Problems regarding the denseness of $a\mathbb{Z}+\mathbb{Z}$

Question

I came across a problem related to the denseness of $a\mathbb{Z}+\mathbb{Z}$. Explicitly, the problem asks:

Let $a$ be irrational. (a) For any $y \in [0,1]$, prove that there are infinitely many $n \in \mathbb{N}$ so that $\{na\} \in [y-\epsilon, y+\epsilon]$.

(b) Prove that there exists $b \in [0,1]$ such that there are infinitely many $n \in \mathbb{N}$ with $\{na\} \in [b-9^{-n}, b+9^{-n}]$.

Where $\{x\}$ denotes the fractional part of $x$. This is a contest problem (Bulgaria EGMO TST). Being always interested in analysis, I tried solving the problem. This is my attempt:

Let $\{x\}$ denote the fractional part of $x$. We choose $n$ big enough so $\frac{1}{n} < \epsilon$. For $k=0,1,...,n$, we let $$ x_k = \{ka \} $$ Divide $[0,1)$ into $n$ equal intervals, then two $x_i, x_j (i<j)$ falls in the same interval, so $|\{(i-j)a\}| < \frac{1}{n} < \epsilon$. That is, there exists $p = j-i < n$ and $q$ so that $$ |pa - q | < \epsilon.$$ This proves the Dirichlet's approximation theorem.

In fact, an implication of which is that $\{na\}$ is dense in $\mathbb{R}$, as $a \mathbb{Z} + \mathbb{Z}$ as an additive subgroup of $\mathbb{R}$ has no least positive element. In particular, for any $y \in (0, 1)$ and $\epsilon > 0$, there are $na$ whose fractional part is in $[y-\epsilon, y+\epsilon]$. If $y$ is never a fractional part of a multiple of $a$ (that is, $y \notin a \mathbb{Z} +\mathbb{Z}$), then we choose $\epsilon \to 0$ to achieve infinite of such $n$. If $y$ is a fractional part of multiple of $a$, then we can choose $|y' - y| < \frac{\epsilon}{2}$ such that $y' \notin a\mathbb{Z}+\mathbb{Z}$ (otherwise $[y-\frac{\epsilon}{2}, y+\frac{\epsilon}{2}] \subseteq a \mathbb{Z} + \mathbb{Z}$, which is absurd by cardinality), then again choosing $\epsilon \to 0$ gives infinite of such $n$. For $y=0, 1$, similar discussions work. This finishes (a).

But I'm stuck for (b). Namely, to construct $b \in [0,1]$ so that $\{na\} \in [b-9^{-n}, b+9^{-n}]$ for infinitely many $n \in \mathbb{N}$. The negation of this statement is that any $b \in [0,1]$ satisfy that $\{na\} \in [b-9^{-n}, b+9^{-n}]$ for only finitely many $n \in \mathbb{N}$. So there's $N_b$ satisfying $n \geq N_b \implies \{na\} \notin [b-9^{-n}, b+9^{-n}]$. We let such $N_b$ be minimal. Note that, then, $N_b$ don't have a uniform bound for $b \in [0,1]$. So for $k \in \mathbb{N}$ we can find $b$ s.t. $N_b > k$. This implies $\{na\} \in [b-9^{-n}, b+9^{-n}]$ for some $n < N_b$ (which isn't useful as $N_b$ could be big). I tried using Bolzano-Weierstrass, and make no progress since.

The aops website mentioned theorem of nested intervals (compact sets?), but I don't quite get it. Any other hint would be appreciated.

Answer 1

Since $\{ n\alpha \}$ is dense in $[0,1]$, Part $(a)$ is almost obvious.

For Part $(b)$, let's assume:

$$B_1=[\{m_1\alpha\} -\frac{1}{9^{m_1}}, \{m_1\alpha\} +\frac{1}{9^{m_1}}],$$

where $m_1 \in \mathbb N.$ It's easy to choose $m_1$ such that $B_1 \subset [0,1]$ as we will see a similar reasoning in the following.

Take $b^* \in B_1$ and $\epsilon >0 $ such that:

$$[b^*-\epsilon,b^*-\epsilon] \subset B_1=[\{m_1\alpha\} -\frac{1}{9^{m_1}}, \{m_1\alpha\} +\frac{1}{9^{m_1}}].$$

By Part $(a)$, there are infinitely many $n$ such that $\{na\} \in [b^*-\frac{\epsilon}{2},b^*-\frac{\epsilon}{2}].$ Pick such a sufficiently large $n$, say $m_2$, such that:

$$[\{m_2\alpha\} -\frac{1}{9^{m_2}}, \{m_2\alpha\} +\frac{1}{9^{m_2}}] \subset [b^*-\epsilon,b^*-\epsilon] .$$

Hence, we have:

$$B_2=[\{m_2\alpha\} -\frac{1}{9^{m_2}}, \{m_2\alpha\} +\frac{1}{9^{m_2}}] \subset [\{m_1\alpha\} -\frac{1}{9^{m_1}}, \{m_1\alpha\} +\frac{1}{9^{m_1}}]=B_1.$$

Doing the same procedure infinitely many times, we will obtain the chain:

$$[0,1] \supset B_1 \supset B_2 \supset B_3 \supset ... \ .$$

Let's consider $\bigcap_{n=1}^{\infty} B_n$, which is not empty. Now, it's easy to see that every element of $\bigcap_{n=1}^{\infty} B_n$ works as the desired $b$ in the question.