**Theorem** The function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=\begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \not \in \mathbb{Q} \end{cases}$(known as [[Rational Indicator Function (Dirichlet Function)|Dirichlet function]]) is [[Discontinuous Function|discontinuous]] at every point in its domain. **Proof (using [[Discontinuous Function (Epsilon-Delta Definition)]])** Take $c \in \mathbb{Q}$ and let $\epsilon=1$. Then for any $\delta>0$ there is an irrational $x$ with $|x-c|<\delta$ since [[Between two Different Real Numbers exists an Irrational Number]] and $|f(x)-f(c)|=|0-1|=1 \geq \epsilon$. Similarly if $c \not \in \mathbb{Q}$, again take $\epsilon= \frac{1}{2}$ and find a rational $x$ with $|x-c|<\delta$ since [[Any real number can approximated arbitrarily closely by a rational number]] so $|f(x)-f(c)|=1 \geq \epsilon.$