Is ${\displaystyle \left\{{\frac{\varphi(n)}{n}},\;\;n\notin\mathbb{P}\right\}}$ dense in $(0,1)$? Sierpiński and Schinzel proved
${\displaystyle \left\{{\frac{\varphi(n)}{n}},\;\;n\in\mathbb{N}\right\}}$ is dense in the
interval $(0,1)$ where $n$ is a natural number
and $\varphi(n)$ is Euler's totient function.
https://en.wikipedia.org/wiki/Euler's_totient_function#Ratio_of_consecutive_values
Is the set still dense in the same interval with the added restriction ${\displaystyle \left\{{\frac{\varphi(n)}{n}},\;\;n\notin\mathbb{P}\right\}}$ where $\mathbb{P}$ denotes the set of prime numbers?
 A: Yes. Note that
$$\left\{{\frac{\varphi(n)}{n}},n\in\mathbb{N}\right\} = \left\{{\frac{\varphi(n)}{n}},n\in\Bbb{N} \setminus \mathbb{P}\right\} \cup \left\{\frac{n-1}{n} : n \in \Bbb{P}\right\}.$$
The latter set admits only a single accumulation point: $1$. So, if we take a sequence from $\{\varphi(n)/n : n \in \Bbb{N}\}$ that converges to $x \in (0, 1)$, then only finitely many sequence terms can belong to $\{(n-1)/n : n \in \Bbb{P}\}$, so the (infinite) tail of the sequence must belong to $\{\varphi(n)/n : n \in \Bbb{N} \setminus \Bbb{P}\}.$
A: In general, if $A$ is a set dense in $(0,1)$ and $a_n\in A$ is a sequence such that $a_n\to 1,$ then $A\setminus \{a_n\}_{n=1}^{\infty}$ is dense in $(0,1).$
This can be fairly easily proved.

You don't really need $a_n\to 1,$ just that $a_n$ converges. But in your case, $a_n=\frac{\phi(p_n)}{p_n},$ converges to $1.$
As I noted in comments, $\frac{\phi(p^k)}{p^k}=\frac{\phi(p)}{p},$ so actually, you aren't even removing any values just by remove the primes $n$. But if you remove the prime powers $n,$ you can use the above argument.
