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Starting with an arbitrary point in a finite metric space, repeatedly select a point maximizing its distance to the selected set. Maintain each point's distance to its nearest selected center; each new center updates all those distances, giving $O(k|V|)$ distance evaluations for $k$ centers.

For the metric k-center problem, let $R$ be the final covering radius and add a point realizing $R$ to the $k$ selected centers. Every pair among these $k+1$ points has distance at least $R$: at each selection the current maximum distance is at least its final value. If the optimum radius is $r^{\ast }$, two of these points belong to the same optimal radius-$r^{\ast }$ closed ball, so the triangle inequality gives $R\le 2r^{\ast }$. The case $R=0$ is already optimal. This proves a factor-two approximation algorithm.

The metric k-center problem chooses at most $k$ centers from a finite metric space to minimize the largest distance from a point to its nearest center. For $1\le k\le |V|$, exactly $k$ centers gives the same optimum: extra centers never increase the radius. The restriction that centers belong to $V$ matters; allowing centers anywhere in an ambient metric space is a different problem.

The basic packing argument is short. If a radius-$r$ solution uses $k$ centers, its $k$ closed balls cover $V$. Two points assigned to the same center have distance at most $2r$ by the triangle inequality. Thus $k+1$ points with pairwise distances greater than $2r$ certify that no radius-$r$ solution exists. Greedily selecting a new center only outside all existing radius-$2r$ closed balls therefore uses at most $k$ centers whenever a radius-$r$ solution exists, and its final covering radius is at most $2r$. The farthest-first traversal achieves the same approximation ratio without knowing $r$.

Assume the standard nonempty instance convention $1\le k\le n$, and set $d(v,v)=0$. Start with no centers. Pick any city still farther than $2r^{\ast }$ from every selected center, add it as a center, and mark every city within distance $2r^{\ast }$ of it as covered. Stop when all cities are covered.

Distinct chosen centers have pairwise distance greater than $2r^{\ast }$. Fix an optimal set of $k$ centers and assign each city to one optimal center within distance $r^{\ast }$. Two cities assigned to the same center have distance at most $2r^{\ast }$ by the triangle inequality. Thus no two chosen centers can be assigned to that same optimal center. The pigeonhole principle shows that the greedy procedure selects at most $k$ centers and has radius at most $2r^{\ast }$. If fewer than $k$ are selected, add arbitrary distinct cities; this cannot worsen the radius.

With the supplied radius, a straightforward implementation updates coverage in $O(n)$ time per selected center, hence takes $O(nk)$ distance tests. The argument also handles $r^{\ast }=0$. It is the packing bound underlying the metric k-center problem.