According to Wikipedia, the Independent Set problem is a special case of the Set Packing problem. But, it seems to me that these problems are equivalent.
The Independent Set search problem is: given a graph $G(V,E)$ and an integer $n$, find $n$ vertices no two of which are adjacent.
The Set Packing search problem is: given a finite collection $C$ of finite sets and an integer $n$, find $n$ sets that are pairwise disjoint.
I think they are equivalent based on the following bidirectional reduction:
→: Given an independent set problem on a graph $G(V,E)$, create a collection of $C$ of sets, where for each vertex $v \in V$ there is a set $S_v \in C$ containing all edges adjacent to $v$. Now, every set packing in $C$ corresponds to a set of vertices no two of which have an edge in common, i.e., this is an independent set in $G$ of the same size.
←: Given a set packing problem on a collection $C$, create a graph $G(V,E)$ where for every set $S \in C$ there is a vertex $v_S \in V$, and there is an edge between $v_{S_1}$ and $v_{S_2}$ iff the sets $S_1$ and $S_2$ intersect. Now, every independent vertex set in $G$ corresponds to a set of sets from $C$ no two of which intersect, i.e., this is a set packing in $C$ of the same size.
My question is: is my reduction correct? If so, are these problem equivalent? Is it possible to use approximation algorithms for one problem on the other problem?