# How to show function is uniformly continuous over metric spaces. [duplicate]

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 $$d_\mathbb R(\inf_Ad(x,a),\inf_Ad(y,a)))<\epsilon \implies d_X(x,y)<\delta$$

• The most opportune situation we can find ourselves when attempting to show $f : X \to Y$ is uniformly continuous is when $d_Y(f(x_1),f(x_2)) \leq d_X(x_1,x_2)$. Have you investigated whether that occurs here? Apr 24, 2023 at 8:19
• You can have a look at similar past questions, such as: Continuity of the function $x\mapsto d(x,A)$ on a metric space Apr 24, 2023 at 9:55

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?

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 $$|d(x,A)-d(y,A)|\leq d(x,y).$$