0
$\begingroup$

Suppose $(X, \rho)$ is a metric space. For a nonempty set $E \subset X$, define $f: X \rightarrow \mathbb{R}$ by $f(x) = \text{dist}(x,E) = \text{inf} \{\rho(x,y): y \in E \}$. State whether $f$ is uniformly continuous and prove it.

I am thinking that it is uniformly continuous because if $x$ and $y$ are close together, they should also be about the same distance from $E$. And I think that this should be true no matter which $x$ and $y$ are chosen (So that it is uniformly continuous, not merely continuous). But I am not sure how to go about the proof. Specifically I am somewhat at a loss as to how to deal with the quantity

$$ | \text{inf}\{ \rho(x,z): z \in E \} - \text{inf}\{ \rho(y,z): z \in E \}|$$

Any help would be appreciated.

  • 0
    Please share some thoughts, ideas, some work you tried, some context, something ... – 2017-01-13
  • 0
    Have you tried playing with the definition of uniform continuity? In the case of uniform continuity, we get to pick our $\epsilon$ first, as opposed to continuity where we fix a point first. – 2017-01-13

2 Answers 2

0

First of all, the metric space $(Y,d)$ is pointless ...

Now, let us prove that the map $X\to\mathbb{R},x\mapsto\mathrm{dist}(x,E)$ is Lipschitz-continuous (with constant $1$) and therefore uniformly continuous.

Given $x,y\in X$, we have for all $a\in E$ :

$\mathrm{dist}(x,E)\le\rho(x,a)\le\rho(x,y)+\rho(y,a)$

Therefore : $\mathrm{dist}(x,E)-\rho(x,y)\le\rho(y,a)$, and taking the infimum we get $\mathrm{dist}(x,E)-\rho(x,y)\le\mathrm{dist}(y,E)$

So that :

$$\mathrm{dist}(x,E)-\mathrm{dist}(y,E)\le\rho(x,y)$$

and, switching $x$ and $y$ :

$$\mathrm{dist}(y,E)-\mathrm{dist}(x,E)\le\rho(y,x)$$

that is (a distance beeing a symmetric map) :

$$-(\mathrm{dist}(x,E)-\mathrm{dist}(y,E))\le\rho(x,y)$$

Finally :

$$\boxed{\vert\mathrm{dist}(x,E)-\mathrm{dist}(y,E)\vert\le\rho(x,y)}$$

0

My analysis is pretty rusty, and I would like to give this a try.

I think the idea is as follows:

Let $\epsilon>0$. Let $x,y \in X$ such that $\rho(x,y)<\delta.$

We want to show that $d(f(x),f(y))<\epsilon$; I'm assuming that $d$ is the standard euclidean length.

Consider the following. For any $e\in E$.

(1) $dist(x,E)\leq \rho(x,y)+dist(y,E)$
(2) $dist(y,E)\leq \rho(y,x)+dist(x,E)$

The above equations show us that $dist(x,E)-dist(y,E)\leq \rho(x,y)$ and $d(y,E)-d(x,E)\leq \rho(x,y)$. Which means that $|dist(x,E)-dist(y,E)|\leq \rho(x,y)<\delta$. This shows us that $\epsilon=\delta$ regardless of our choice of $x,y \in X$.