$\mathbf {The \ Problem \ is}:$ Give an example of a metric space $X$, a point $x_0$, and a set $A \subset X$ so that $\operatorname{dist}(x_0,A) = 1$ where $\operatorname{dist}(x_0,A) =\operatorname{inf}\{d(x_0,y) : y \in \bar{A}\}$ with the criterion that $d(x,x_0) \neq 1$ for every $x \in \bar{A}.$
$\mathbf {My \ approach} :$ Actually, I couldn't think much about the problem but I thought about the Cantor Set $K$ as an example but $1 \in K$, so I think we have to produce examples in $\mathbb R^2$ . It's better to provide examples such that it can be generalised in $\mathbb R^n$ .
$\mathbf {Bonus} :$ Can you provide hints along the same line of two non-singleton sets $A$ and $B$ such that $\operatorname{dist}(A,B) = 1$ where $\operatorname{dist}(A,B) =\operatorname{inf}\{d(a,b) : a \in \bar{A} , b \in \bar{B} \}$ but $d(a,b) \neq 1$ for any $a \in \bar{A}, b \in \bar{B} .$
Best Answer
Let $X= \{2\} \cup (0,1)$ with the usual metric. $A=(0,1),x_0=2$ is an example for the first question.
Let $X$ be the real line with the usual metric. $A=\{2^{n}:n\geq 1\}$ and $B=\{2^{n}+1+\frac 1 {2n}: n\geq 1\}$ is an example of the second question. Note that $n \neq m$ implies $|2^{n}-(2^{m}+\frac 1 {2m})| \geq |2^{n}-2^{m}|-\frac 1 {2m} \geq 2 -\frac 1 2 >1$ so $d(a,b) \neq 1$ for $ \in A, b \in B$. Since $|2^{n}-({2^{n}}+1+\frac 1 {2n})| \to 1$ as $ n \to \infty$ it follows that $d(A,B)=1$.