Article title: Structural cohesion and embeddedness in two-mode networks
Authors: Benjamin Cornwell and Jake Burchard
Journal: The Journal of Mathematical Sociology
Bibliometrics: Volume 43, Issues 4, pages 179–194
DOI: 10.1080/0022250X.2019.1606806
Explanation of correction
Theorem 3.1 and Corollary 3.1.1 in the article are false. While Theorem 3.1 correctly states that , the reverse inequality is not necessarily true (a family of counterexamples can be produced to show this). It should be noted that these statements, while false, are nevertheless tangential to the main emphasis of the paper, which is that cohesion in two-mode networks should be studied without one-mode projections, and that this can be done using both what we call “two-sided” and “one- sided” approaches. We have replaced Theorem 3.1 and Corollary 3.1.1 with the following new, correct theorems and accompanying text.
New Text
Theorem 3.1. Let be a two-side -connected two-mode network with disjoint sets of nodes and , let ’s regular one-mode connectivity be , and let be the minimum degree of . Then, .
Proof. We prove the result by contradiction. If it were the case that , then , so or . However, by definition if or nodes are removed from , or will become disconnected, respectively, and therefore so will . This leads to a contradiction given that is defined as the minimum number of nodes needed to be removed to disconnect . Therefore, .
Now assume that . This implies that , i.e. and . such that , where is the set of nodes in . Since is bipartite, or , where is the neighborhood of node , i.e. the subgraph of induced by the nodes adjacent to . Without loss of generality, assume . Then, the subgraph induced by \is disconnected, and since , , which is a contradiction. Therefore .Theorem 3.2. For a two-side -connected network with disjoint sets of nodes and , such that , where is the number of node-independent paths between and .
Proof. We prove the result by contradiction. Assume that there does not exist such a pair of nodes . Then , or, since , or . Without loss of generality, assume that . By definition, it is possible to remove a set of nodes to disconnect . Consider two nodes which become disconnected after removing . Since and are in the same set, and there are node-independent paths between them, then there must be at least one node from set on each of those paths. If we are only allowed to remove nodes from set , at least nodes from must be removed to disconnect and . Since , we have contradicted our earlier statement that nodes from could be removed to disconnect and . Therefore, such that .
Applying Theorem 3.2 to Figure 5, since the network in Figure 5 is two-side 1-connected, such that . In other words, there exists some pair of nodes which only have one node independent path between them. For example, nodes and : all paths between them must go through , so there is only one node independent path between them.
Additional new text
In the Conclusion section of the paper the sentence “(2) One can identify a group that contains a mix of actors from both node sets that remain connected via multiple pathways despite the removal of some specified number of actors from both node sets” should instead read “(2) One can identify a group in which both sets of nodes are robust to the removal of nodes from the other set”.
Acknowledgments
The authors apologize for any inconvenience caused. We thank Paul Dreyer (Senior Mathematician, RAND) for bringing this error to our attention.