Problem 108. (a) Use the definition of continuity to show that the function Jz, ifr is rational o, r is irrational D() - is continuous at 0.

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#108 part a
# Continuity: What It Isn't and What It Is

### Problem 108

(a) Use the definition of continuity to show that the function

\[ 
D(x) = 
\begin{cases} 
1, & \text{if } x \text{ is rational} \\ 
0, & \text{if } x \text{ is irrational} 
\end{cases}
\]

is continuous at 0.

(b) Let \( a \neq 0 \). Use the definition of continuity to show that \( D \) is not continuous at \( a \). (Hint: You might want to break this up into two cases where \( a \) is rational or irrational. Show that no choice of \( \delta > 0 \) will work for \( \epsilon = |\frac{a}{4}| \). Note that Theorem 9 of Chapter 11 will probably help here.)

### 6.2 Sequences and Continuity

There is an alternative way to prove that the function

\[ 
D(x) = 
\begin{cases} 
1, & \text{if } x \text{ is rational} \\ 
0, & \text{if } x \text{ is irrational} 
\end{cases}
\]

is not continuous at \( a \neq 0 \). We will examine this by looking at the relationship between our definitions of convergence and continuity. The two ideas are actually quite closely connected, as illustrated by the following very useful theorem.

**Theorem 15.** The function \( f \) is continuous at \( a \) if and only if it satisfies the following property:

For all sequences \( (x_n) \), if \( \lim x_n = a \) then \( \lim f(x_n) = f(a) \).

Theorem 15 says that in order for \( f \) to be continuous, it is necessary and sufficient that any sequence \( (x_n) \) converging to \( a \) must force the sequence \( (f(x_n)) \) to converge to \( f(a) \). A picture of this situation is below though, as always, the formal proof will not rely on the diagram.

### Diagram Explanation

The diagram shows a graph illustrating the function \( f(x) \) and the concept of continuity through sequences. The x-axis represents various sequence values \( x_1, x_2, x_3
Transcribed Image Text:# Continuity: What It Isn't and What It Is ### Problem 108 (a) Use the definition of continuity to show that the function \[ D(x) = \begin{cases} 1, & \text{if } x \text{ is rational} \\ 0, & \text{if } x \text{ is irrational} \end{cases} \] is continuous at 0. (b) Let \( a \neq 0 \). Use the definition of continuity to show that \( D \) is not continuous at \( a \). (Hint: You might want to break this up into two cases where \( a \) is rational or irrational. Show that no choice of \( \delta > 0 \) will work for \( \epsilon = |\frac{a}{4}| \). Note that Theorem 9 of Chapter 11 will probably help here.) ### 6.2 Sequences and Continuity There is an alternative way to prove that the function \[ D(x) = \begin{cases} 1, & \text{if } x \text{ is rational} \\ 0, & \text{if } x \text{ is irrational} \end{cases} \] is not continuous at \( a \neq 0 \). We will examine this by looking at the relationship between our definitions of convergence and continuity. The two ideas are actually quite closely connected, as illustrated by the following very useful theorem. **Theorem 15.** The function \( f \) is continuous at \( a \) if and only if it satisfies the following property: For all sequences \( (x_n) \), if \( \lim x_n = a \) then \( \lim f(x_n) = f(a) \). Theorem 15 says that in order for \( f \) to be continuous, it is necessary and sufficient that any sequence \( (x_n) \) converging to \( a \) must force the sequence \( (f(x_n)) \) to converge to \( f(a) \). A picture of this situation is below though, as always, the formal proof will not rely on the diagram. ### Diagram Explanation The diagram shows a graph illustrating the function \( f(x) \) and the concept of continuity through sequences. The x-axis represents various sequence values \( x_1, x_2, x_3
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