The equidistant set of a closed set 1 Let $F$ be a closed set in $\mathbb{R}^m$ with $m\geq2$. For $r>0$, let $F_r$ be the set of points $x$ in $\mathbb{R}^m$ such that dist($x,F)=r$ (distance between $x$ and $F$). Notice that if $F$ is not connected, then $F_r$ mais also be disconnected. Now suppose for some $r>0$, the set $F_r$ has a bounded component; that is for some $R>0$ a component of $F_r$ is contained in the ball centered at the origin with  radius $R$. This is clear from a picture that then for all $0<\rho<r$ the set $F_\rho$ has also a bounded component. But how would you  prove this?
 A: This is mostly a simplified version of the counterexample given by Moishe Kohan in the comments - my contribution is just illustrating it and choosing nice coordinates that make computation easy.

This is false. Consider the set $F\subseteq \mathbb R^2$ consisting of the union of:

*

*A closed ray $(x,0)$ for $x\geq 1$.


*Line segments connecting $(1,0)$ to $(1,-1)$ to $(-1, -1)$ to $(-1, 1)$ to $(0,1)$.


*A closed ray $(0,y)$ for $y\geq 1$.

It can be readily observed that $\{(0,0)\}$ is a connected component of $F_1$ - and, indeed, is the only point of $F_1$ within $(-1,1)\times (-1,1)$. More generally, $F_r$ has a bounded component whenever $\frac{1}{\sqrt{2}} < r \leq 1$.
We can see this visually, illustrated at an $r$ slightly less than $1$ for more clarity:

However, if $r \leq \frac{1}{\sqrt{2}}$, we can describe $F_r$ as two unbounded components, consisting of line segments, rays, and circular arcs - though I'll leave the precise listing and verification of this claim to the reader, as it's entirely a mechanical proof. The set $F_r$ would be as illustrated in the following picture:

The fact that $F_1$ has a bounded component but $F_{1/2}$ does not contradicts your desired theorem.
A: How would I prove it?
Lemma 1. $F\subseteq(F_r)_r$.
Lemma 2. If $F_r=\bigsqcup_j{A^j}$ is a decomposition into connected components, let $B^j=A^j_r\cap F$.  Then each $B^j$ is relatively clopen and $F=\bigsqcup_j{B^j}$.
Lemma 3. With the same notation, $A^j$ is bounded iff $B^j$ is.  (Use Lemma 1.)
Lemma 4. With the same notation, $F_{\rho}=\bigcup_j{B^j_{\rho}}$, and each $B^j_{\rho}$ is relatively closed.
Lemma 5. When $\rho<r$, the decomposition in Lemma 4 is disjoint.
I leave the proofs (and how to apply Lemmas 3 & 5 to your problem) to you.
These lemmas apply in any metric space, not just $\mathbb{R}^{\geq2}$, although they get a little tricky to apply to your problem in the discrete metric space….
A: Let $0< \rho <r.$ Let $F_r$ has a bounded component $C_r \subseteq F_r$. From the definition of $F_r$ it is clear that $0<\rho <r \implies diam(C_\rho)<diam(C_r),$ hence $C_{\rho}$ are bounded for any $\rho \in (0,r).$
Note that $C_\rho$ might be one of many components which is a subset of the closet set bounded by $C_r.$
