Since π interchanges c and d, π(f(U)) ≠ f(U).

Now suppose \(A\) to be partitioned into a (necessarily infinite) mutually disjoint set \(P\) of pairs. Take \(G\) to be the group of permutations of \(A\) which fix all the pairs in \(P\). Then \(P\in Sym(V)\); it can now be shown that \(Sym(V)\) contains no choice function on \(P\). For suppose \(f\) were a choice function on \(P\) and \(f \in Sym(V)\). Then \(f\) has a finite support which may be taken to be of the form \(\{a_{1}, \ldots, a_{n}, b_{1},\ldots,b_{n}\} with each pair \{a_{i}, b_