(15.1, 15.2, 15.4) Determine which of the following series converge. Justify your answers.
This series diverges, as the terms do not converge to zero ( for all ; this subsequence does not converge)
This series converges, by the alternating series test.
This series diverges, for example by comparison with the series
This series converges, for example by comparison with the series
(17.4) Prove that the function is continuous on its natural domain . Hint: Think about Example 5 in §8.
Proof. In Example 5, §8 it is proved that for any convergent sequence of nonnegative real numbers, .
This is the definition of continuity of the function .
(17.5)
Prove that is a continuous function for any .
Use induction, and that is continuous on a set if both and are.
Prove that every polynomial function is continuous.
Use induction, and that and are continuous on a set if both and are, for any real number .
(17.13)
Let for rational numbers and for irrational numbers. Show that is discontinous at every in .
Proof. Consider , and let be an irrational number. Then the sequence is a sequence of irrational numbers converging to . But the sequence converges to 0 even though . Hence is not continuous at .
Now, consider any irrational number. By density of the rationals, there exists a rational number for which . Observe that converges to , and that converges to 1 while . Hence is not continuous at .
Let for rational numbers and for irrational numbers. Show that is continuous at and at no other point.
Proof. Much of the reasoning in (a) holds, with minor changes. The meaningful thing to see here is why is continuous at .
Let be any sequence of real numbers converging to 0. If only finitely many of are irrational, it is easy to see that converges to 0, as all are identically 0 for large .
Suppose there are an infinite number of rational and let these define the sequence . Observe both that and and that . So , and we conclude that is continuous at 0.
Let be a set of real numbers. An element is interior to if for some we have
The set is open in if every point in is interior to .
Let be a function, and for any subset , let be the set;
Show that is continuous if and only if, for every open set , is also an open set.
Proof. () Suppose is continuous and let be any open set in (viewed as the range of !) Let . Notice that as , and is open, there is some for which, for any , implies that .
Now, as is continuous, there exists a number so that whenever and , ; in other words, ! So .
We have just shown that for all there exists so that implies that . That is, we have just shown that is open, as desired.
() Suppose that for all open sets , we know that is open.
Observe that any open interval is an open set. (Let and observe that satisfies the correct condition)
Let , let (the domain of !) and consider the open interval which is an open set in the range of .
We have that is itself open by assumption. In other words, there exists so that for any , if and , . So in fact, , meaning that .
Hence is continuous by the definition of continuity.