A simple question about density in the interval $[0.1,1)$

Question

Theorem: Let $\mathbb{P}$ be the set prime numbers and $S$ is a set that has been made as below: put a point on the beginning of each member of $\Bbb{P}$ like $0.2$ or $0.19$ then $S=\{0.2,0.3,0.5,0.7,...\}$ is dense in the interval $(0.1,1)$ of real numbers.

Now I want ask below conjecture is true?

Conjecture: For each subinterval of $[0.1,1)$ like $(a,b)$ then $\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ such that $t\cdot 10^k\in \Bbb P$.

I thank you in advance.

Answer 1

Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$.

The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.

We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$.

Now comes the prime number theorem:

$$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$

where $\pi(n) = \# \{ p \leqslant n : p \text{ is prime} \}.$ By the above we have

$$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$

hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted.

Answer 2

You can simply use the Theorem you stated, along with some basic real analysis notions as follows:

Since $S$ is dense in $(0.1,1)$ then it's dense in every subinterval $(a,b)$.

So for every $ε>0$ and $x \in (a,b)$ there is a $t \in S$ such that $|t-x| <ε$.

Also there is a (sufficiently big) $m \in N$ such that $|b-a| > \frac{1}{10^m}$. Let this $m$ be the infimum of all the m's such that this holds.

Now let $ε= \frac{1}{10^m}$ with $m\in N$ and $x$ be a number with infinite digits $x=0.d_1 d_2 ... d_n... d_m ...$ in $(a,b)$. Then a $t=0.p_1 p_2 ... p_n$ will exist with $$|0.p_1 p_2 ... p_n-0.d_1 d_2 ... d_n ... d_m ...| <1/10^m$$

Therefore $t$ must have AT LEAST $m$ digits, otherwise the inequality can't hold. Note that this $t$ will always exist, due to $S$ being dense in $(a,b)$. So there is a $t$ with $k$ digits, $k \ge m$. Therefore $t \times 10^k$ is a prime.

EDIT:We know this for at least one $k$, not all of them though. We need something that will give us another lower bound for $m$, such that for numbers bigger that $m$ there is a prime with $k \ge m$ digits, $\forall k$. We reduced the conjecture to the above statement.