Suppose that (1) is fulfilled and let be a sequence in converging to . We have to show that
holds. To show this, let
be given. Due to (1) there exists a
fulfilling the estimation property (from the definition of continuity) and because of the convergence of to there exists a natural number such that for all
holds. By the choice of we have
so that the image sequence converges to .
Suppose now that (2) is fulfilled. We assume that is not continuous. Then there exists an
such that for all
there exist elements
such that their distance to is at most , but such that the distance of their value to is larger than . This holds in particular
This means that for every natural number
there exists a
This sequence converges to , but the image sequence does not converge to , since the distance of its members to is always at least . This contradicts condition (2).