Suppose I am working on a compact subset $S$ of $\mathbb{R}^N$, and I have a closed subset, $A\subset S$. Let $A+\epsilon$ denote the $\epsilon$-thickening of $A$ in $S$ given by:
$$A+\epsilon = \{s \in S: d(s,A)\leq \epsilon\}$$
And let $A-\epsilon$ denote the $\epsilon$-thinning of $A$ in $S$ given by:
$$A-\epsilon = A\setminus\text{Int}(A^c+\epsilon)$$
Here $d$ is the Euclidean distance and Int is the interior. Assume $\epsilon > 0 $ and $A+\epsilon$ and $A-\epsilon$ are non-empty.
I want to find a function $f:\mathbb{R}^N\rightarrow \mathbb{R}^N$ such that $f$ is strictly increasing in each coordinate (i.e. $z>y \implies f(x_1,...,z,...x_N) > f(x_1,..,.y,...x_N)$ and continuous, such that $f(A)=A+\epsilon$. Is this always possible? What if I substitute $A+\epsilon$ for $A-\epsilon$? How can I show this holds?
So far I think I need something like this: Let $a \in A-\epsilon$ and define $f$ as:
$$f(x)=x+\epsilon \frac{x-a}{\|x-a\|}$$
But I'm a bit confused.