Question

prove that the function f given by

f(x)={ 2,if x is irrational

-2,if x is rational

is discontinuous, for all x€R,using the sequential definition of continuity

Answer 1

ANSWER.

By the definition of continuity from condition $\lim _{ n\rightarrow \infty }{ { x }_{ n } } =a$ it follows $\lim _{ n\rightarrow \infty }{ { f(x }_{ n })=f(a) } \\ \\$ . Therefore , if $\lim _{ n\rightarrow \infty }{ { x }_{ n } } =\lim _{ n\rightarrow \infty }{ { y }_{ n } } =a$ then $\lim _{ n\rightarrow \infty }{ { f(x }_{ n })=\lim _{ n\rightarrow \infty }{ { f(\ y }_{ n })\ \ } }$

Let $a$ be a real number and $n$ be a natural number. Between the numbers $a-1/n , a+1/n$ we choose a rational number $x_n$ and irrational number $y_n$ . Since $a-\frac { 1 }{ n } <{ x }_{ n }<a+\frac { 1 }{ n } \quad ,a-\frac { 1 }{ n } <{ y }_{ n }<a+\frac { 1 }{ n }$ , and $\lim _{ n\rightarrow \infty }{ \left( a-\frac { 1 }{ n } \right) = } \lim _{ n\rightarrow \infty }{ \left( a+\frac { 1 }{ n } \right) =a }$, then $\lim _{ n\rightarrow \infty }{ { x }_{ n }= } \lim _{ n\rightarrow \infty }{ { y }_{ n }=a }$ . For sequences $\left( { x }_{ n } \right) \ in\ Q,\ \left( { y }_{ n } \right) \ in\ R\setminus Q$ we have $f\left( { x }_{ n } \right) =-2,\ f\left( { y }_{ n } \right) = 2$ for all $n\in N$ . So

$\lim _{ n\rightarrow \infty }{ { f(x }_{ n })=-2\neq } 2=\ \lim _{ n\rightarrow \infty }{ { f(\ y }_{ n })\ \ } .$

Hence, $f$ is not continuous at $a$ , and as $a$ was arbitrary , that $f$ is not continuous at any $a\in R.$