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Suppose I have the following asymptotic: $$\{n \in \mathbb{Z} \cap [1, X] : f(n) \text{ is good}\} \sim cX$$ as $X \rightarrow \infty$ for some absolute constant $c$ where "good" means some desired property. Does this mean there are infinitely many $n$ such that $f(n)$ is good?

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up vote 2 down vote accepted

Let $A = \{n\in\Bbb N: f(n) \text{ is good}\}$ be the set of all good points. I assume that you have a typo with $X$ instead of $x$ and your assumption is in fact $$ \#(A\cap [1,x]) \sim cx, \quad x\to\infty $$ which is by the definition $$ \lim_{x\to\infty }\frac{ \#(A\cap [1,x])}{cx} = 1. \tag{1} $$ In case $A$ is finite, you have that the numerator in $(1)$ is a bounded function of $x$, and thus the limit would be $0$. Thus, $A$ is infinite.

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