Proof
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
the estimate
-

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
for every
,
.
This means that for every natural number
,
there exists a
with
-
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).