# Show that $\psi(x)$ isn't continuous at $x=1$

[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! :)

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

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.

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.