Question: Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function that satisfies $f(q+1/n)=f(q)$ for every $q\in\mathbb{Q}$ and for every $n\in\mathbb{N}$. Show that $f$ must be a constant function.
Solution: Substituting $q-1/n$ for $q$ in $$f(q+1/n)=f(q),$$ we have $$f(q)=f(q-1/n)$$ for all $q\in\mathbb{Q}$ and for all $n\in\mathbb{N}$. Let the original relation be denoted by $(*)$ and the derived relation be denoted by $(**)$.
Now select any positive rational number $q=r/s,$ where $r,s\in\mathbb{N}$. Thus, by repeated application of $(**)$, we have $$f\left(\frac{r}{s}\right)=f\left(\frac{r-1}{s}\right)=f\left(\frac{r-2}{s}\right)=\cdots=f\left(\frac{0}{s}\right)=f(0).$$
Next select any negative rational number $q=r/s$, where $r\in\mathbb{Z}_{\le 1}$ and $s\in\mathbb{N}.$ Thus, by repeated application of $(*),$ we have $$f\left(\frac{r}{s}\right)=f\left(\frac{r+1}{s}\right)=f\left(\frac{r+2}{s}\right)=\cdots=f\left(\frac{0}{s}\right)=f(0).$$
Thus, for all $q\in\mathbb{Q}$, we have $f(q)=f(0)$.
Next select any irrational number $q'$. We know that there exists a convergent sequence $(x_n)_{n\ge 1}$ of rational numbers such that it converges to $q'$. Now since $f$ is continuous on $\mathbb{R}$, implies that it is continuous at $q'$. Thus, by the sequential definition of limit we can conclude that the sequence $f(x_n)$ converges to $f(q').$ Now note that $f(x_n)=f(0)$, for all $n\in\mathbb{N}$. This implies that $f(x_n)$ converges to $f(0)$, which in turn implies that $f(q')=f(0)$. Now since $q'$ is arbitrary, therefore, $f(q')=f(0)$ for all irrational numbers $q'$.
Thus, we can conclude that $f(x)=f(0)$ for all $x\in\mathbb{R}$, that is $f$ is a constant function.
Is this solution correct and rigorous enough and is there any other way to solve the problem?
from Hot Weekly Questions - Mathematics Stack Exchange
Sanket Biswas
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