Chapter 04 Functional Limits and Continuity

4.2 Functional Limits

4.2.3

Review the definition of Thomae’s function $t(x)$ from Section 4.1.

a. Construct three different sequences $(x_n)$, $(y_n)$ and $(z_n)$ , each of which converges to 1 without using the number 1 as a term in the sequence.

Solution: consider

$$x_n = \left\{ 1 + (1/2)^n \right\} \\ y_n = \left\{ 1 + (1/3)^n \right\} \\ z_n = \left\{ 1 + (1/4)^n \right\} \\ $$

b. Now, compute $\lim t(x_n)$, $\lim t(y_n)$, and $\lim t(z_n)$.

Solution:

$$\lim t(x_n) = (1/2)^n \\ \lim t(y_n) = (1/3)^n \\ \lim t(z_n) = (1/4)^n \\ $$

c. Make an educated conjecture for $\lim_{x→1} t(x)$, and use Definition 4.2.1B to verify the claim. (Given $\epsilon > 0$, consider the set of points $\{x ∈ R : t(x) ≥ \epsilon \}$. Argue that all the points in this set are isolated.)

Proof: For any $x$, if $x$ is irrational, then $t(x) = 0 < \epsilon$. If $x$ is rational, then assume $x = p/q$ and $1/N < \epsilon$. Then for any $n$, $p/q$ lies between $[k_n/n, (k_n+1)/n]$. For $n = 1, \dots, N$, we obtain $2n$ points. Let

$$\delta = \min \left\{ |\frac{k_n}{n} - x| \ne 0, |\frac{k_{n}+1}{n} - x| \ne 0, n = 1 \dots N \right\}$$

And for any rational $y \in V_{\delta }(x), y \neq x$, it's denominator must be bigger than $N$, so $t(y) < \epsilon$. So $x$ is an isolated point.

4.2.4

Consider the reasonable but erroneous claim that

$$\lim_{x→10} \frac{1}{[[x]]} = \frac{1}{10}.$$

a. Find the largest $δ$ that represents a proper response to the challenge of $\epsilon = 1/2$.

Solution:

$$1/10 + 1/2 = 3/5 \\ 1/10 - 1/2 = -2/5 \\ $$

So in this case if $\delta =8$, then $10 - 8 = 2$, $f(2) = 1/2 < 3/5$. And $10 + 8 = 18, f(18) = 1/17 < 3/5$.

b. Find the largest $δ$ that represents a proper response to $\epsilon = 1/50$.

Solution: $\delta = 1$. $1/10 - 1/9 = 1/90, 1/10 - 1/10 = 0$.

(c) Find the largest $\epsilon$ challenge for which there is no suitable $\delta$ response possible.

Solution: $1/90$.

4.2.6

Decide if the following claims are true or false, and give short justifications for each conclusion.

(a) If a particular $δ$ has been constructed as a suitable response to a particular $\epsilon$ challenge, then any smaller positive $δ$ will also suffice.

Solution: Yes. Assume $\delta_0 < \delta$ then $V_{\delta_0}(x) \subset V_{\delta}(x)$, so any smaller positive $δ$ will also suffice.

(b) If $\lim_{x→a} f(x) = L$ and $a$ happens to be in the domain of $f$, then $L = f(a)$.

False: Consider Thomae’s function, $\lim_{x→1} f(x) = 0$, but $f(1) = 1$.

(c) If $\lim_{x→a} f(x) = L$, the If $\lim_{x→a} 3[f(x)-2]^2 = 3(L-2)^2$.

True: Corollary 4.2.4 (Algebraic Limit Theorem for Functional Limits).

(d) If $\lim_{x→a} f(x) = 0$, then $\lim_{x→a} f(x)g(x) = 0$ for any function $g$ (with domain equal to the domain of $f$.)

False: consider $f(x) = x - 1, g(x) = \frac{1}{x-1}$, then $\lim_{x→1} f(x)g(x) = 1$.

4.2.7

Let $g : A \rightarrow \mathbb{R}$ and assume $f$ is a bounded function on $A$ in the sense that there exists $M > 0$ satisfying $|f(x)| ≤ M$ for all $x ∈ A$. Show that if $\lim_{x→c} g(x) = 0$, then $\lim_{x→c} g(x)f(x) = 0$ as well.

Proof: Given $\epsilon$, we can find $\delta$ such that $\forall x \in V_{\delta}(c), g(x) < \frac{\epsilon}{M}$, then

$$|f(x)g(x) - 0| = |f(x)g(x)| = |f(x)| |g(x)| < M \frac{epsilon}{M} = \epsilon$$

So $\lim_{x→c} g(x)f(x) = 0$.

4.2.8

Compute each limit or state that it does not exist. Use the tools developed in this section to justify each conclusion.

(a)

$$\lim_{x \to 2} \frac{|x-2|}{x-2}$$

Solution: dose not exist. Consider

$$x_n = 2 + \frac{1}{n} \\ y_n = 2 - \frac{1}{n}$$

Then $\lim_{n \to \infty} x_n = 1$ and $\lim_{n \to \infty} y_n - -1$. Based on Corollary 4.2.5, the limit does not exist.

(b)

$$\lim_{x \to \frac{7}{4}} \frac{|x-2|}{x-2}$$

Solution: -1.

(c)

$$\lim_{x \to 0} (-1)^{[[1/x]]}$$

Solution: consider

$$x_n = \frac{1}{2n}\\ y_n = \frac{1}{2n+1}$$

Then $\lim_{n \to \infty} x_n = 1$ and $\lim_{n \to \infty} y_n - -1$. Based on Corollary 4.2.5, the limit does not exist.

(d)

$$\lim_{x \to 0} \sqrt[3]{x} (-1)^{[[1/x]]}$$

Solution: 0. $\lim_{x \to 0} \sqrt[3]{x} = 0$. and $|(-1)^{[[1/x]]}| \leq 1$. Based on exercise 4.2.7, it's 0.

4.2.9 (Infinite Limits)

Definition: $\lim_{x \to c} f(x) = \infty$ means that for all $M > 0$ we can find a $δ > 0$ such that whenever $0 < |x − c| < δ$, it follows that $f(x) > M$.

(a) Show $\lim_{x \to 0} 1/x^2 = \infty$ in the sense described in the previous definition.

Proof: Given $M > 0$, let $\delta = 1/\sqrt{M}$, then when $0 < |x| < 1/\sqrt{M}$, $1/x^2 > M$.

(b) Now, construct a definition for the statement $\lim_{x \to \infty} f(x) = L$. Show $\lim_{x \to \infty} 1/x = 0$.

Solution: for all $\epsilon > 0$, we can find a $N$ such that whenever $x > N$, $f(x) \in V_{\epsilon}(L)$.

To prove $\lim_{x \to \infty} 1/x = 0$, given $\epsilon > 0$, let $N = 1/\epsilon$.

(c) What would a rigorous definition for $\lim_{x \to \infty} f(x) = \infty$ look like? Give an example of such a limit.

Solution: It means that for all $M > 0$ we can find a $N > 0$ such that whenever $x > N$, it follows that $f(x) > M$. One example is $f(x) = x$.

4.2.10 (Right and Left Limits)

(a) Give a proper definition in the style of Definition 4.2.1 for the right-hand and left-hand limit statements:

$$\lim_{x \to a^+} f(x) = L \text{ and} \lim_{x \to a^-} f(x) = L$$

Solution: $\lim_{x \to a^+} f(x) = L$ means for all $\epsilon > 0$, we can find $\delta$, such that whenever $0 < x - a < \delta$, $|f(x) - L| < \epsilon$.

(b) Prove that $\lim_{x \to a} f(x) = L$ if and only if both the right and left-hand limits equal $L$.

Proof:

• $\Rightarrow$ if $\lim_{x \to a} f(x) = L$, then for any $\epsilon > 0$, we can find $\delta$, such that whenever $|x-a| < \delta$, $|f(x)- L| < \epsilon$. Then, we find $\delta$ such that whenever $0 < x - a < \delta$, $|f(x) - L| < \epsilon$. So $\lim_{x \to a^+} f(x) = L$.
• $\Leftarrow$ if $\lim_{x \to a^+} f(x) = L$ and $\lim_{x \to a^-} f(x) = L$. Then given $\epsilon > 0$, we can find $\delta^+, \delta^-$ such that as long as $0 < x - a < \delta^+$ or $0 < a - x < \delta^-$, $|f(x) - L| < \epsilon$. Let $\delta = \min \left\{ \delta^+, \delta^- \right\}$.

4.2.11 (Squeeze Theorem).

Let $f, g, h$ satisfy $f(x) \leq g(x) \leq h(x)$ for all $x$ in some common domain $A$. if $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} h(x) = L$ at some limit point $c$ of $A$, show $\lim_{x \to c} g(x) = L$.

Proof: Given any sequence $(x_n) \rightarrow c$, $f(x_n) \leq g(x_n) \leq h(x_n)$, so $\lim_{n \to \infty} g(x_n) = L$. According to Theorem 4.2.3, $\lim_{x \to c} g(x) = L$.

4.3 Continuous Functions

4.3.3

(a) Supply a proof for Theorem 4.3.9 using the $\epsilon–δ$ characterization of continuity.

Proof: Let $d = f(c)$. Since $g$ is continuous at $d$, then given any $\epsilon > 0$, we can find $V_{\delta }(d)$, as long as $x \in V_{\delta }(d)$, $g(x) \in V_{\epsilon }(g(d))$.

And since $f$ is continuous at $c$, then we can find $\zeta > 0$, as long as $x \in V_{\zeta }(c)$, $f(x) \in V_{\delta }(d)$. And in turn, $g(f(x)) \in V_{\epsilon}(g(f(c))$. So $g \circ f$ is continuous at $c$.

4.3.4

Assume $f$ and $g$ are defined on all of $\mathbf{R}$ and that $\lim_{x \to p} f(x) = q$ and $\lim_{x \to q} g(x) = r$.

(a) Give an example to show that it may not be true that

$$\lim_{x \to p} g(f(x)) = r.$$

Let $f$ be the Modified Dirichlet Function, i.e.

$$f(x) = \begin{cases} x &\text{if } x \in \mathbf{Q} \\ 0 &\text{if } x \not\in \mathbf{Q} \\ \end{cases}$$

Let $g(x)$ be

$$g(x) = \begin{cases} 1 &\text{if } x = 0 \\ 0 &\text{if } x \neq 0 \\ \end{cases}$$

Assume $(x_n) \rightarrow 0$ and $x_n$ are all irrational numbers. Then $g(f(x_n)) \rightarrow 1$.

On the other hand, $(y_n) \rightarrow 0$ and $y_n$ are all rational numbers, so $g(f(y_n)) \rightarrow 0$.

$\lim_{x \to p} g(f(x))$ does not exist at all.

(b) Show that the result in (a) does follow if we assume $f$ and $g$ are continuous.

Proof: This is Theorem 4.3.9.

(c) Does the result in (a) hold if we only assume $f$ is continuous? How about if we only assume that $g$ is continuous?

Proof: if only $f$ is continuous, then (a) might not hold. Let

$$f(x) = x \sin \frac{1}{x} \\ g(x) = \begin{cases} 1 &\text{if } x = 0 \\ 0 &\text{if } x \neq 0 \\ \end{cases}$$

Then $f(x)$ is continuous. Consider $(x_n) = \frac{1}{(2n \pi )}$, then $(x_n) \rightarrow 0$, and $\lim_{n \to \infty} g(f(x_n)) = \lim_{n \to \infty} 1 = 1$.

Now consider $(y_n) = \frac{1}{(2n + 1/2) \pi}$, then $(y_n) \rightarrow 0$, and $\lim_{n \to \infty} g(f(y_n)) = \lim_{n \to \infty} 0 = 0$. So $\lim_{x \to 0} g(f(x)) = 0$ does not exist.

Now if only $g$ is continuous. Let $\lim_{x \to q} g(x) = r = g(q)$. Then if $x \in V_{\delta}(q)$, then $g(x) \in V_{\epsilon}(r)$. Since $\lim_{x \to p} f(x) = q$, then we can find $\sigma$, such that $x \in V_{\sigma}(p)$, $f(x) \in V_{\delta}(q)$.

4.3.5

Show using Definition 4.3.1 that if $c$ is an isolated point of $A ⊆ \mathbf{R}$, then $f : A → \mathbf{R}$ is continuous at $c$.

Proof: Since $c$ is an isolated point of $A$, there is a $\delta$, such that $V_{\delta_0}(c) \cap A = \left\{ c \right\}$. Given any $\epsilon > 0$, as long as $|x - c| < \delta_0$, $|f(x) - f(c)| = |f(c) - f(c)| = 0 < \epsilon$ $\square$.

4.3.6

Provide an example of each or explain why the request is impossible.

(a) Two functions $f$ and $g$, neither of which is continuous at $0$ but such that $f(x)g(x)$ and $f(x) + g(x)$ are continuous at $0$.

Solution: Consider

$$f(x) = \begin{cases} 1 &\text{if } x = 0 \\ 0 &\text{if } x \neq 0 \\ \end{cases} \\ $$

and

$$g(x) = \begin{cases} 1 &\text{if } x \neq 0 \\ 0 &\text{if } x = 0 \\ \end{cases} \\ $$

Then $f(x)g(x) = 0$ and $f(x) + g(x) = 1$ are continuous everywhere.

(b) A function $f(x)$ continuous at 0 and $g(x)$ not continuous at $0$ such that $f(x) + g(x)$ is continuous at $0$.

Solution: it's not possible. Let $h(x) = f(x) + g(x)$. If $h(x), f(x)$ are continuous. Then $h(x) - f(x)$ will be continuous.

(c) A function $f(x)$ continuous at $0$ and $g(x)$ not continuous at $0$ such that $f(x)g(x)$ is continuous at $0$.

Solution: Let $f(x) = 0$, and

$$g(x) = \begin{cases} 1 &\text{if } x \neq 0 \\ 0 &\text{if } x = 0 \\ \end{cases} \\ $$

$\square$

(d) A function $f(x)$ not continuous at $0$ such that $f(x) + \frac{1}{f(x)}$ is continuous at $0$.

$$f(x) = \begin{cases} x + 2 &\text{if } x \neq 0 \\ \frac{1}{2} &\text{if } x = 0 \\ \end{cases}$$

$\square$

(e) A function $f(x)$ not continuous at 0 such that $[f(x)]^3$ is continuous at $0$.

Solution: this is not possible.

If $f(0) = 0$, then since $f^3(x)$ is continuous, we can find $\delta$ such that $|f^3(x)| < \epsilon^3$, then $|f(x)| < \epsilon$.

If $f(0) = c \neq 0$, then

$$|f(x) - f(0)| = |\frac{f^3(x) - f^3(0)}{f^2(x) + f(x)f(0) + f^2(0)}| \\ = |\frac{f^3(x) - f^3(0)}{(f(x) + f(0)/2)^2+3/4f^2(0)}| \\ \leq |\frac{f^3(x) - f^3(0)}{3/4f^2(0)}|$$

So $|f(x) - f(0)| < \epsilon$.

4.3.7

(a) Referring to the proper theorems, give a formal argument that Dirichlet’s function from Section 4.1 is nowhere-continuous on R.

Proof:

$$g(x) = \begin{cases} 1 &\text{if } x \in \mathbf{Q}\\ 0 &\text{if } x \not\in \mathbf{Q}\\ \end{cases}$$

Given any $x$, one sequence $(a_n) \rightarrow x$, where they are all rational numbers. Given $(b_n) \rightarrow x$, where they are all irrational numbers. According to Corollary 4.3.3, Dirichlet’s function from is nowhere-continuous on $\mathbf{R}$. $\square$

(b) Review the definition of Thomae’s function in Section 4.1 and demonstrate that it fails to be continuous at every rational point.

Proof: Given $(b_n) \rightarrow x$, where they are all irrational numbers.

(c) Use the characterization of continuity in Theorem 4.3.2 (iii) to show that Thomae’s function is continuous at every irrational point in $\mathbf{R}$. (Given $ε > 0$, consider the set of points $\{x ∈ R : t(x) ≥ ε\}$.)

Proof: For any irrational point, and any natural number $n$, we can find integer $m$, such that $m/n < x < (m+1)/n$. Let $\epsilon < 1/N$, and find $\delta$ such that none of the $2N$ points are in the $V_{\delta}(x)$. $\square$

4.3.8

Decide if the following claims are true or false, providing either a short proof or counterexample to justify each conclusion. Assume throughout that $g$ is defined and continuous on all of $\mathbf{R}$.

(a) If $g(x)≥0$ for all $x<1$ ,then $g(1)≥0$ as well.

Proof: True. Given $(x_n) \rightarrow 1$. Since $g(x_n) \geq 0$, then $g(1) = \lim_{n \to \infty} g(x_n) \geq 0$. $\square$.

(b) If $g(r)=0$ for all $r∈ \mathbf{Q}$ ,then $g(x)=0$ for all $x∈R$.

Proof: True. Given $(x_n) \rightarrow a$, where $a$ is an irrational number. $g(a) = \lim_{n \to \infty} g(x_n) = 0$.

4.3.10

Observe that if $a$ and $b$ are real numbers, then

$$\max \left\{ a, b \right\} = \frac{1}{2} [(a+b)+|a-b|].$$

1. show that if $f_1, f_2, \dots, f_n$ are continuous functions, then $g(x) = \max \left\{ f_1(x), f_2(x), \dots f_n(x) \right\}$ is a continuous function.

Proof: If $f(x)$ is a continuous function and given $c$, and $\epsilon > 0$, we can find $\delta > 0$, if $x \in V_{\delta }(c)$, then $|f(x) - f(c)| < \epsilon$. Since $||f(x)| - |f(c)|| \le |f(x) - f(c)| < \epsilon$, so $|f(x)|$ is also continuous function.

Then $g(x) = \max \left\{ f_1(x), f_2(x) \right\}= \frac{1}{2} [(f_1(x) + f_2(x)) + |f_1(x) - f_2(x)|]$ is also continuous.

Then we can use induction to see it's true for any $n$.

• another approach

For $c$, assume $g(c) = \max \left\{ f_1(c), f_2(c), \dots f_n(c) \right\} = f_1(c)$. Assume $(x_n) \rightarrow c$, and assume $(x_{n_k})$ is a subsequence and $g(x_{n_k}) = f_2(x_{n_k})$. Since $f_2$ is continuous, we have $(f_2(x_{n_k})) \rightarrow f_2(c)$. Since $\max \left\{ f_1(c), f_2(c), \dots f_n(c) \right\} = f_1(c)$, we have $f_1(c) \geq f_2(c)$. Since $f_1(x_{n_k}) \leq f_2(x_{n_k})$ and $(f_1(x_{n_k})) \rightarrow f_1(c)$, we have $f_1(c) \leq f_2(c)$. So $f_1(c) = f_2(c)$. This means $g(x_n) \rightarrow g(c)$. So $g(x)$ is continuous.

1. Let’s explore whether the result in (a) extends to the infinite case. For each $n ∈ \mathbf{N}$, define $f_n$ on $\mathbf{R}$ by

$$f_n(x) = \begin{cases} 1 &\text{if } |x| \geq 1/n \\ n|x| &\text{if } |x| < 1/n \\ \end{cases}$$

Now explicitly compute $h(x) = \sup \left\{f_1(x), f_2(x), f_3(x), \dots\right\}$.

Solution:

$$h(x) = \begin{cases} 1 &\text{if } x \neq 0 \\ 0 &\text{if } x = 0 \\ \end{cases}$$

So $h(x)$ is not continuous at $0$.

4.3.11 (Contraction Mapping Theorem).

Let $f$ be a function defined on all of $\mathbf{R}$, and assume there is a constant $c$ such that $0 <c< 1$ and

$$\left| f(x) - f(y) \right| \leq c|x-y|$$

for all $x, y \in \mathbf{R}$.

i. Show that $f(x)$ is continuous on $\mathbf{R}$.

Proof: Given $\epsilon$, let $\delta = \epsilon / c$. Then $\left| f(x) - f(y) \right| \leq c|x-y| < c (\epsilon / c) = \epsilon$.

ii. Pick some point $y_1 ∈ \mathbf{R}$ and construct the sequence

$$(y_1, f(y_1), f(f(y_1)), \dots)$$

In general, if $y_{n+1} = f(y_n)$, show that the resulting sequence $(y_n)$ is a Cauchy sequence. Hence we may let $y = \lim y_n$.

Proof: First, we prove $(y_n)$ is bounded. Assume $|y_2 - y_1| = M$, then $|y_2| \leq \left| y_1 \right| + M$.

$$\left| y_3 - y_2 \right| \leq c |y_2 - y_1| = cM \\ |y_3| \leq |y_2| + cM \leq |y_1| + (1 + c)M \\ |y_n| \leq |y_1| + \frac{1}{1 - c}M = K$$

So assume $n > m$,

$$|y_n - y_m| = c^{m-1} | y_{n-m+1} - y_1| \leq c^{m-1} 2K$$

So $(y_n)$ is Cauchy sequence.

iii. Prove that $y$ is a fixed point of $f$ (i.e., $f(y) = y$) and that it is unique in this regard.

Proof: Assume $f(y) \neq y$. Since $(y_n) \rightarrow y$, we can find $y_1$ and $y_2 = f(y_1)$, which are close enough to $y$. And $|y_1 - y| < \epsilon$, $|y_2 - y| < \epsilon$ but $|y_2 - f(y)| > \epsilon$. Then $\epsilon < |y_2 - f(y)| = |f(y_1)-f(y)| \leq c|y_1-y| < |y_1 - y| < \epsilon$, we have a contradiction.

iv. Finally, prove that if $x$ is any arbitrary point in , then the sequence $(x, f(x), f(f(x)),...)$ converges to $y$ defined in (2).

Proof: Assume $(x_n) \rightarrow x$ and $(y_n) \rightarrow y$, then $|x - y| \le |x - x_n | + |x_n - y_n | + | y_n - y | < \epsilon + c^n|x - y| + \epsilon$ and can be arbitrarily small. So $x = y$.

4.3.12

Let $F ⊆ R$ be a nonempty closed set and define $g(x) = \inf\{|x − a| : a ∈ F\}$. Show that $g$ is continuous on all of $\mathbf{R}$ and $g(x) \neq 0$ for all $x \not\in F$.

Proof:

• If $c \in F$, then $g(c) = 0$, $|g(x) - g(c)| = |g(x)| \leq |x - c|$. So $g$ is continuous at $c$.
• If $c \not\in F$, then we can find $x_1, x_2 \in F$ such that $c \in (x1, x2) \cap F = \emptyset$. So $g(c) = \min \left\{ c - x1, x2 - c \right\} \neq 0$. Choose $\delta < \min \left\{ c - x_1, x_2 - c \right\}$, then $|g(x) - g(c)| \leq |x - c|$, so $g(x)$ is also continuous at $c \not\in F$.

4.3.13

Let $f$ be a function defined on all of $\mathbf{R}$ that satisfies the additive condition $f (x + y) = f (x) + f (y)$ for all $x, y ∈ \mathbf{R}$.

(a) Show that $f(0)=0$ and that $f(−x)=−f(x)$ for all $x∈\mathbf{R}$.

Proof: $f(0) = f(0+0) = f(0) + f(0)$, so $f(0) = 0$. $f(x) + f(-x) = f(x-x) = f(0) = 0$, so $f(-x) = -f(x)$.

(b) Let $k=f(1)$. Show that $f(n)=kn$ for all $n∈N$,and then prove that $f(z) = kz$ for all $z ∈ Z$. Now, prove that $f(r) = kr$ for any rational number $r$.

Proof: let $r = p/q$. $f(1) = f(1/q + \dots+1/q) = qf(1/q)$, so $f(q) = f(1)/q$. So $f(r) = kr$.

(c) Show that if $f$ is continuous at $x = 0$, then $f$ is continuous at every point in $\mathbf{R}$ and conclude that $f(x) = kx$ for all $x ∈ R$.

Proof: $|f(x) - f(y)| = |f(x-y)|$, since $f(0)$ is continuous at 0, then $|f(x-y)| < \epsilon$. So $f$ is continuous at every point in $\mathbf{R}$. We can find a rational sequence converges to $x$, so $f(x) = kx$.

4.3.14

(a) Let $F$ be a closed set. Construct a function $f : \mathbf{R} \mapsto \mathbf{R}$, such that the set of points where $f$ fails to be continuous is precisely $F$. (The concept of the interior of a set, discussed in Exercise 3.2.14, may be useful.)

Solution: consider

$$f(x) = \begin{cases} 0 &\text{if } x \not\in F\\ 1 &\text{if } x \in F \cap \mathbf{Q}\\ -1 &\text{if } x \in F \cap \overline{\mathbf{Q}}\\ \end{cases}$$

If $x \in E^o$, then we can find $(x-\epsilon , x+\epsilon) \subset F$, since both $\mathbf{Q}$ and

(b) Now consider an open set $O$. Construct a function $g : R → R$ whose set of discontinuous points is precisely $O$. (For this problem, the function in Exercise 4.3.12 may be useful.)

Solution: consider $F = \mathbf{R} - O$ and $g(x)$ is in Exercise 4.3.12.

$$f(x) = \begin{cases} 0 &\text{if } x \in O \cap I \\ g(x) &\text{otherwise} \\ \end{cases}$$

First, $f(x)$ is continuous at $c \in F$. It can be proved the same way in 4.3.12.

Next if $x \in O$ is rational, then $g(x) > 0$. But we can find a sequence of irrational numbers $(x_n) \rightarrow x$ and $f(x_n) = 0$. So $\lim_{n \to \infty} f(x_n) = 0 \neq f(x)$.

Finally if $x \in O$ is irrational, and assume $x_1 < y_1 < x < y_2 < x_2$ where $x_1, x_2 \in F$, $y_1, y_2 \in O \cap \mathbf{Q}$. Given any $y_1 < y < y_2 \in \mathbf{Q}$, $g(y) > \min \left\{ y_1-x_1, x_2-y_2\right\} > 0$. Since $f(x) = 0$, then $f(x)$ is not continuous at $x$.

4.4 Continuous Functions on Compact Sets

4.4.4

Decide whether each of the following statements is true or false, justifying each conclusion.

(a) If $f$ is continuous on $[a,b]$ with $f(x) > 0$ for all $a ≤ x ≤ b$, then $1/f$ is bounded on $[a, b]$ (meaning $1/f$ has bounded range).

Solution: True. Since $f$ is continuous on $[a,b]$. then $f(x)$ attains maximum and minimum value, say $c$ and $d$ which are all $> 0$. Then $1/f$ is bounded by $[1/c, 1/d]$.

(b) If $f$ is uniformly continuous on a bounded set $A$, then $f(A)$ is bounded.

Solution: True. Assume otherwise, we can find a sequence $(x_n)$ such that $\lim_{n \to \infty} f(x_n) = \infty$. Since $A$ is bounded, we can assume $(x_n) \rightarrow x$. Given $\epsilon_0$ and any $\delta$, we can find $\delta' < \delta/2$, such that $x_1, x_2 \in V_{\delta /2}(x)$ and $|f(x_1) - f(x_2)| > \epsilon$. So $f$ is not uniformly continuous. We have a contradiction.

(c) If $f$ is defined on $\mathbf{R}$ and $f(K)$ is compact whenever $K$ is compact, then $f$ is continuous on $\mathbf{R}$.

Solution: False. Consdier Dirichlet function. $f(K) = \{0, 1\}$ is always compact.

4.4.6

Give an example of each of the following, or state that such a request is impossible. For any that are impossible, supply a short explanation for why this is the case.

(a) A continuous function $f : (0,1) → R$ and a Cauchy sequence $(x_n)$ such that $f(x_n)$ is not a Cauchy sequence.

Solution: Consdier $f(x) = 1/x$ and $x_n = 1/n$.

(b) A uniformly continuous function $f : (0, 1) → R$ and a Cauchy sequence $(x_n)$ such that $f(x_n)$ is not a Cauchy sequence;

Solution: This is impossible. Given $\epsilon$, we can find $\delta$, such that $|x_n - x_m| < \delta$, $|f(x_n) - f(x_m)| < \epsilon$. We can also find $N$, as long as $n, m > N$ then $|x_n - x_m| < \delta$。 Then we have $|f(x_n) - f(x_m)| < \epsilon$.

(c) A continuous function $f : [0,∞) → \mathbf{R}$ and a Cauchy sequence $(x_n)$ such that $f (x_n)$ is not a Cauchy sequence.

Solution: This is impossible. Since $(x_n)$ is Cauchy sequence, then it's bounded. Assume $x_n \in [0, M]$, then $f$ is uniformly continuous. So $f (x_n)$ must be Cauchy.

4.4.7.

Prove that $f (x) = \sqrt[]{x}$ is uniformly continuous on $[0, ∞)$.

Proof: If $x_1, x_2 \geq 1$, then

$$|\sqrt[]{x_1} - \sqrt[]{x_2}| = |(x_1 - x_2)/(\sqrt[]{x_1} + \sqrt[]{x_2})| \leq |(x_1 - x_2)/2|$$

So $f(x)$ is uniformly continuous on $[1, \infty ]$. It's also uniformly continuous on $[0,1]$ since $[0,1]$ is a closed set. Similar to Exercise 4.4.5, $f(x)$ is uniformly continuous on $[0, ∞)$.

4.4.8

Give an example of each of the following, or provide a short argument for why the request is impossible.

(a) A continuous function defined on $[0, 1]$ with range $(0, 1)$.

Solution: this is impossible. According to Theorem 4.4.1, $f(K)$ must be compact. But $(0,1)$ is not.

(b) A continuous function defined on $(0, 1)$ with range $[0, 1]$.

Solution: Consider $f(x) = |\sin 2 \pi x|$

(c) A continuous function defined on $(0, 1]$ with range $(0, 1)$.

Solution: Consdier

$$f(x) = \frac{1}{2} + \frac{(1-x)}{2} \sin \frac{1}{x}$$

4.4.9

(Lipschitz Functions). A function $f : A → R$ is called Lipschitz if there exists a bound $M > 0$ such that

$$|\frac{f(x) - f(y)}{x - y}| \leq M$$

for all $x \neq y ∈ A$. Geometrically speaking, a function $f$ is Lipschitz if there is a uniform bound on the magnitude of the slopes of lines drawn through any two points on the graph of $f$.

(a) Show that if $f : A → \mathbf{R}$ is Lipschitz, then it is uniformly continuous on $A$.

Proof: Given $\epsilon$, we can find $\delta = \epsilon / M$.

(b) Is the converse statement true? Are all uniformly continuous functions necessarily Lipschitz?

Solution: False. Consider $f(x) = x^2 : \mathbf{N} \rightarrow \mathbf{R}$. It's uniformly continuous. But it's not Lipschitz.

4.4.10

Assume that $f$ and $g$ are uniformly continuous functions defined on a common domain $A$. Which of the following combinations are necessarily uniformly continuous on $A$:

(a) $f(x) + g(x)$

Solution: yes.

(b) $f(x)g(x)$

Solution: No. Consider $f(x) = g(x) = x$.

(c) $f(x)/g(x)$

Solution: No. Consider $f(x) = x, g(x) = 1/x, A = [1, \infty ]$.

(d) $f(g(x))$

Solution: Yes. Given $\epsilon$, since $f$ is uniformly continuous, we can find $\zeta$, as long as $|y_1 - y_2| < \zeta$, $|f(y_1) - f(y_2)| < \epsilon$. Since $g$ is also uniformly continuous, we can find $\delta$, as long as $|x_1 - x_2| < \delta$, $|g(x_1) - g(x_2)| < \zeta$, so $|f(g(x_1)) - f(g(x_2))| < \epsilon$.

4.4.11 (Topological Characterization of Continuity)

Let $g$ be defined on all of $\mathbf{R}$. If $B$ is a subset of $\mathbf{R}$, define the set $g^{−1}(B)$ by

$$g^{−1}(B)= \left\{ x \in \mathbf{R} : g(x) \in B \right\}$$

Show that $g$ is continuous if and only if $g^{−1}(O)$ is open whenever $O ⊆ \mathbf{R}$ is an open set.

Proof: $\Rightarrow$ Assume $g$ is continuous, and $O \subseteq \mathbf{R}$ is open. $x \in g^{−1}(O)$, so $y = f(x) \in O$. Since $O$ is open, we can find $V_{\epsilon}(y) \subseteq O$. Since $g$ is continuous, we can find $U_{\delta}(x)$ such that $f(U) \subseteq O$, so $U \subseteq g^{−1}(O)$. Which means $g^{−1}(O)$ is open.

$\Leftarrow$: Assume $f(x) = y$, then $V_{\epsilon}(y)$ is open, so $x \in g^{-1}(V_{\epsilon}(y))$ is open. Then we can fine $V_{\delta }(x) \subset g^{-1}(V_{\epsilon}(y))$, so $f(V_{\delta }(x)) \subseteq V_{\epsilon}(y)$. So $g$ is continuous.

4.4.12.

Review Exercise 4.4.11, and then determine which of the following statements is true about a continuous function defined on R:

(a) $f^{-1}(B)$ is finite whenever $B$ is finite.

Solution: False. Consider a constant function: $f(x) = 0$.

(b) $f^{−1}(K)$ is compact whenever $K$ is compact.

Solution: False. Consider $f(x) = \sin x$. When $K = [-1, 1]$, then $f^{−1}(K) = \mathbf{R}$ which is not compact.

(c) $f^{−1}(A)$ is bounded whenever $A$ is bounded.

Solution: Same example from (b).

(d) $f^{−1}(F)$ is closed whenever $F$ is closed.

Solution: False. Consider $f(x) = \ln x$, and $F = \mathbf{R}$. Then $f^{−1}(F) = (0, +\infty)$ which is open.

4.4.13 (Continuous Extension Theorem)

(a) Show that a uniformly continuous function preserves Cauchy sequences; that is, if $f : A → R$ is uniformly continuous and $(x_n) ⊆ A$ is a Cauchy sequence, then show $f(x_n)$ is a Cauchy sequence.

4.5 The Intermediate Value Theorem

4.5.1

Show how the Intermediate Value Theorem follows as a corol- lary to Theorem 4.5.2.

Proof: Let $C = f([a, b])$, since $[a, b]$ is connected and $f$ is continuous, then according to Theorem 4.5.2 $C$ is connected also. Since $f(a), f(b) \in C$, then $[f(a), f(b)] \subseteq C$, then $L \in C$. Then there exists a point $c ∈ (a,b)$ where $f(c) = L$. $\square$

4.5.2

Provide an example of each of the following, or explain why the request is impossible.

(a) A continuous function defined on an open interval with range equal to a closed interval.

Solution: Consider $\sin x$ defined on $(-100 \pi , +100 \pi )$. The range of it is $[-1, +1]$

(b) A continuous function defined on a closed interval with range equal to an open interval.

Solution: If the closed interval is bounded, then it's compact. Then the range has to be compact, which is closed and bounded. So it's impossible for the range to be open.

On the other hand, if the closed interval is unbounded, e.g. $[0, +\infty)$, then consider $x \sin x$. Then range is $\mathbf{R}$, which is open.

(c) A continuous function defined on an open interval with range equal to an unbounded closed set different from $\mathbf{R}$.

Solution: Consider $f(x) = \frac{1}{x^2 - 1}$ defined on $(-1, 1)$. The range is $(-\infty, -1]$.

(d) A continuous function defined on all of $\mathbf{R}$ with range equal to $\mathbf{Q}$.

Solution: this is impossible. According to theorem 4.5.2, $f(\mathbf{R})$ is connected. But $\mathbf{Q}$ is not connected.

4.5.3

A function $f$ is increasing on $A$ if $f(x) ≤ f(y)$ for all $x < y$ in $A$. Show that if $f$ is increasing on $[a, b]$ and satisfies the intermediate value property (Definition 4.5.3), then $f$ is continuous on $[a, b]$.

Proof: Let $c \in (a, b)$ and any $\epsilon$ such that $f(a) \leq f(c) - \epsilon < f(c) \leq f(c) + \epsilon \leq f(b)$. Since $f(x)$ satisfies intermediate value property, we can find $c_0, c_1$ such that $f(c_0) = f(c) - \epsilon$ and $f(c_1) = f(c) + \epsilon$. Since $f$ is increasing, we have $c_0 < c < c_1$.

Let $\delta = \min \left\{ c - c_0, c_1 - c \right\}$. If $x \in V_{\delta}(c)$, $x \in (c_0, c_1)$ and $f(x) \in (f(c_0), f(c_1))$. So $\left| f(c) - f(x) \right| < \epsilon$.

$\square$

4.5.4

Let $g$ be continuous on an interval $A$ and let $F$ be the set of points where $g$ fails to be one-to-one; that is,

$$F = \left\{ x \in A: f(x) = f(y) \text{ for some } y \neq x \text{ and } y \in A \right\}$$

Show $F$ is either empty or uncountable.

Proof: If $F$ is not empty, then let $a, b \in F$ and $a < b$.

If $f(x)$ is the same for all $x \in [a, b]$, then $F \supseteq [a, b]$, which is uncountable.

If $c \in [a, b]$ and $f(c) \neq f(a)$. Assume $f(c) > f(a)$, then using (IVF), for every $k \in (f(a), f(c))$ we can find $d \in (a, c)$ and $e \in (c, b)$, such that $f(d) = f(e) = k$. Since $(f(a), f(c))$ is uncountable, then $F$ is uncountable. $\square$.

4.5.5

(a) Finish the proof of the Intermediate Value Theorem using the Axiom of Completeness started previously.

Proof: continue with where we left. We first show $f(c) \not > 0$. If $f(c) = a > 0$, then since $f(x)$ is continuous, there is a $V_{\epsilon}(c)$, such that as long as $x \in V_{\epsilon}(c)$, $f(x) > 0$. Then $c$ is not $\sup K$. So $f(c) \not > 0$.

Next we show $f(c) \not < 0$. The same argument above can be applied to this scenario.

So $f(c) = 0$. $\square$

(b) Finish the proof of the Intermediate Value Theorem using the Nested Interval Property started previously.

Proof: $(a_n) \rightarrow c$. Since $f$ is continuous, $f(a_n) \rightarrow f(c)$. Since $f(a_n) < 0$, then $f(c) \leq 0$. For the same reason, $f(c) \geq 0$. So $f(c) = 0$. $\square$

4.5.6

Let $f : [0, 1] → \mathbf{R}$ be continuous with $f(0) = f (1)$.

(a) Show that there must exist $x, y ∈ [0, 1]$ satisfying $|x − y| = 1/2$ and $f(x) = f(y)$.

Proof: define $g(x) = f(x - \frac{1}{2}), x \in [\frac{1}{2}, 1]$. And $h(x) = f(x) - g(x)$.

If $f(1/2) = f(0)$ we already found the desired $x, y$.

If $f(1/2) > f(0)$, then $h(1/2) > 0, h(1) < 0$. Since $h(x)$ is continuous, according to IVF, we can find $c$ such that $h(c) = 0$.

The same argument can be applied when $f(1/2) < f(0)$. $\square$

(b) Show that for each $n ∈ N$ there exist $x_n,y_n ∈ [0,1]$ with $|x_n − y_n| = 1/n$ and $f(x_n) = f(y_n)$.

Proof: Define $h(x) = f(x) - f(x - \frac{1}{n}), x \in [\frac{1}{n}, 1]$.

If $f(\frac{1}{n}) > f(0)$, we can find $1 \leq k \lt n$, such that $h(\frac{k}{n}) > 0$ and $h(\frac{k+1}{n}) < 0$. Then we find $c \in (\frac{k}{n}, \frac{k+1}{n})$, such that $h(c) = 0$.

The same argument can be applied when $f(1/n) < f(0)$. $\square$

(c) If $h ∈ (0,1/2)$ is not of the form $1/n$, there does not necessarily exist $|x − y| = h$ satisfying $f(x) = f(y)$. Provide an example that illustrates this using $h = 2/5$.

Solution: Consider the following piecewise linear function.

$$f(0) = 0 \\ f(\frac{1}{5}) = -2 \\ f(\frac{2}{5}) = +1 \\ f(\frac{3}{5}) = -1 \\ f(\frac{4}{5}) = +2 \\ f(1) = 0 \\ $$

Think about this way, the $g(x) = f(x - h)$ is a $f(x)$ right shift by $h$. If $g(x)$ and $f(x)$ does not have intersection, then there does not exist $|x − y| = h$ satisfying $f(x) = f(y)$.

So we can construct a zig-zag piecewise linear function. Assume $\frac{1}{n+1} < h < \frac{1}{n}$ and $nh + \delta = 1$ and $\delta < h$.

$$f(0) = 0 \\ f(\delta) = -n \\ f(h) = +1 \\ f(h + \delta ) = -(n-1) \\ f(2h) = +2 \\ \cdots \\ f(nh) = n \\ f(nh + \delta) = f(1) = 0 \\ $$

$\square$

4.5.7

Let $f$ be a continuous function on the closed interval $[0,1]$ with range also contained in $[0, 1]$. Prove that f must have a fixed point; that is, show $f(x) = x$ for at least one value of $x ∈ [0,1]$.

Proof: If $f(0) = 0$, then we found it. If $f(1) = 1$, we also found it. Assume $f(0) > 0, f(1) < 1$, consider $g(x) = f(x) - x$, which is continuous and $g(0) > 0, g(1) < 0$. Then according to IVT, $\exists c$ such that $g(c) = 0$. Then $f(c) = c$.

4.5.8 (Inverse functions).

If a function $f : A → R$ is one-to-one, then we can define the inverse function $f^{−1}$ on the range of $f$ in the natural way: $f^{−1}(y) = x$ where $y = f(x)$. Show that if $f$ is continuous on an interval $[a, b]$ and one-to-one, then $f^{−1}$ is also continuous.

Proof: Assume $f(a) < f(b)$, we first prove $f$ is an increasing function. Assume otherwise, then we can find $c < d$ with $f(c) > f(d)$. * If $f(a) < f(c)$, then we find $a < c < d$ with $f(a) < f(c) > f(d)$. * If $f(a) > f(c)$, then we find $c < d < b$ with $f(c) > f(d) < f(b)$.

In both cases, we will find $f$ is not one-to-one. So $f$ has to be increasing function.

If $x < y$ and $f(c) = x, f(d) = y$, then $c < d$. That means $f^{-1}(x) < f^{-1}(y)$. So $f^{-1}$ is also increasing.

And $f^{-1}$ satisfies intermediate value property. Based on exercise 4.5.3, $f^{-1}$ is also continuous. $\square$

4.6 The Intermediate Value Theorem

4.6.1.

Using modifications of these functions, construct a function $f : \mathbf{R} → \mathbf{R}$ so that

(a) $D_f = \mathbf{Z}^c$

Solution: we can modify the modified Dirichlet's function

$$f(x) = \begin{cases} x & x \text{ is rational}\\ k & x \text{ is irrational and } x \in (\frac{2k-1}{2}, \frac{2k+1}{2}) \\ \end{cases}$$ $\square$

(b) $D_f = \{ x: 0 < x \leq 1\}$

Solution: Consdier

$$f(x) = \begin{cases} 0 & x < 0 \text{ or } x > 1\\ h(x) & x \in [0, 1]\\ \end{cases}$$ $\square$

4.6.2.

Given a countable set $A = \{a_1,a_2,a_3,...\}$, define $f(a_n) = 1/n$ and $f(x)=0$ for all $x \not ∈A$. Find $D_f$.

Solution: $D_f = A$. If $b \not \in A$, then $f(b) = 0$. If $b$ is a limit point of $A$, then given $\epsilon$, we can find $N$ such that $n > N, 1/n < \epsilon$. Then we can find $\delta$, such that for $n \leq N$, $a_n \not \in V_{\delta}(b)$.

Then for any $x \in V_{\delta}(b)$, $f(x) < \epsilon$. So $f$ is continuous at $b$.

It's easier to see $f$ is continuous at $b$ if $b$ is not a limit point of $A$.

For any $a_n$ there always exists $x \in V_{\delta}(a_n)$ but $x \not \in A$ for any $\delta$, so $f$ is not continuous at $a_n$. $\square$

4.6.3

State a similar definition for the left-hand limit

$$\lim_{x \to c^-} f(x) = L$$

Solution: if for all $ε > 0$ there exists a $δ > 0$ such that $|f(x)−L| < ε$ whenever $-δ < x−c < 0$.

Equivalently, in terms of sequences, $\lim_{x \to c^-} f(x) = L$ if $\lim f(x_n) = L$ for all sequences $(x_n)$ satisfying $x_n < c$ and $\lim (x_n) = c$.

4.6.4

Supply a proof for Theorem 4.6.3: Given $f : A → R$ and a limit point $c$ of $A$, $\lim_{x \to c} f(x) = L$ if and only if

$$\lim_{x \to c^+} f(x) = L \text{ and } \lim_{x \to c^-} f(x) = L.$$

Proof: $\Rightarrow$ Given any sequence $(x_n)$ satisfying $x_n > c$ and $\lim (x_n) = c$, $f$ is continuous, then $\lim_{x \to c} f(x_n) = L$. That proves $\lim_{x \to c^+} f(x) = L$.

Same thing when $x_n < c$.

$\Leftarrow$: Given any sequence $(x_n)$ such that $\lim (x_n) = c$. Then we can divide it into 2 subsequences $(x^+_n) > c$ and $(x^-_n) < c$.

We have the following: * $\lim (x^+_n) = c$, so $\lim f(x^+_n) = L$. * $\lim (x^-_n) = c$, so $\lim f(x^-_n) = L$.

That means $\lim f(x_n) = L$. $\square$

4.6.5.

Prove that the only type of discontinuity a monotone function can have is a jump discontinuity.

Proof: Assume $f$ is a monotonous increasing function and it's not continuous at $c$. * Let $A = \{f(x)|x < c\}$. $f(x)$ is upper bounded by $f(c)$. Let $L^- = \sup A$, then $L^- = \lim f(x^-_n)$. * Let $B = \{f(x)|x > c\}$. $f(x)$ is lower bounded by $f(c)$. Let $L^+ = \inf B$, then $L^+ = \lim f(x^+_n)$.

That proves $f(x)$ can only have jump discontinuity.

4.6.6.

Construct a bijection between the set of jump discontinuities of a monotone function f and a subset of $\mathbf{Q}$. Conclude that $D_f$ for a monotone function $f$ must either be finite or countable, but not uncountable.

Proof: For any discontinuited point $c$, let $a = \lim_{x \to c^+} f(x)$ and $b = \lim_{x \to c^-} f(x)$. Since $a \not = b$, we can find a rational number $q \in (a, b)$. That means this mapping is surjective.

Assume $c_0 < c_1$, since $f$ is monotone increasing, we have

$$\lim_{x \to c^-_0} f(x) < \lim_{x \to c^+_0} f(x) \le \lim_{x \to c^-_1} f(x) < \lim_{x \to c^+_1} f(x)$$

So basically $(a_{c_i}, b_{c_i})$ does not intersect with each other. Then we find a one-to-one mapping between the set of jump discontinuities and a subset of $\mathbf{Q}$. So this is bijection.

We proved this problem. $\square$

4.6.7.

In Section 4.1 we constructed functions where the set of discontinuity was $\mathbf{R}$ (Dirichlet’s function), $\mathbf{R} \setminus \{0\}$ (modified Dirichlet function), and $\mathbf{Q}$ (Thomae’s function).

(a) Show that in each of the above cases we get an $F_σ$ set as the set where the function is discontinuous.

Proof:

• Dirichlet’s function: $\mathbf{R}$ it self is a closed set.
• modified Dirichlet function: $\mathbf{R} \setminus \{0\}$
  • Consider $A_n = (-\infty, -1/n] \cup [1/n, \infty)$. $A_n$ is closed. $\cup_{n = 1}^{\infty}A_n = \mathbf{R} \setminus \{0\}$.
• $\mathbf{Q}$ (Thomae’s function): A set contains a single rational number is a closed set. And $\mathbf{Q}$ is countable.

(b) Show that the two sets of discontinuity in Exercise 4.6.1 are $F_σ$ sets.

• $\mathbf{Z}^c$: Consider

$$A_n = \{\cup_i [i+1/n, (i+1)-1/n] | n > 2\}$$

$A_n$ is countable union of closed set, so $A_n$ is a closed set. Also $\cup A_n = \mathbf{Z}^c$.

• $D_f = \{ x: 0 < x \leq 1\}$

Consider

$$A_n = \{(-\infty, -1/n]\} \\ B = [1, \infty)$$

So $D_f = B \cup (\cup A_n)$.

4.6.8.

Prove that, for a fixed $α > 0$, the set $D_f^α$ is closed.

Proof: Let $x$ be a limit point of $D_f^α$. Given any $\delta$, we can find $a \in U_{\delta}(x)$ such that $a \in D_f^α$, since $U_{\delta}(x)$ is open, we can find $U_{\zeta}(a) \subset U_{\delta}(x)$. Since $a \in D_f^α$, we can find $y, z \in U_{\zeta}(a)$ with $|f(y) − f(z)| \ge α$. Since $y, z \in U_{\delta}(x)$, $x$ is also not $α$-continuous. $\square$

4.6.9.

If $α < α'$, show that $D_f^{α'} ⊆ D_f^{α}$.

Proof: If $x \in D_f^{α'}$, then given $\delta$, we can find $y, z \in V_{\delta}(x)$ such that $|f(y) - f(z)| > \alpha' > \alpha$. So $x \in D_f^{α}$. Then $D_f^{α'} ⊆ D_f^{α}$. $\square$

4.6.10.

Let $α > 0$ be given. Show that if $f$ is continuous at $x$, then it is $α$-continuous at $x$ as well. Explain how it follows that $D_f^α ⊆ D_f$.

Proof: If $f$ is continuous at $x$, then we can find a $U_{\delta}(x) = (x−δ, x+δ)$ such that $|y-x| < \alpha /2$ and $|z-x| < \alpha /2$. Then $|y-z| < \alpha$. So $f$ is $α$-continuous at $x$ as well.

Then given $x \in D_f^{α'}$, $f$ has to be discontinuous at $x$, so $x \in D_f^{α}$. Then $D_f^{α'} ⊆ D_f^{α}$.

4.6.11.

Show that if $f$ is not continuous at $x$, then $f$ is not $α$-continuous for some $α$ > 0. Now explain why this guarantees that

$$D_f = \displaystyle \cup_{n = 1}^{\infty} D_f^{\alpha_n}$$

where $α_n = 1/n$.

Proof: If $f$ is not continuous at $x$, then exists $\epsilon$, given any $\delta$, we can find $y \in U_{\delta}(x)$ such that $|f(y)-f(x)| >= \epsilon$. So $f$ is not $\epsilon$-continuous at $x$. We can find $n$ such that $1/n < \epsilon$.

According to 4.6.9, $x \in D_f^{\epsilon} ⊆ D_f^{1/n}$. Then it follows:

$$D_f = \displaystyle \cup_{n = 1}^{\infty} D_f^{\alpha_n}$$

$\square$