Find a function continuous at a single point of an interval, but discontinuous at all other points of the interval.
Consider the interval for any
, and define
for all .
We claim is continuous at 0, but discontinuous at every other point of the interval.
Proof. First, we show is continuous at 0. To do this we must show that for all
there exists a
such that
whenever
. So, let
be given. Then, choose
. Then we have,
since from the definition of
. Thus,
is continuous at 0.
Now, we must show is discontinuous at every other point of the interval. Let
be any nonzero point of the interval.
We consider two cases:
Case #1: is rational. Then
. Since
by assumption, we consider
. Since the irrationals are dense in the reals (see Exercise #9 of Section I.3.12 here) we know that for any
there is some irrational number
such that
. But then,
. This means
is not continuous at
since we have found an
such that there is no
we can choose such that
whenever
.
Case #2: is irrational. Since
is irrational we know
. Consider
. Since the rationals are dense in the reals (see Exercise #6 of Section I.3.12 here) we know that for any
there exists some rational number
such that
. But then,
. This means
is not continuous at
. (Again, because we have shown that there exists an
such that for every
there is some number
such that
with
(Note: There are much nicer ways to do this using sequences, but Apostol doesn’t develop those techniques, so we have to work straight from the epsilon-delta definitions.)