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[Edited] Let $\varphi : \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $\varphi(x)=0, \forall x\in\mathbb{Q}$ and, $\varphi(x)=2, \forall x\in\mathbb{R}−\mathbb{Q}$. Let $\psi(x):\mathbb{R} \rightarrow \mathbb{R}$ be a function such that $\psi(x) = x\varphi(x), \forall x \in\mathbb{R}.$

I need to show that the function $\psi(x)$ isn't continuous in $x=1$.

I know I have basically two options here: 1) Prove by definition using $\varepsilon$ and $\delta$. And 2) using sequences.

  1. What I did until now: $\mid \psi(x) - \psi(1)\mid \geq \varepsilon \Leftrightarrow \mid x\cdot\varphi(x) - 1\cdot \varphi(1) \mid \geq \varepsilon \Leftrightarrow \mid x\cdot\varphi(x) - 0 \mid \geq \varepsilon \Leftrightarrow \mid x\cdot\varphi(x) \mid \geq\varepsilon$. Now I thought about two cases:

1.1) If $x\in\mathbb{Q}$: $\mid x\cdot0 \mid \geq\varepsilon \Leftrightarrow \mid 0 \mid\geq\varepsilon$. And here I got officially stucked and can't see how to go any further to find a $\delta$.

1.2) If $x\in\mathbb{R}-\mathbb{Q}$: $\mid x\cdot2 \mid\geq\varepsilon \Rightarrow \mid x-1 \mid \geq\frac{\varepsilon-2}{2}$. Here $\delta = \frac{\varepsilon-2}{2}$.

And then I would be taking $\delta$ as the minimum between 1.1) and 1.2) $\delta$'s. But then I couldn't find a $\delta$ in 1.1).

  1. I have no idea what sequences I should use to prove it.

Anyway, as you can see, I'm kinda stucked in both ways.

Any hint would be fully and deeply appreciated. Thanks in advance! :)

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  • $\begingroup$ @ThomasAndrews yeah, I've mistyped it. Thank you for the heads up! $\endgroup$
    – geep
    Mar 14, 2022 at 0:30

2 Answers 2

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You can easily prove it by using sequential criteria of continiuty.

If a function is continious at a point c then for any sequence (Xn) converging to c ,f(Xn) must converge to f(c).

Now, since Q & R-Q both are dense in R ,we can easily construt two sequence (first one of rational terms & second one for irrational terms) converging to any point c in R.

Here, we can use same concept. take a sequence (Xn) of rational terms converging to 1 but f(Xn) converges to 0 & take a sequence (Yn) of irrational terms converging to 1 but f(Yn) converges to 2.

Hence, by sequential criteria ,function is discontinious at x=1.

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Rewriting slightly, define $$ \psi(x) = \begin{cases} 0 & \text{if $x \in \mathbb{Q}$}\\ 2x & \text{if $x \in \mathbb{R} \setminus \mathbb{Q}$}. \end{cases} $$ Using $\epsilon-\delta$, it suffices to show that if for some $\epsilon > 0$ and every $\delta > O$ there is an $x \in \mathbb{R}$ such that $|x - 1| < \delta$ and $|\psi(x) - \psi(1)| > \epsilon$.

Suppose $\epsilon = 1/2$ and suppose we have found a $\delta > 0$ that works. We show that it actually can't work. By $\epsilon-\delta$, if $\delta$ works, then so does every $0 < \delta’ < \delta$. Hence we may assume $\delta < 1/2$. Now $\epsilon-\delta$ states that

$$ \forall x (|x-1| < \delta \Rightarrow |\psi(x) - \psi(1)| < 1/2). $$

But since the irrationals are dense in $\mathbb{R}$, they meet the open interval $$\{ y \in \mathbb{R} \mid |y - 1| < 1/2 \} = (1/2,3/2).$$ So let $x \in (\mathbb{R} \setminus \mathbb{Q}) \cap (1/2,3/2)$. Then, by definition, $\psi(x) = x\varphi(x) = 2x > 1$ since $x > 1/2$. So there exists $x$ such that $|x-1| < \delta$ yet $|\psi(x) - \psi(1)| = |\psi(x)| > 1 > \epsilon$, a contradiction.

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