$E=\left \{ f(n)-\lfloor f(n) \rfloor ,n \in \mathbb{N} \right \} $

Question

Let $f$ be a strictly increasing function such that:

$\lim_{x\to+\infty} f(x)= +\infty$

and

$\lim_{x\to+\infty} [f(x+1)-f(x)]= 0.$

Prove that $E=\left \{ f(n)-\lfloor f(n) \rfloor ,n \in \mathbb{N} \right \} $ is dense in $[0,1]$.

What I have tried to do is:

for all $a \in \mathbb{R}$

for some $n \in \mathbb{N}$ we have $|f(n+1)-f(n)|<a$

How can I prove for all $x \in [0,1]$ there exist an $n$ such that:

$|x-(f(n)-\lfloor f(n) \rfloor )|<a\ ?$

Please help me with this question.

Answer 1

Let $q$ be any positive integer, and let $N$ be such that $|f(n+1) - f(n)| < \frac{1}{q}$ for all $n \geq N$. Then we claim:

Claim 1. For each integer $p$ with $p > qf(N)$, there exists $n \geq N$ so that $f(n) \in [\frac{p}{q}, \frac{p+1}{q})$.

Proof. Let $n = \min \{ n \in \mathbb{N} : qf(n) \geq p \}$. Since $f$ is unbounded, $n$ is well-defined. We also observe:

• Since $qf(n) \geq p > q f(N)$, we know that $n > N$ and hence $n-1 \geq N$.
• $qf(n-1) < p$ by the definition of $n$.

Then

$$ 0 \leq qf(n) - p < q[f(n) - f(n-1)] < q \cdot \frac{1}{q} = 1 $$

and therefore $f(n) \in [\frac{p}{q}, \frac{p+1}{q})$.

Claim 2. For each $x \in [0, 1]$, there exists $\hat{x} \in E$ so that $|x - \hat{x}| \leq \frac{1}{q}$.

Proof. Let $r \in \{0, \ldots, q-1\}$ be such that $x \in [\frac{r}{q}, \frac{r+1}{q}]$. Then choose $p \in \mathbb{Z}$ so that $p > qf(n)$ and $p \equiv r \pmod{q}$. Then invoke Claim 1 to choose $n \in \mathbb{N}$ so that $f(n) \in [\frac{p}{q}, \frac{p+1}{q})$. We claim that $\hat{x} = f(n) - \lfloor f(n)\rfloor$ satisfies $|x - \hat{x}| \leq \frac{1}{q}$.

Indeed, by writing $m = \lfloor p/q\rfloor$, we get $ p = mq + r$ and $\lfloor f(n) \rfloor = m$. Hence,

$$ \hat{x} = f(n) - m \in [\tfrac{p-mq}{q}, \tfrac{p+1-mq}{q}) = [ \tfrac{r}{q}, \tfrac{r+1}{q}), $$

from which the desired claim follows.

Conclusion. Since Claim 2 is true for arbitrary integer $q \geq 1$, it follows that $E$ is dense in $[0, 1]$.