




Outline: The limit of a sequence all of whose members are at least d must be at least d. If f is continuous, the set of points x for which f(x) d must therefore contain all its limit points, and, similarly, so must the set of points y for which f(y) d.
If f has skipped over the value d, so that we had f(x_{1}) = a,
f(y_{1}) = b, a > d > b and for no x is f(x) = d, these two sets must form a partition of the interval between x_{1} and y_{1} into two disjoint sets, each of which contains all its limit points.
But no such partition is possible:
Suppose y_{1 }< x_{1 }. Let z be the smallest number in this interval all of whose neighborhoods contains a number x with f(x) > d.
z is then a limit point of the set of numbers x for which f(x) > d. This implies f(z) > d. But then z is not y_{1 }, and we can find a set of numbers r all between y_{1} and z that converge to z and have
f(r) < d; This contradicts the continuity of f.