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.)