1 f (x) = 1 if x is irrational if x is rational

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 63RE
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Consider the functionf : R → R given piecewise by
f (2) = { 0
1
f (x) =
if æ is irrational
if x is rational
Now let c be any real number in R.
Case 1: If c is rational, choose a sequence (xn :n € N) of irrational numbers converging to c (which is
allowed by Proposition 4.6.11). So æn → c, f(xn) = 1 for all n, and f(c) = 0. But since f(æn) = 1 for
all n, we cannot have f(xn) → 0 (a sequence of l's cannot converge to 0!). That is, it is impossible that
f(2n) → f(c), so by Theorem 4.6.7, f cannot be continuous at c.
What is Case 2?
Transcribed Image Text:Consider the functionf : R → R given piecewise by f (2) = { 0 1 f (x) = if æ is irrational if x is rational Now let c be any real number in R. Case 1: If c is rational, choose a sequence (xn :n € N) of irrational numbers converging to c (which is allowed by Proposition 4.6.11). So æn → c, f(xn) = 1 for all n, and f(c) = 0. But since f(æn) = 1 for all n, we cannot have f(xn) → 0 (a sequence of l's cannot converge to 0!). That is, it is impossible that f(2n) → f(c), so by Theorem 4.6.7, f cannot be continuous at c. What is Case 2?
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