Answer to Question #203165 in Real Analysis for Rajkumar

Question

Prove that the function f defined by

f(x) = -2, if is rational

f(x) = 2, if is irrational

is discontinuous,∀ x ∈ R, using the sequential definition of continuity.

Answer 1

Given

Let $f: \R \to \R$ denote the  function:

$\forall x \in \R: f( x) = \begin {cases} -2 & : x \in Q \\ 2 & : x \in Q^c \end {cases}$

where $Q$  denotes the set of rational numbers.

Then $f$  is discontinuous at every$x \in \R$

suppose -2=c and 2=d {for easily proof for generalization}

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Proof

Discontinuity for rational numbers:-

Let $\epsilon = \frac {| c-d| } 2$

let $x \in Q$

Let $\delta \in \R_{>0}$  be arbitrary.

Let $y∈Q$  such that $|x−y|<δ$

$Without \space loss \space of \space generality, let y>x.$

$From \space Between \space two \space Rational \space Numbers \space exists \space Irrational \space Number:\\ ∃z∈R∖Q:x<z<y$

$and \space so:\\ |f(x)−f(z)|=|c−d|>ϵ\\ Similarly if y<x:\\ ∃z∈R∖Q:y<z<x:|f(x)−f(z)|>ϵ$

$and \space by \space definition \space of \space continuity \space f \space is \space discontinuous \space at \space x.\\$

again now

Discontinuity for irrational numbers:-

$Let \space x∈R∖Q.\\ Let \space δ∈R>0 \space be \space arbitrary.\\ Let \space y∈R∖Q \space such \space that \space |x−y|<δ.\\ Without \space loss \space of \space generality, \space let y>x.\\ From \space Between \space two \space Real \space Numbers \space exists \space Rational \space Number:\\ ∃z∈Q:x<z<y\\ and \space so:\\ |f(x)−f(z)|=|c−d|>ϵ\\ Similarly \space if \space y<x:\\ ∃z∈Q:y<z<x:|f(x)−f(z)|>ϵ\\ and \space by definition \space of \space continuity \space f \space is \space discontinuous \space at \space x.\\ □\\ f \space has \space been \space shown \space to \space be \space discontinuous \space at \space all \space x∈R \\ \space whether \space x \space is \space rational \space or \space irrational. \space \\ Hence \space the \space result.\\$