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I'm having some trouble in the following proof:

Let $f:\mathbb{R\to R}$ be given by: $$ f(x) = \begin{cases} 1,&x\in\mathbb Q \\ 0,&x\notin\mathbb Q \end{cases} $$

Prove that $f$ is not continuous for any value in $\mathbb R$.

Can anyone please help me with this?

Thank you!

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1  
are you looking for a hint or a solution? What did you try so far? –  Ittay Weiss Jan 9 '13 at 21:59
    
Try with the sequential characterization of continuity, the density of $\mathbb{Q}$ in $\mathbb{R}$, and the density of $\mathbb{R}\setminus \mathbb{Q}$ in $\mathbb{R}$. –  1015 Jan 9 '13 at 22:00
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Do you know the definition of what it means for $f$ to be continuous at some value of $R$? –  MJD Jan 9 '13 at 22:01
    
@IttayWeiss For a solution. –  Gabriel Bianconi Jan 9 '13 at 22:45
    
@MJD I've studied the definition in terms of limit (lim x->c f(x) = f(c)). –  Gabriel Bianconi Jan 9 '13 at 22:46

4 Answers 4

up vote 6 down vote accepted

Let $x\in \Bbb R$, and $\epsilon=1/2$. If $f$ is continuous, then there should be a $\delta$ region around $x$ which is sent entirely inside $(f(x)-\epsilon, f(x)+\epsilon)$.

But what do you know about rationals and irrationals in intervals? Draw a picture for the two cases ($f(x)=1$ and $f(x)=0$) and you should see what's going on.

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Let $x\in\mathbb R$. Show that there is a sequence of rational numbers $q_n\to x$ and another sequence of irrational numbers $r_n\to x$.

We have that $\lim_{n\to\infty} q_n=\lim_{n\to\infty} r_n$. But what can you say about $\lim_{n\to\infty} f(q_n)$ and $\lim_{n\to\infty} f(r_n)$?

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Recall that if a function is continuous, then it is also sequentially continuous.

Now consider $x \in \mathbb{R}$.

If $x \in \mathbb{Q}$, consider the sequence $x_n = x - \dfrac1{n\sqrt2}$. Clearly, $x_n \in \mathbb{R} \backslash \mathbb{Q}$ and $x_n \to x$.

Hence, $f(x_n) = 0$ for all $x_n$, whereas $f(x) = 1$.

Do a similar argument when $x \in \mathbb{R} \backslash \mathbb{Q}$. Consider the sequence $x_n = \dfrac{\lfloor 10^n x \rfloor}{10^n}$. Clearly, $x_n \in \mathbb{Q}$ and $x_n \to x$.

Hence, $f(x_n) = 1$ for all $x_n$, whereas $f(x) = 0$.

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Let $x \in \mathbb{R}$. Let $0 < \epsilon < 1$. By the density of $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$, for any $\delta > 0$, we can find a rational $q \in (x - \delta, x + \delta)$ and an irrational $r \in (x - \delta, x + \delta)$. If $x \in \mathbb{Q}$, then $|f(x) - f(r)| = 1$. Otherwise, $|f(x) - f(r)| = 1$. Hence, $f$ is not continuous at $x$.

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Your answer was great too! However, I decided to accept rschwieb's answer because he posted it earlier. Thank you very much Brian! –  Gabriel Bianconi Jan 9 '13 at 22:48

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