanks guys this very helpful – marry Jan 16 '15 at 0:22