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Show that $f(x)$ is not continuous at every point $x_0$ not at the origin.

$f(x) = \begin{cases} 2x, & \text{x rational} \\ -2x, & \text{x irrational} \end{cases}$

My working so far:

Using the converse statement,

$\exists\epsilon>0$ such that $\forall\delta>0, \exists x_\delta$ such that $|x_\delta-x_0|<\delta$ and $|f(x_\delta)-f(x_0)|\ge\epsilon$

Then, $|2x_\delta-2x_0|<2\delta??$......and now I'm stuck. How do I go about finding $\epsilon$ and $x_\delta$ ?

I understand there are multiple questions on epsilon-delta proofs but a lot of them simply say choose $\epsilon$ as _ without really explaining. Also the fact that the proof isn't for a specific point is throwing me off.

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As you asked, let's use the definition: we fix $x_0\neq 0$ and we have to find $\varepsilon>0$ which fits the requirement.

Suppose $x_0$ is rational: by the density of $\mathbb{R}\setminus\mathbb{Q}$ in $\mathbb{R}$ we can always find $x_{\delta}$ such that

  • $|x_0 - x_{\delta}|<\delta$, and
  • $x_{\delta}\in \mathbb{R}\setminus\mathbb{Q}$

then $$| f(x_0) - f(x_{\delta})| = |2x_0 - (-2x_{\delta})| = 2|x_0 + x_{\delta}|$$ and using $x_{\delta} = (x_{\delta} - x_0) + x_0 $ we get $$ | f(x_0) - f(x_{\delta})| \geq 2(2|x_0| - |x_0 - x_{\delta}|) \geq 4|x_0| $$ Thus, taking $\varepsilon = 2|x_0|$ we have the desired result. The same reasoning applies if we take $x_0$ irrational.

The only problem is when $x_0 = 0$: in that case the modulus is $0$ so we cannot use the previous estimate, indeed you have to prove the continuity at $x_0 = 0$. It is easy to prove continuity at $0$ using sequential continuity, i.e. showing that for every sequence $\{a_n\}$ such that $a_n \to 0$ as $n\to +\infty$ we get $f(a_n) \to f(0)$ as $n\to \infty$.

The same strategy could also be used to prove that $f$ is not continuous if $x_0\neq 0$, as Surb pointed out.

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Let $(a_n)\subset \mathbb Q$ and $(b_n)\subset \mathbb R\backslash \mathbb Q$ two sequences that converge to $a\in\mathbb R\backslash \{0\}$.

$$\lim_{n\to\infty }f(a_n)=2a\neq -2a=\lim_{n\to\infty }f(b_n),$$ and thus $f$ is not continuous.

Notice that $(a_n)$ and $(b_n)$ exist by density of $\mathbb Q$ and $\mathbb R\backslash \mathbb Q$ in $\mathbb R$.

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  • $\begingroup$ Ah, so I don't need to use the epsilon-delta proof? $\endgroup$ – user226067 Mar 24 '15 at 14:52
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In fact, $f(x)$ is continuous at $x_0$ if and only if

for every sequence ${x_n}{\to}x_0$, $f(x_n){\to}f(x_0)$.

In fact, $f(x)$ is not continuous at $x_0$ if and only if

there exist a sequence ${x_n}{\to}x_0$, but $f(x_n){\nrightarrow}f(x_0)$.

If $x_0{\in}{\mathbb Q}$, take $x_n=x_0-{\frac{1}{\sqrt2n}}$. Then ${x_n}{\to}x_0$, but $f(x_n){\nrightarrow}f(x_0)$.

If $x_0{\notin}{\mathbb Q}$, since $\mathbb Q$ is dense in $\mathbb R$, we can choose a sequence {$x_n$} in $\mathbb Q$ and ${x_n}{\to}x_0$, but $f(x_n)\to2x_0$ while $f(x_0)=-2x_0$ .

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