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.)
Define the function:
Prove there is no such that
Proof. We prove this by contradiction. Suppose there does exist some such that . From the definition of limit this means that for all there exists a such that
First, we claim that for any such we must have . This must be the case since if then , so we may choose such that . But then, since for all , we have
(Where we used part (i) this exercise (Section I.4.9, Exercise #1) for the inequality .) This contradicts our choice of , so must be less than or equal to 1.
Next, suppose . Then choose . To obtain our contradiction we must show that there is no such that
By the Archimedean property of the real numbers we know that for any , there exists a positive integer with such that . (The Archimedean property guarantees us an integer, but then the inequality will hold for any larger integer, so we can choose one with the property of being .) But,
Then, from the definition of and since and we have,
This contradicts that if then for every we have for all . In other words, we found an greater than 0 (in particular, we found ) such that no matter how small we choose there exists an such that is smaller than , but is bigger than . This exactly contradicts the definition of . Hence, there can be no such number
This proof shows that there is now way to define so that is continuous at 0. We see this since for to be continuous at 0 we must have . But if we try to define for any , we will fail to achieve continuity since the proof shows for any we might want to choose.
Consider the function
Determine what happens to as . Define so that will be continuous at , if possible.
First, we sketch the graph of on the intervals and .
To determine what happens to as we can take the limit, (recalling that ),
Since is not defined at it is not continuous there. However, since the limit exists we could redefine by
Then, this new (which has the same values as the original function everywhere the original function is defined, but has the additional feature of being defined at ) is continuous at since it is defined there and .
Note what we have done here. We had a function which was undefined at the point . We created a new function (which we also called , which is a bit confusing) that took the same values as the original function for all , and took the value 1 when . In this way we “removed” the discontinuity of the original at the point , but we should be aware that this is actually a new function (since it has a different domain than the original).
Find the points at which is continuous and at which is continuous.
and , are continuous everywhere we know (by Theorem 3.2) that is continuous everywhere is not zero. We proved in this exercise (Section 2.8, #1 (b) of Apostol) that
Thus, is continuous for
and by Theorem 3.2 we then have is continuous everywhere that is not zero. We proved in the same exercise (Section 2.8, #1(a) of Apostol) that
Hence, is continuous for
For a constants , define:
If are fixed, find all values for such that is continuous at .
By the definition of continuity of a function at a point, we know that is continuous at means is defined at , and .
Since and are defined for all , we know that is defined at for all . So, we must then find values of such that
From the definition of , we know
Then, we evaluate the limit as through values greater than (since the limit as through values less than is since is a continuous function, and for values less than , ),
Thus, for to be continuous at we must have,
If , then we have