Maliyah Robles

2022-07-07

Construct an increasing function $f$ on $R$ that is continuous at every irrational number and is discontinuous at every rational number.

Solution: Let ${r}_{n}$ be a sequence with distinct terms whose range is $\mathbb{Q}$. Let $f:\mathbb{R}\to \mathbb{R}$ be given by

$f(x)=\sum _{{r}_{n}<x}\frac{1}{{2}^{n}}$

Solution: Let ${r}_{n}$ be a sequence with distinct terms whose range is $\mathbb{Q}$. Let $f:\mathbb{R}\to \mathbb{R}$ be given by

$f(x)=\sum _{{r}_{n}<x}\frac{1}{{2}^{n}}$

Alexzander Bowman

Beginner2022-07-08Added 19 answers

Let

${f}_{0}(x)=\{\begin{array}{ll}0& \text{if}{r}_{0}\u2a7ex\\ 1& \text{if}{r}_{0}x.\end{array}$

It's increasing. Besides, it is discontinuous at ${r}_{0}$ and only at ${r}_{0}$.

Now, let

${f}_{1}(x)=\{\begin{array}{ll}0& \text{if}{r}_{1}\u2a7ex\\ \frac{1}{2}& \text{if}{r}_{1}x.\end{array}$

It's increasing and it is discontinuous at ${r}_{1}$ and only at ${r}_{1}$. So, ${f}_{0}+{f}_{1}$ is increasing and it is discontinuous at ${r}_{0}$ and at ${r}_{1}$ and only at those points.

More generally, for each $n\in {\mathbb{Z}}_{+}$, let

${f}_{n}(x)=\{\begin{array}{ll}0& \text{if}{r}_{n}\u2a7ex\\ \frac{1}{{2}^{n}}& \text{if}{r}_{n}x.\end{array}$

Then $f$ is increasing, since it is equal to $\sum _{n=0}^{\mathrm{\infty}}{f}_{n}$. And it is not hard to see that it is discontinuous at $x$ if and only if $x\in \{{q}_{n}\mid n\in {\mathbb{Z}}_{+}\}=\mathbb{Q}$ (this follows from the fact that the convergence of the series $\sum _{n=0}^{\mathrm{\infty}}{f}_{n}$ is uniform, by the Weierstrass $M$-test). The reason why I told you in the comments that it should be $\frac{1}{{2}^{n}}$ rather than $\frac{1}{2n}$ was so that the expression $\sum _{n=0}^{\mathrm{\infty}}{f}_{n}$ makes sense, that is, so that it converges, for every $x\in \mathbb{R}$.

${f}_{0}(x)=\{\begin{array}{ll}0& \text{if}{r}_{0}\u2a7ex\\ 1& \text{if}{r}_{0}x.\end{array}$

It's increasing. Besides, it is discontinuous at ${r}_{0}$ and only at ${r}_{0}$.

Now, let

${f}_{1}(x)=\{\begin{array}{ll}0& \text{if}{r}_{1}\u2a7ex\\ \frac{1}{2}& \text{if}{r}_{1}x.\end{array}$

It's increasing and it is discontinuous at ${r}_{1}$ and only at ${r}_{1}$. So, ${f}_{0}+{f}_{1}$ is increasing and it is discontinuous at ${r}_{0}$ and at ${r}_{1}$ and only at those points.

More generally, for each $n\in {\mathbb{Z}}_{+}$, let

${f}_{n}(x)=\{\begin{array}{ll}0& \text{if}{r}_{n}\u2a7ex\\ \frac{1}{{2}^{n}}& \text{if}{r}_{n}x.\end{array}$

Then $f$ is increasing, since it is equal to $\sum _{n=0}^{\mathrm{\infty}}{f}_{n}$. And it is not hard to see that it is discontinuous at $x$ if and only if $x\in \{{q}_{n}\mid n\in {\mathbb{Z}}_{+}\}=\mathbb{Q}$ (this follows from the fact that the convergence of the series $\sum _{n=0}^{\mathrm{\infty}}{f}_{n}$ is uniform, by the Weierstrass $M$-test). The reason why I told you in the comments that it should be $\frac{1}{{2}^{n}}$ rather than $\frac{1}{2n}$ was so that the expression $\sum _{n=0}^{\mathrm{\infty}}{f}_{n}$ makes sense, that is, so that it converges, for every $x\in \mathbb{R}$.

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