∀k ∈ ℕ, k ∈ W ⇔ k ∉ QkWe know that W ∈ Q(ℕ), as all of its elements are in ℕ. But we also know that W cannot be in our list, as it differs from each of the Qk's due to the presence or absence of the integer k. A contradiction, and the result is established. QED
Note: The argument is similar to the diagonalization argument that is used to establish that there are more reals numbers than integers.We can generalize the above result to all infinite sets, countable or otherwise. Consider the set A of all sets. By definition, all subsets of ℘(A) are in A. But ℘(A) contains more elements than does A, a contradiction.