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I am supposed to prove the continuity of function: $$\varphi (x) = \left\{\begin{align} 1 &, \quad x\in\mathbb{Q} \\ -1 &, \quad x\in\mathbb{R}\setminus \mathbb{Q} \end{align}\right.$$ Can anyone help me?

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I'll use the "definition" of continuity (using $\varepsilon-\delta$) to prove that $\varphi$ is discontinuous; I will also use the fact that $\mathbb Q$ is dense (and hence $\mathbb R\setminus\mathbb Q$).

The definition tells you that $\varphi$ is continuous at a point $x\in\mathbb R$ if and only if

$$\forall\varepsilon>0\;\exists\delta>0\;\forall t\in\mathbb R(|t-x|<\delta\;\Longrightarrow\;|\varphi(t)-\varphi(x)|<\varepsilon);$$

thus, $\varphi$ is discontinuous at a point $x\in\mathbb R$ if and only if

$$\exists\varepsilon>0\;\forall\delta>0\;\exists t\in\mathbb R(|t-x|<\delta\;\wedge\;|\varphi(t)-\varphi(x)|\geqslant\varepsilon).$$

So, take any $0<\varepsilon\leqslant2$ and let $x\in\mathbb R.$ If $x$ is rational, and if $\delta>0,$ take any irrational $t\in(x-\delta,x+\delta)$ (whose existence follows from the density of irrationals in $\mathbb R$). Then $|\varphi(t)-\varphi(x)|=2\geqslant\varepsilon.$ A similar process applies if $x$ is rational

Carlos Israel Jalpa Rico
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  • Why do you use that epsilon is less or equal than 2? I mean, why 2? –  Oct 02 '19 at 15:40
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    @MartinN. $2 = \max |f(x) - f(y)|$. This is used in the second last statement. "Then $|\varphi(t) - \varphi(x)| = 2 \geqslant \varepsilon$." – Sera Gunn Oct 02 '19 at 15:42
  • @MartinN. Because that's the distance between the two possible distinct values of $\varphi.$ By how it is defined, intuitively, that's what you have to focus on – Carlos Israel Jalpa Rico Oct 02 '19 at 15:43
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    @CarlosIsraelJalpaRico Thank you so miuch –  Oct 02 '19 at 15:56