Inapproximability of the $k$-center problem

Question

The $k$-center problem:

Given:

1. Undirected, complete graph $G=(V,E)$,
2. a distance function $d:E\rightarrow\mathbb{R}$ such that $d_{ii}=0, d_{ij}=d_{ji}$ for each pair for vertices $i,j\in V$, and that the distances obey triangle inequality,
3. A positive integer $k$.

Find a set $S\subseteq V, |S|=k$ of $k$ cluster centers with the objective to minimize the maximum distance of a vertex to its cluster center.

We know that the problem is NP-hard and a simple greedy algorithm gives 2-approximation. I was going through a proof from the book "The Design of Approximation Algorithms" for a theorem that states that an approximation algorithm with factor less than 2 would imply P=NP. This is done using a reduction from the dominating set problem.

Theorem: There is no $\alpha$-approximation algorithm for the $k$-center problem with $\alpha<2$ unless P=NP.

Proof: Given an instance of the dominating set problem, we can define an instance of the $k$-center problem by setting the distance between adjacent vertices to 1 and non-adjacent vertices to 2. Then there is a dominating set of size $k$ if and only if the optimal radius for this $k$-center instance is 1. Furthermore, any $\alpha$-approximation with $\alpha<2$ must always produce a solution of radius 1 if such a solution exists.

Question: I think I understand the proof. However, I am confused with the setting of distance between non-adjacent vertices to 2. What if we set it to 3 instead of 2? Would that mean the $\alpha<3$ is not possible unless P=NP? This is not true as we have a 2-approximation algorithm. The proof should not work for values greater than 2. However, I am not able to figure out which step of the proof breaks when we set distance between non-adjacent vertices to 3.

Dominating Set Problem: Given an undirected graph $G=(V,E)$ and an integer $k$, we must decide if there exists a set $S\subseteq V$ of size $k$ such that each vertex is either in $S$ or adjacent to a vertex in $S$.

Answer 1

If the weight for adjacent vertices is $w_{a}$, using any other $w_{na} > 2 w_{a}$ for non-adjacent ones would break the triangle equality, which is a fundamental assumption to prove the 2-approximation algorithm. In fact, think about three vertices $i,j,k$ in the $k$-center instance such that $i$ and $j$ are adjacent, $j$ and $k$ are adjacent but $i$ and $k$ are not. This would break $d_{ij}+d_{jk}=2w_{a}\geq d_{ik}= w_{na}$ because $w_{na}\geq 2 w_{a}$.

The constant numbers 1 and 2 are just two example of constants $w_{a}$ and $w_{na}$ that satisfy $0 < w_{a} < w_{na} \leq 2 w_{a}$.