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Why is following sequence of functions discontinuous everywhere?? $$f_n(x)=\left\{\begin{matrix} f(x)-\frac{1}{n},x \in \mathbb{Q}\\ f(x)+\frac{1}{n}, x \notin \mathbb{Q} \end{matrix}\right.$$
where $f:\mathbb{R} \to \mathbb{R}$ a continuous function

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    $\begingroup$ What do you mean by the sequence being continuous or not? Each of the $f_n$ functions itself is plainly discontinuous everywhere, but I don't think that the sequence of all $f_n$s can _itself_ be said to be "not continuous". On the contrary, it converges uniformly to $f$. $\endgroup$ – hmakholm left over Monica Apr 6 '14 at 13:41
  • $\begingroup$ Why is each of the $f_n$ functions itself plainly discontinuous everywhere?? $\endgroup$ – evinda Apr 6 '14 at 13:44
  • $\begingroup$ x @evinda: Because in every interval it has function values that are at least $2/n$ away from each other. $\endgroup$ – hmakholm left over Monica Apr 6 '14 at 13:50
  • $\begingroup$ @HenningMakholm How can I show it with the definition,using sequences? $\endgroup$ – evinda Apr 6 '14 at 13:51
  • $\begingroup$ @evinda are you able to see that the function $$ \chi_\mathbb{Q}(x) = \begin{cases} 1 & x \in \mathbb{Q}\\ 0 & \text{otherwise} \end{cases} $$ is discontinuous everywhere? $\endgroup$ – Omnomnomnom Apr 6 '14 at 13:54
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Suppose that $f_n(x)$ is continuous at a point $x_0$. Then for every sequence $\{x_k\}$ for which $x_k \to x_0$, it must be that $\lim_{k \to \infty}f_n(x_k) = f_n(x_0)$.

If $x_0$ is an irrational number, choose a sequences of rationals $\{q_k\}$ for which $q_k \to x_0$, and note that $\lim_{k \to \infty}f_n(q_k) \neq f_n(x_0)$. Similarly, if $x_0$ is a rational number, choose a sequence of irrationals $\{p_k\}$ for which $p_k \to x_0$, noting that $\lim_{k \to \infty}f_n(p_k) \neq f_n(x_0)$.

In either case, we note that $f_n$ cannot be continuous at $x_0$ for any $x_0$.

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  • $\begingroup$ I understand..thanks a lot!!! $\endgroup$ – evinda Apr 6 '14 at 15:21
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Let $\chi_{\mathbb Q}$ denote the characteristic function of the rationals. Since $$f_n(x)=f(x)-\frac 1n+\frac 2n\chi_{\mathbb Q}(x),$$ it suffices to show that the characteristic function of the rationals is not continuous at any point.

Tis comes from the fact that each open non empty interval contains rationals numbers and irrational ones.

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  • $\begingroup$ Couldn't we also show it,by using the definition? So,we take a sequence of rational numbers $q_k$ and a sequence of irrations numbers $a_k$..But why is then : $lim_{k \to +\infty} {q_k}=lim_{k \to +\infty} {a_k}=x$ ? $\endgroup$ – evinda Apr 6 '14 at 13:58
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    $\begingroup$ Yes, we can. We have to choose these sequences convergent to $x$. $\endgroup$ – Davide Giraudo Apr 6 '14 at 13:59
  • $\begingroup$ But,why is it possible that both sequences converge to $x$?? Do we know it from the density of rational and irrational numbers?? $\endgroup$ – evinda Apr 6 '14 at 14:01
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    $\begingroup$ @evinda if you know that both the rational numbers and irrational numbers are dense, then it certainly follows that we can find both a sequence $q_k \to x$ and $a_k \to x$ as you've described. $\endgroup$ – Omnomnomnom Apr 6 '14 at 14:06

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