complexity theory – Inapproximability of graph problems on a restricted setting

I am considring the following problem $mathcal{P}$.

$mathcal{P}$: Given an undirected graph $G$, and an integer $k$, find a set of vertices $S subseteq V(G)$, with $|S| = k$, such that the number of edges in the subgraph induced by $S$ is minimized.

which is clearly NP-hard, as the answer is 0 iff there is an independent set of size $k$ in $G$. So I am interested in studying whether the problem can be approximated assuming $mathrm{P} neq mathrm{NP}$ (assumption implicit from now on).

Now, for the restricted setting, imagine $c > 0$ is an arbitrary constant, and then define $cmathcal{P}$ as the same problem but with the restriction that $|V(G)| geq c cdot k$. The question I am wondering is whether, given the following claims 1) and 2), it is true that 1) implies 2).

  1. There is a constant $rho > 1$ such that $mathcal{P}$ cannot be approximated by a factor better than $rho$.
  2. There is a constant $rho’ > 1$ such that for any constant $c > 0$ the problem $cmathcal{P}$ cannot be approximated by a factor better than $rho’$.

I would appreciate any help, or pointers to problems when something like that is proven.

approximation – Polynomial variable of inapproximability after reduction

I proved the inapproximability of a problem that, given a multigraph $G = (V, E)$ and a set of vertices $U subseteq V$ tries to maximize a score $f(U)$ whose value depends on the edges of the graph, so it takes values in ${0, dots, m}$, $m$ being the number of edges in the graph.

As the result was obtained by (exactly) reducing from the Maximum Independent Set which is known for being inapproximable within $n^{1- epsilon}$, I am not sure how this is translated to my problem: Is it inapproximable within $n^{1- epsilon}$ or within $m^{1- epsilon}$ (as, in a certain sense, it counts the number of edges)?

(In the proof we build a multigraph in which each edge contributing to the score is associated with an independent vertex in the original graph and vice versa)