Lipschitz continuous image of a complete subset

Question

A metric space $(A,d)$ is said to be Atsuji space if set of limit points $A'$ is compact and for each $\epsilon>0$, $A\setminus \bigcup\limits_{a\in A'}B(a,\epsilon)$ is uniformly discrete, where $B(a,\epsilon)$ is open ball centered at $a$, radius $\epsilon$. Some properties of an Atsuji space $X$ are:

1. An Atsuji space is always complete.
2. For any two disjoint nonempty closed subsets $A,B$ of $X$, there is $r>0$ such that $\big[\bigcup\limits_{a\in A}B(a,r)\big]\cap \big[\bigcup\limits_{b\in B}B(b,r)\big]=\emptyset$.
3. Any closed subset of $X$ is Atsuji.

My query is described as follows:

Let $(X,d)$ be a metric space, and $A$ be an Atsuji subset in $X$. For an open set $U$ in $X$ with $A\subset U$, we consider the map $f(\cdot)=d(\cdot,X\setminus U):A\to [0,\infty)$. Will the subset $f(A)\subset \mathbb R$ contain its infimum?

What I tried is: If the infimum of $f(A)$ is not in $f(A)$, then $f(A)$ is not complete in $\mathbb R$. So there is a Cauchy sequence $\{x_n\}$ in $f(A)$ which is not convergent in $f(A)$, and after it I am stuck; because $f$ is Lipschitz I don't get any good information about the set $\bigcup\{f^{-1}(x_n):n\in\mathbb N \}$.

Answer 1

I think the answer is no. So something to try do is explicitly construct such a set $A = \bigcup\{f^{-1}(x_n) : n\in \mathbb{N}\}$ on which $f$ won't attain its minimum. Then you just need to try find such an $A$ that is Atsuji.

Here's one example:

Let $L$ be the lower half-plane $L = \{(x,y)\in \mathbb{R}^2 : y\leq 0\}$. Let $A$ be the set $A = \{(n, \frac{1}{n}) : n\in \mathbb{N}\}$. Now choose our metric space $X$ to be $X = A\cup L$ with the usual Euclidean metric on $\mathbb{R}^2$, and choose $U$ to be the open set $\bigcup\{B((n, \frac{1}{n}), \frac{1}{n})\ : n \in \mathbb{N} \}$. We have that $A$ is Atsuji, and is contained in $U$, which is disjoint from $L$. But we can still get points of $A$ arbitrarily close to $L$. Specifically, you get $f(A) = \{f((n, \frac{1}{n})) : n \in \mathbb{N}\} = \{\frac{1}{n} : n \in \mathbb{N}\}$ which does not contain its infimum, $0$.