# Harrison Chapman

## Math 317: Homework 6

Due: Friday, October 18, 2019

Exercises:

1. $$\displaystyle\sum { \left[ \cos \left( \frac {n \pi}{2} \right)\right]^n }$$

2. $$\displaystyle\sum_{n=2}^{\infty} { \frac{(-1)^n}{n - \sqrt{n}} }$$

3. $$\displaystyle\sum_{n=2}^\infty { \frac {1}{\sqrt n \log n} }$$

4. $$\displaystyle\sum { \frac {\log n}{n^3} }$$

1. This series diverges, as the terms do not converge to zero. ($$a_{2k} = 1$$ for all $$k \in \mathbb N$$; 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 $$\sum {1/n}$$.

4. This series converges, for example by comparison with the series $$\sum {1/n^{3/2}}$$.

2. This question is worth double points.
1. Prove that $$x^m$$ is a continuous function for any $$m \in \mathbb N$$.

2. Prove that every polynomial function $$\sum_{i=0}^n{a_i x^i}$$ 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 $$f(x) = 1$$ for rational numbers $$x$$ and $$f(x) = 0$$ for irrational numbers. Show that $$f$$ is discontinous at every $$x$$ in $$\mathbb R$$.

2. Recall the function $$f(x)$$ from part (a). Let $$g(x)$$ be the function with domain $$\mathbb Q$$ defined as $$g(x) = f(x)$$ for all $$x \in \mathbb Q$$. Show that $$g$$ is continuous on its domain.

3. Let $$h(x) = x$$ for rational numbers $$x$$ and $$h(x) = 0$$ for irrational numbers. Show that $$h$$ is continuous at $$0$$ and at no other point.

1. Proof. Consider $$q \in \mathbb Q$$, and let $$\alpha$$ be an irrational number. Then the sequence $$(q + \alpha/n)$$ is a sequence of irrational numbers converging to $$q$$. But the sequence $$(f(q + \alpha/n))$$ converges to 0 even though $$f(q) = 1$$. Hence $$f$$ is not continuous at $$q$$.

Now, consider $$\alpha$$ any irrational number. By density of the rationals, there exists a rational number $$q_n$$ for which $$\alpha - 1/n < q_n < \alpha$$. Observe that $$q_n$$ converges to $$\alpha$$, and that $$(f(q_n))$$ converges to 1 while $$f(\alpha) = 0$$. Hence $$f$$ is not continuous at $$\alpha$$. $$\Box$$

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 $$h$$ is continuous at $$0$$.

Let $$(x_n)$$ be any sequence of real numbers converging to 0. If only finitely many of $$x_n$$ are irrational, it is easy to see that $$(h(x_n))$$ converges to 0, as all $$h(x_n)$$ are identically 0 for large $$n$$.

Suppose there are an infinite number of rational $$x_n$$ and let these define the sequence $$(x_{n_k})$$. Observe both that $$\lim h(x_{n_k}) = \lim x_{n_k} = 0$$ and $$\limsup h(x_n) \le \limsup h(x_{n_k}) = 0$$ and that $$\liminf h(x_n) \ge \liminf h(x_{n_k}) = 0$$. So $$\lim h(x_n) = 0 = h(0)$$, and we conclude that $$h$$ is continuous at 0. $$\Box$$

3. Let $$E \subseteq \mathbb R$$ be a set of real numbers. An element $$s_0 \in E$$ is interior to $$E$$ if for some $$r > 0$$ we have

$\{ s \in \mathbb R : |s - s_0| < r\} = (s_0 - r, s_0 + r) \subset E.$

The set $$E$$ is open in $$\mathbb R$$ if every point in $$E$$ is interior to $$E$$.

Let $$f: \mathbb R \to \mathbb R$$ be a function, and for any subset $$V \subset \mathbb R$$, let $$f^{-1}(V) \subset \mathbb R$$ be the set;

$f^{-1}(V) = \{ x \in \mathbb R : f(x) \in V \}.$

Show that $$f$$ is continuous if and only if, for every open set $$U \subset \mathbb R$$, $$f^{-1}(U)$$ is also an open set.

Proof. ($$\Rightarrow$$) Suppose $$f$$ is continuous and let $$U$$ be any open set in $$\mathbb R$$ (viewed as the range of $$f$$!) Let $$x \in f^{-1}(U)$$. Notice that as $$f(x) \in U$$, and $$U$$ is open, there is some $$\epsilon > 0$$ for which, for any $$y \in \mathbb R$$, $$|y - f(x)| < \epsilon$$ implies that $$y \in U$$.

Now, as $$f$$ is continuous, there exists a number $$\delta > 0$$ so that whenever $$t \in \mathbb R$$ and $$|t - x| < \delta$$, $$|f(t) - f(x)| < \epsilon$$; in other words, $$f(t) \in U$$! So $$t \in f^{-1}(U)$$.

We have just shown that for all $$x \in f^{-1}(U)$$ there exists $$\delta > 0$$ so that $$|t - x| < \delta$$ implies that $$t \in f^{-1}(U)$$. That is, we have just shown that $$f^{-1}(U)$$ is open, as desired.

($$\Leftarrow$$) Suppose that for all open sets $$U \subseteq \mathbb R$$, we know that $$f^{-1}(U)$$ is open.

Observe that any open interval $$(a,b)$$ is an open set. (Let $$x \in (a,b)$$ and observe that $$r = \min{|x-a|, |x-b|}) > 0$$ satisfies the correct condition)

Let $$\epsilon > 0$$, let $$x \in \mathbb R$$ (the domain of $$f$$!) and consider the open interval $$I = (f(x) - \epsilon, f(x) + \epsilon) = \{y \in \mathbb R : |f(x) - y| < \epsilon\}$$ which is an open set in the range of $$f$$.

We have that $$f^{-1}(I)$$ is itself open by assumption. In other words, there exists $$\delta > 0$$ so that for any $$x \in f^{-1}(I)$$, if $$t \in \mathbb R$$ and $$|t - x| < \delta$$, $$t \in f^{-1}(I)$$. So in fact, $$f(t) \in I$$, meaning that $$|f(t) - f(x)| < \epsilon$$.

Hence $$f$$ is continuous by the $$\epsilon,\delta$$ definition of continuity. $$\Box$$