The Dirichlet function is defined by \(f(x)=\left\{ \begin{array}{l@{\quad}l} 1 & \hbox{if }x \hbox{ is rational} \\[3pt] 0 & \hbox{if }x \hbox{ is irrational} \end{array} \right. .\)

Prove \(\lim\limits_{x\rightarrow c}\) \(f( x) \) does not exist for any \(c.\)

Solution We use a proof by contradiction. That is, we assume that \(\lim\limits_{x\rightarrow c} f(x)\) exists and show that this leads to a contradiction.

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Assume \(\lim\limits_{x\rightarrow c}f(x)=L\) for some number \(c\). Now if we are given \(\epsilon =\dfrac{1}{2}\) (or any smaller positive number), then there is a positive number \(\delta \), so that \begin{equation*} \hbox{whenever }\quad 0\lt\vert x-c\vert \lt\delta\qquad \hbox{ then }\quad \vert f( x) -L \vert \lt\dfrac{1}{2} \end{equation*}

Suppose \(x_{1}\) is a rational number satisfying \(0\lt \vert x_{1}-c\vert \lt\delta \), and \(x_{2}\) is an irrational number satisfying \(0\lt\vert x_{2}-c \vert \lt\delta \). Then from the definition of the function \(f\), \begin{equation*} f (x_{1}) =1\qquad \hbox{and}\qquad f( x_{2}) =0 \end{equation*}

Using these values in the inequality \(\vert f(x)-L\vert \lt\epsilon \), we get \begin{equation*} \begin{array}[t]{rcl} \vert f(x_{1})-L\vert &=&\vert 1-L\vert \lt\dfrac{1}{2}\\[5pt] -\dfrac{1}{2}&\lt&1-L\lt\dfrac{1}{2} \\[5pt] -\dfrac{3}{2}&\lt&-L\lt-\dfrac{1}{2} \\[5pt] \dfrac{1}{2}&\lt&L\lt\dfrac{3}{2} \end{array} \quad \hbox{ and }\quad \begin{array}[t]{rcl} \vert f(x_{2})-L\vert &=& \vert 0-L\vert \lt\dfrac{1}{2} \\[5pt] -\dfrac{1}{2}&\lt&-L\lt\dfrac{1}{2} \\[5pt] \dfrac{1}{2}&\gt&L\gt-\dfrac{1}{2} \\[5pt] -\dfrac{1}{2}&\lt&L\lt\dfrac{1}{2} \end{array} \end{equation*}

From the left inequality, we have \(L\gt\dfrac{1}{2}\), and from the right inequality, we have \(L\lt\dfrac{1}{2}\). Since it is impossible for both inequalities to be satisfied, we conclude that \(\lim\limits_{x\rightarrow c}f(x)\) does not exist.