Dirichlet's Function

Let $g (x)$ be:

g (x) = {\begin{cases} 1 & x \in Q \\ 0 & x \notin Q \end{cases}

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This is discontinuous because for any $c \in R$ we can construct a sequence of rational numbers $(q_{n}) \subseteq Q$ where $q_{n} \to c$ and also $(x_{n}) \subseteq (Q)^{c}$ where $x_{n} \to c$ . Notice that:

g (q_{n}) = 1 \to 1, g (x_{n}) = 0 \to 0

These two sequences converge to $c$ but converge to different values, so then the function cannot have a limit at $c$ and thus isn't continuous there.