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- Apr 14, 2013

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Hey!!

Let $(X, d)$ be a metric space. For $A \subseteq X$ und $x \in X$ we define $d_A : X \rightarrow \mathbb{R}$ by \begin{equation*}d_A(x):=\inf\{d(x,y)\mid y\in A\}\end{equation*}

I want to prove the below statements:

I have done the following:

Let $(X, d)$ be a metric space. For $A \subseteq X$ und $x \in X$ we define $d_A : X \rightarrow \mathbb{R}$ by \begin{equation*}d_A(x):=\inf\{d(x,y)\mid y\in A\}\end{equation*}

I want to prove the below statements:

- $A$ is closed iff for all $x\in X$ with $d(x,A)=0$ it holds that $x\in A$.
- The map $d_A:X\rightarrow \mathbb{R}$ is continuous.
- Let $A,B\subseteq X$ be disjunctive, closed subsets. Then there is a continuous function $f:X\rightarrow [0,1]$ such that $$f(x)=0\iff x\in A\ \text{ and } \ f(x)=1\iff x\in B$$

I have done the following:

- Since $A$ is closed, it holds that $\forall x\in X\setminus A : \exists \epsilon >0 : B_{\epsilon}(x)\cap A=\emptyset$.

A ball $B_{\epsilon}(x)$ is the set of all points $y\in X$ satisfying $d(x,y)<\epsilon$.

Is the definition correct so far? How could we continue?

$$$$

- Let $a\in A$.

We have that $d_A(x)=\inf\{d(x,y)\mid y\in A\}\leq d(x,a)$. Applying the triangle inequality we get $d(x,a)\leq d(x,y)+d(y,a)$. That means that we have $d_A(x)\leq d(x,y)+d(y,a)$.

We take the infimum over $a\in A$ and we get: $\inf_ad_A(x)\leq \inf_a\left (d(x,y)+d(y,a)\right ) \Rightarrow d_A(x)\leq d(x,y)+\inf_ad(y,a)$.

Since $(X,d)$ is a metric space, we have that $d$ is a metric and the property of symmetry is satisfied, and so we have $d(x,y)=d(y,x)$.

Therefore we get $d_A(x)\leq d(x,y)+\inf_ad(a,y)\Rightarrow d_A(x)\leq d(x,y)+\inf_ad(y,a) \Rightarrow d_A(x)\leq d(x,y)+d_A(y) \Rightarrow d_A(x)-d_A(y)\leq d(x,y)$

Let $\varepsilon>0$. We choose $\delta:=\varepsilon$. Then, if $d(x,y)<\delta$, we have that $|d_A(x)-d_A(y)|\leq d(x,y)<\varepsilon$.

This means that $d_A$ is continuous.

Is everything correct?

$$$$

- Could you give me a hint for this one? i got stuck right now.

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