There exists a point $y_0 \in A $ with $d(y_0,x) = d(x,A)$ if choose the correct option

Question

let A be a subset of $\mathbb{R}^P$ and $ x \in \mathbb{R}^P$ denotes $d(x,A) = \inf \{ d(x,y) : y \in A\}$.There exists a point $y_0 \in A $ with $d(y_0,x) = d(x,A)$ if

choose the correct option

$a)$ $A$ is any closed non emepty subset of $\mathbb{R}^P$

$b)$$ A$ is any non empty subset of $\mathbb{R}^P$

$c)$$ A$ is any non empty compact subset of $\mathbb{R}^P$

$d)$$ A$ is any non empty bounded subset of $\mathbb{R}^P$

My attempt : i thinks option a) will correct because if A is closed then $x \in \bar A= A,$ that is $d(y_0,x) = d(x,A)=0$

I don't know that other option pliz help me

Answer 1

a) It is correct. Take $a\in A$ and consider the closed ball $\overline{B_{d(x,a)}(x)}$. Then $a\in\overline{B_{d(x,a)}(x)}\cap A$ and, since $\overline{B_{d(x,a)}(x)}\cap A$ is closed and bounded, it is compact. So, there is a $a'\in\overline{B_{d(x,a)}(x)}\cap A$ such that$$d(x,a')=\min\{d(x,c)\,|\,c\in\overline{B_{d(x,a)}(x)}\cap A\}.$$And if $c\in A\setminus\left(\overline{B_{d(x,a)}(x)}\cap A\right)$, then $d(x,c)>d(x,a)\geqslant d(x,a')$. So, $d(x,a')=d(x,A)$.

b) It is false. take $x=(0,0)$ and $A=(0,1)\times\{0\}$.

c) It is correct, since a) holds.

d) It is false. Consider the same example as inmy answer to b).

Answer 2

a) and c) are true and b) and d) are false. For a) choose $\{x_k\} \subset A$ such that $d(x,x_k)$ converges to $d(x,A)$. Then $\{x_k\}$ is bouunded, so it has a convergent subsequence. If the limit of the subsequence is $y$ the $ y \in A$ and $d(x,y)=d(x,A)$. C) follows from a). For b) and d) Take $x=0$ and $A=(0,1)$ in $\mathbb R$. Thanks to Jose carlos Santos for pointing out an error in my earlier answer.