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### Proof that a continuous function has no gaps

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(x1) = a,

f(y1) = b, a > d > b and for no x is f(x) = d, these two sets must form a partition of the interval between x1 and y1 into two disjoint sets, each of which contains all its limit points.

But no such partition is possible:

Suppose y1 < x1 . 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 y1 , and we can find a set of numbers r all between y1 and z that converge to z and have

f(r) < d; This contradicts the continuity of f.