a set $Z\subset{\mathbb{R}}^d$ with $|Z|=\ell$ such that:

1.  \[connect\_property1\] $Y$ is $(\ell,\operatorname{opt}_k)$-coverable;

2.  \[connect\_property2\] $\operatorname{cost_{rad}}({\mathcal{C}}_k)\leq\operatorname{cost_{rad}}({\mathcal{P}}_\ell)$;

3.  \[connect\_property3\] For all $n\in{\mathbb{N}}$ with $\ell+1\leq n\leq|Y|$, every cluster in ${\mathcal{P}}_n$ is $(Z,\operatorname{opt}_k)$-connected to another cluster in ${\mathcal{P}}_n$.

Here, the collection ${\mathcal{P}}_1,\ldots,{\mathcal{P}}_{|Y|}$ denotes the hierarchical clustering computed by Algorithm \[rad\_algo\] on input $Y$.

To define $Y,Z$, and $\ell$ we consider the $(k+1)$-clustering computed by Algorithm \[rad\_algo\] on input $X$. We know that $X=\bigcup_{A\in {\mathcal{C}}_{k+1}}A$ is $(k,\operatorname{opt}_k)$-coverable. Let $E\subseteq{\mathcal{C}}_{k+1}$ be a minimal subset such that $\bigcup_{A\in E}A$ is $(|E|-1,\operatorname{opt}_k)$-coverable, i.e., for all sets $F\subseteq {\mathcal{C}}_{k+1}$ with $|F|<|E|$ the union $\bigcup_{A\in F}A$ is not $(|F|-1,\operatorname{opt}_k)$-coverable. Since a set $F$ of size $1$ cannot be $(|F|-1,\operatorname{opt}_k)$-coverable, we get $|E|\geq2$.

Let $Y:=\bigcup_{A\in E}A$ and $\ell:=|E|-1$. Then, $\ell\leq k$ and $Y$ is $(\ell,\operatorname{opt}_k)$-coverable. This establishes property \[connect\_property1\].

It follows that there exists a set $Z\subset{\mathbb{R}}^d$ with $|Z|=\ell$ and $Y\subset\bigcup_{z\in Z}\operatorname{B}_{\operatorname{opt}_k}^d(z)$. Furthermore, we let ${\mathcal{P}}_1,\ldots,{\mathcal{P}}_{|Y|}$ be the hierarchical clustering computed by Algorithm \[rad\_algo\] on input $Y$.

Since $Y$ is the union of the clusters from $E\subseteq{\mathcal{C}}_{k+1}$, each merge step between the computation of ${\mathcal{C}}_{|X|}$ and ${\mathcal{C}}_{k+1}$ merges either two clusters $A,B\subset Y$ or two clusters $A,B\subset X\setminus Y$. The merge steps inside $X\setminus Y$ have no influence on the clusters 