Fix $0 < \alpha < 1$ a real. Let $S\_\alpha$ the set of integers $n \geq 1$ such that be $\phi(n)>\alpha n$. For $x>0$, let $S\_\alpha(x)$ be the number of positive integers $n$ less han $x$ 
such that $n \in S\_\alpha$.



> 
> Does $S\_\alpha$ has a natural density (that is, does $\lim\_{x \rightarrow \infty} S\_\alpha(x)/x$ exist)? If so, is that density $0$, $1$ or in between? If not, same questions for $\liminf\_{x \rightarrow \infty} S\_\alpha(x)/x$ and $\limsup\_{x \rightarrow \infty} S\_\alpha(x)/x$.
> 
> 
> 


Of course, $n \in S\_\alpha$ if and only if $\prod\_{p \mid n} (1-1/p) > \alpha$, where the product is over prime factors of $n$. But I am not sure where to go from here.



Density of integers $n$ whose totient $\varphi(n)$ is larger than $\alpha n$
Yes -- look up the Erdos--Wintner theorem. (The special case you asked about was earlier proved by Schoenberg.) If we call the density $D(\alpha)$, then in fact $D$ is continuous and strictly decreasing on $[0,1]$, with $D(0)=1$ and $D(1)=0$.


