Vrftjkry

Let $(X,d)$ be a metric space and $Asubset X$ and $xin X$. Then

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




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



• 
  

  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.

  
  

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 :

  
  
  

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

  
  
  

  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 :

  
  
  

  $$ bigvert d(x,A) - d(y,A) bigvert leq d(x,y) $$

  
  
  

  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*}
$$