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How can I show that $f_n(x)$ is discontinuous at each point?

$f_n(x)=\left\{\begin{matrix} f(x)-\frac{1}{n} &, x \in \mathbb{Q} \\ f(x)+\frac{1}{n} &,x \notin \mathbb{Q} \end{matrix}\right.$

$f: \mathbb{R} \to \mathbb{R}$ continuous function

My attempt is to say that there are sequences of rational and irrational numbers , $q_k$ and $a_k$, respectively ,with the property $\lim_{k \to +\infty} q_k=\lim_{k \to +\infty } a_k=x$

How can I proceed ?

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  • $\begingroup$ You use the definition of $f$ to define $f$... Anyway consider the sequence you mentioned and show that $\lim_{k \to \infty}f(q_k)\neq\lim_{k \to \infty}f(a_k)$. This is a contradiction to continuity of $f$ at $x.$ $\endgroup$ – Surb Jun 25 '14 at 13:51
  • $\begingroup$ your definition of $f$ does not make sense: you use $f(x)$ to define $f(x)$ (and additionally $n$ is not defined) $\endgroup$ – Denis Jun 25 '14 at 13:51
  • $\begingroup$ @Surb I edited my post,I meant $f_n$.Why do we show that $\lim_{k \to +\infty} f_n(q_k) \neq \lim_{k \to \infty} f_n(a_k)$? $\endgroup$ – user159870 Jun 25 '14 at 13:55
  • $\begingroup$ @Denis I edited my post. $\endgroup$ – user159870 Jun 25 '14 at 13:55
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    $\begingroup$ now $f$ is not defined... $\endgroup$ – Denis Jun 25 '14 at 13:56
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A function $g: \mathbb{R} \to \mathbb{R}$ is continuous at $x_0$ if (by definition) $\lim_{x\to x_0}g(x) = g(x_0)$. You can show that this is true if and only if for every sequence $(x_k)_{k} \subset \mathbb{R}$ such that $\lim_{k \to \infty} x_k = x$ it holds $\lim_{k \to \infty} g(x_k) = g(x)$.

So let $x_0 \in \mathbb{R}$, by density of $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ in $\mathbb{R}$ you can find two sequences $(q_k)_k \subset \mathbb{Q}, (a_k)_k \subset \mathbb{R}\setminus\mathbb{Q}$ such that $\lim_{k \to \infty} q_k = x_0 = \lim_{k \to \infty} a_k$ but then $$\lim_{k \to \infty} f_n(q_k) = \lim_{k \to \infty} f(q_k)-1/n = f(x_0)-1/n\neq f(x_0)+1/n = \lim_{k \to \infty} f(a_k)+1/n = \lim_{k \to \infty} f_n(a_k)$$ this shows that $f$ can't be continuous at $x_0$.

Note that we have used the fact that $f$ is continuous to say $\lim_{k \to \infty} f(q_k)= f(x_0)=\lim_{k \to \infty} f(a_k)$.

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  • $\begingroup$ Ok, I understand. thanks! $\endgroup$ – user159870 Jun 28 '14 at 19:47

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