# How to Prove Characteristic Function of the Rationals is Discontinuous Using Sequences

The characteristic function of the rationals $$\chi_{\mathbb{Q}}(x)=\begin{cases}1&x\in\mathbb{Q}\\0 & x\not\in\mathbb{Q}\end{cases}$$ is discontinuous for all $$x$$. I have seen a proof that $$\chi_{\mathbb{Q}}$$ is discontinuous at all rational numbers $$x_0$$ by constructing a sequence $$\langle S_n\rangle$$ of irrational numbers which converge to the rational number $$x_0$$ as $$n\rightarrow\infty$$, which is possible because both the rationals and the irrationals are dense in $$\mathbb{R}$$. Then $$\lim_{n\rightarrow\infty}{\chi_\mathbb{Q}(S_n)}=0\neq\chi_\mathbb{Q}(x_0)=1,$$ and thus the function cannot be continuous. My question is this: How do we justify that $$\lim_{n\rightarrow\infty}{\chi_\mathbb{Q}(S_n)}=0$$ in a rigorous manner? I understand the definition of a limit of a function using a $$\delta-\epsilon$$ argument, and I understand the definition of a limit of a sequence, but I am confused as to how to apply a $$\delta-\epsilon$$ argument to a limit of a function evaluated at a sequence. NB: I am not looking for an explanation of how to prove that $$\chi_\mathbb{Q}$$ is continuous by a direct calculation of the limit; I am specifically interested in how to do this using sequences.

• Let $\varepsilon \gt 0$. Then for all $n\in \mathbb N$, $|\chi_{\Bbb Q}(s_n)|\lt \varepsilon$.
– cqfd
Apr 12, 2020 at 13:30

For a sequence a $$\delta - \epsilon$$ argument is replaced by a $$N-\epsilon$$ argument. If somebody claims the limit of a sequence is $$L$$, you should be able to give them an $$\epsilon \gt 0$$ and they can find an $$N$$ where all the terms after $$N$$ are within $$\epsilon$$ of $$L$$. The idea is the same as $$\delta - \epsilon$$. The values of the squence can bounce around as much as they want early, but eventually must all be within $$\epsilon$$ of $$L$$.
In this case it is easy because the sequence is identically $$0$$, so for any $$\epsilon$$ you can choose $$N=1$$ and it works.
• So basically we're finding the limit of the sequence $\langle\chi_\mathbb{Q}(S_n)\rangle$ as $n\rightarrow\infty$? Apr 12, 2020 at 13:37
• Yes, that is correct. If you pick all irrationals, all the terms are $0$ Apr 12, 2020 at 13:56