 Possibilities (ii) and (iii) may be ruled out because \(f\) is continuous. This leaves (i) and (iv), in either of which \(f(x) = f(x + \varepsilon)\). So \(f\) is locally, and hence globally, constant, that is, constantly 1 or 0. In the first case \(V = \varnothing\) , and in the second \(U = \varnothing\).