Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Question
Consider the function \( f : \mathbb{R} \to \mathbb{R} \) given piecewise by \[ f(x) = \begin{cases} 1 & \text{if } x \text{ is irrational} \\ 0 & \text{if } x \text{ is rational} \end{cases} \] Now let \( c \) be any real number in \( \mathbb{R} \). **Case 1:** If \( c \) is rational, choose a sequence \( (x_n : n \in \mathbb{N}) \) of irrational numbers converging to \( c \) (which is allowed by Proposition 4.6.1). So \( x_n \to c, f(x_n) = 1 \) for all \( n \), and \( f(c) = 0 \). But since \( f(x_n) = 1 \) for all \( n \), we cannot have \( f(x_n) \to 0 \) (a sequence of 1's cannot converge to 0!). That is, it is impossible that \( f(x_n) \to f(c) \), so by Theorem 4.6.7, \( f \) cannot be continuous at \( c \). **What is Case 2?**
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Transcribed Image Text:Consider the function \( f : \mathbb{R} \to \mathbb{R} \) given piecewise by \[ f(x) = \begin{cases} 1 & \text{if } x \text{ is irrational} \\ 0 & \text{if } x \text{ is rational} \end{cases} \] Now let \( c \) be any real number in \( \mathbb{R} \). **Case 1:** If \( c \) is rational, choose a sequence \( (x_n : n \in \mathbb{N}) \) of irrational numbers converging to \( c \) (which is allowed by Proposition 4.6.1). So \( x_n \to c, f(x_n) = 1 \) for all \( n \), and \( f(c) = 0 \). But since \( f(x_n) = 1 \) for all \( n \), we cannot have \( f(x_n) \to 0 \) (a sequence of 1's cannot converge to 0!). That is, it is impossible that \( f(x_n) \to f(c) \), so by Theorem 4.6.7, \( f \) cannot be continuous at \( c \). **What is Case 2?**
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