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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.

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

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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.

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  • $\begingroup$ So basically we're finding the limit of the sequence $\langle\chi_\mathbb{Q}(S_n)\rangle$ as $n\rightarrow\infty$? $\endgroup$
    – Math Rules
    Apr 12, 2020 at 13:37
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    $\begingroup$ Yes, that is correct. If you pick all irrationals, all the terms are $0$ $\endgroup$ Apr 12, 2020 at 13:56
  • $\begingroup$ Thank you so much, that makes sense and I appreciate the quick responses! I have accepted your answer and tried to upvote it, but my vote won't show publicly since I don't have enough reputation yet. $\endgroup$
    – Math Rules
    Apr 12, 2020 at 14:00

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