two clusters. In order to preserve the logarithmic dependency of the approximation factor on $k$, we show that it is sufficient to analyze Algorithm \[rad\_algo\] on a subset $Y\subseteq X$ satisfying a certain connectivity property. Using this property, we are able to apply a combinatorial approach that relies on the number of merge steps left.

We start by defining the connectivity property that will be used to relate clusters to an optimal $k$-clustering.

Let $Z\subseteq{\mathbb{R}}^d$ and $r\in{\mathbb{R}}$. Two sets $A,B\subseteq{\mathbb{R}}^d$ are called *$(Z,r)$-connected* if there exists a $z\in Z$ with $\operatorname{B}_r^d(z)\cap A\neq\varnothing$ and $\operatorname{B}_r^d(z)\cap B\neq\varnothing$.

Note that for any two $(Z,r)$-connected clusters $A,B$, we have $$\label{connect_eq}
\operatorname{rad}(A\cup B)\leq\operatorname{rad}(A)+2\cdot\operatorname{rad}(B)+2r.$$

Next, we show that for any input set $X$ we can bound the cost of the $k$-clustering computed by Algorithm \[rad\_algo\] by the cost of the $\ell$-clustering computed by the algorithm on a connected subset $Y\subseteq X$ for a proper $\ell\leq k$. Recall that by our convention from the beginning of Section \[rad\_sec\], the clusterings computed by Algorithm \[rad\_algo\] on a particular input set are uniquely determined.

\[connect\_lem\] Let $X\subset{\mathbb{R}}^d$ be finite and $k\in{\mathbb{N}}$ with $k\leq|X|$. Then, there exists a subset $Y\subseteq X$, a number $\ell\in{\mathbb{N}}$ with $\ell\leq k$ and $\ell\leq|Y|$, and 