Department of Mathematics

Colorado State University

Math 317: Homework 6

Due: Friday, October 18, 2019

Reading: Read sections 17, 18, 19

Exercises:

  1. Determine which of the following series converge. Justify your answers.

    1. This series diverges, as the terms do not converge to zero. ( for all ; this subsequence does not converge)

    2. This series converges, by the alternating series test.

    3. This series diverges, for example by comparison with the series .

    4. This series converges, for example by comparison with the series .

  2. This question is worth double points.
    1. Prove that is a continuous function for any .

    2. Prove that every polynomial function is continuous.

    3. A rational function \(g\) is a quotient of polynomials \(p,q\): \(g(x) = p(x)/q(x)\), whose domain is \(\{x \in \mathbb R : q(x) \ne 0\}\). Prove that every rational function is continuous.

    1. Use induction on \(m\), together with how \(f(x) = x\) is continuous and products of continuous functions are continuous.

    2. Use induction on \(n\), together with how \(f(x) = x^m\) is continuous for nonnegative integers \(m\) and that scalar multiples and sums of continuous functions are continuous.

    3. Use that quotients of continuous functions are continuous whenever the denominator is nonzero.

    1. Let for rational numbers and for irrational numbers. Show that is discontinous at every in .

    2. Recall the function from part (a). Let be the function with domain defined as for all . Show that is continuous on its domain.

    3. Let for rational numbers and for irrational numbers. Show that is continuous at and at no other point.

    1. 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 .

    2. Now that the domain is simply \(\mathbb Q\), we only consider rational sequences \((q_n)\) converging to a rational number \(q\). Of course, for all rational numbers \(g(q_n) = 1\) and so \(\lim g(q_n) = \lim 1 = 1 = g(q)\). So \(g\) is actually continuous on its domain.

    3. 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.

  3. 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.