How to show function is uniformly continuous over metric spaces.

Question

Let $X$ be a metric space , $A\subseteq X$.

Let $d(x,A)=\inf_Ad(x,a)$.

Prove $d(\cdot,A):X \to \mathbb{R}$ is uniformly continuous.

$f:X \to Y$ is said to be uniformly continuous if for every $\epsilon>0$ there exists $\delta>0$ so that $d_Y(f({x_1}),f({x_2}))<\epsilon$ for all $x_1,x_2 \in X$ with $d_X(x_1,x_2)<\delta.$

In my problem $Y=\mathbb{R}$.

Hence I have to prove $d_\mathbb R(\inf_Ad(x,a),\inf_Ad(y,a)))<\epsilon \implies d_X(x,y)<\delta$

Honestly I am a little bit confused about this problem , appreciate any help.

Answer 1

Brian Moehring has provided a workable hint that can be used as a good point of attack. Now whenever you are coming across distance from a point to a set, chances are you can work with triangle inequality. Not a principle to blindly adhere, but worth a try.

$\forall x, y\in X, a\in A\subset X, $ by traingle's inequality:

$$ d(x, a) \leq d(x, y) + d(y, a) \implies d(x, A) \leq d(x, y) + d(y, A)\overset{\mathrm{taking \inf_{a\in A}~ of~ both ~sides}}{\implies } d(x, A) \leq d(x, y) + d(y, A). $$ Now can you bring it home?

Answer 2

For any $x,y\in X$ and $a\in A$, note that using triangle inequality $$d(x,a)\leq d(x,y)+d(y,a)\Rightarrow d(x,a)-d(y,a) \leq d(x,y).$$ That is, $$d(x,A)-d(y,A)\leq d(x,y).$$ Similarly, one can prove that $$d(y,A)-d(x,A)\leq d(x,y).$$ Hence $$|d(x,A)-d(y,A)|\leq d(x,y).$$