 This same point can be made in yet another way, involving a relativized form of Russell’s argument. Let \(B\) be any set. By ZA, the set \(R_B = \{x \in B: x \not\in x\}\) exists, but it cannot be an element of \(B\). For if it is an element of \(B\), then we can ask whether or not it is an element of \(R_B\); and it is if and only if it is not. Thus something, namely \(R_B\), is “missing” from each set \(B\). So again, \(V\) is not a set, since nothing can be missing from \(V\). But notice the following subtlety: unlike the previous argument involving the direct application of Aussonderungs to V, the present argument hints at the idea that, while \(V\) is not a set, “V” is not an empty name. The next strategy for dealing with Russell’s paradox capitalizes on this hint. 