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 $% $. 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

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;

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 $y \in \mathbb R$, $% $ implies that $y \in U$.

Now, as $f$ is continuous, there exists a number $\delta > 0$ so that whenever $t \in \mathbb R$ and $% $, $% $; 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 $% $ 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 $% $ 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 $% $, $t \in f^{-1}(I)$. So in fact, $f(t) \in I$, meaning that $% $.

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