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Let $(X,d)$ be a metric space and $A\subset X$ and $x\in X$. Then

  • $x\to d(x,A)$ is a uniformly continuous function.

  • If $\partial A=\{x\in X\,:\,d(x,A)=0\}\cap\{x\in X\,:\,d(x,X-A)=0\}$, then $\partial A$ is closed for any $A\subset X$.

  • If $A,B$ are subsets of $X$ then $d(A, B)=d(B,A)$.

    If the function is uniformly continuous then $d(x,y)<\delta $ implies $d(d(x,A)-d(y,A))<\epsilon$ , I can not handle the last expression. Difficulty continues for 3rd choice also. At least give me some hints. And for the 2nd choice I think it is true, as it is intersection of two closed sets. As finite intersection of closed sets closed $\partial A$ closed. And logic for former two sets closed is that the sets are preimage of closed set (singleton $\{0\}$ is closed as $\mathbb R$ is $T_1$) under continuous map. I am not very much sure about my ideas and want a verification.

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user121418
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1 Answers1

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Let $(X,d)$ be a metric space and $A$ be any non-empty subset of $X$. Then, for any $x \in X$, let $d(x,A)$ be :

$$ d(x,A) = \inf \limits_{z \in A} d(x,z) $$

  • It is true that the mapping $x \, \longmapsto \, d(x,A)$ is uniformly continuous as it is lipschitz continuous (with a lipschitz constant equal to 1). (If you don't know about "lipschitz continuity", have a look here.) Let us prove that :

    $$ \forall (x,y) \in X^{2}, \, \big\vert d(x,A) - d(y,A) \big\vert \leq d(x,y) $$

    Let $(x,y) \in X^{2}$ and let $z \in A$. It follows from the triangular inequality for $d$ that :

    $$ d(x,z) \leq d(x,y) + d(y,z) \tag{1} $$

    Note that, by definition : $\forall z \in A, \, d(x,A) \leq d(x,z)$. Using this idea a first time gives :

    $$ d(x,A) \leq d(x,z) \leq d(x,y) + d(y,z) $$

    Using the same idea again leads to :

    $$ d(x,A) - d(y,A) \leq d(x,y) \tag{2} $$

    By symmetry ($x$ and $y$ play symmetric roles here), we have :

    $$ d(y,A) - d(x,A) \leq d(x,y) \tag{3} $$

    Eventually, $(2)$ and $(3)$ write :

    $$ d(y,A) - d(x,y) \leq d(x,A) \leq d(y,A) + d(x,y) $$

    which is exactly :

    $$ \big\vert d(x,A) - d(y,A) \big\vert \leq d(x,y) $$

    As a conclusion, $x \, \longmapsto \, d(x,A)$ is lipschitz continuous on $X$, so it is uniformly continuous on $X$.

  • You are right, $\partial A$ is closed as it is the intersection of two closed sets. Each of the two sets are closed as they are the preimage of $\left\{ 0 \right\}$ (which is closed in $(\mathbb{R},\vert \cdot \vert)$) under the continuous mappings $x \, \longmapsto \, d(x,A)$ and $x \, \longmapsto \, d(x,X \smallsetminus A)$.
  • Let $A$ and $B$ be two non-empty subsets of $X$. Since $d$ is a distance, $d(x,y)=d(y,x)$ for all $(x,y) \in X^{2}$. So, $d(A,B) = d(B,A)$ follows easily. By definition :

$$ \begin{align*} d(A,B) &= \inf \limits_{x \in A} d(x,B) \\[2mm] &= \inf \limits_{x \in A} \inf \limits_{y \in B} d(x,y) \\[2mm] &= \inf \limits_{x \in A} \inf \limits_{y \in B} d(y,x) \\[2mm] &= \inf \limits_{y \in B} \inf \limits_{x \in A} d(y,x) \\[2mm] &= \inf \limits_{y \in B} d(y,A) \\[2mm] &= d(B,A) \\ \end{align*} $$

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