Real Analysis Homework: Show Uniform Continuity & Density in R

[MIT2014]

Homework Statement

Let X$$\subset$$R be nonempty. Let f:R$$\rightarrow$$R be defined by f(x)=inf_{t E X}|t-x|.

Show:
1. f is uniformly continuous
2. f$$\equiv$$0 if and only if X is dense in R

Homework Equations

none

The Attempt at a Solution

I am clueless as to how to go about this. Help!

 

[micromass]

For the first part, you'll need to show

$$\forall \epsilon >0: \exists \delta>0: \forall x,a: |x-a|<\delta~\Rightarrow |f(x)-f(a)|<\epsilon$$

So we'll need to do something with |f(x)-f(a)|. This is what we do (make sure you understand all steps):

$$|f(x)-f(a)|=|\inf_{t\in X}{|t-x|}-\inf_{t\in X}{|t-a|}|\leq|\inf_{t\in X}{|t-x|-|t-a|}|\leq \inf_{t\in X}{|x-a|}=|x-a|$$

So take $$\delta=\epsilon$$


Another way you can proof 1, is to pick $$x_0\in X$$ and then notice that $$0\leq f(x)\leq |x-x_0|$$. Then f is uniform continuous since it is sandwiched between uniform continuous functions...

 

[MIT2014]

Thanks for the answer! It makes a lot of sense.

For the second part, I have a general idea of how to solve the problem. If X is dense in R, then inf|t-x| will always be zero. However, how can this be expressed with a formal proof?

 

[micromass]

Take X dense and take x arbitrarly. For every $$\epsilon>0$$, there exists t in X such that $$|t-x|<\epsilon$$. Thus $$\inf_{t\in X}{|t-x|}<\epsilon$$. This holds for every epsilon, so the value must be zero.