I'm trying to arrive at estimate 1.17 (page 21) of Koukoulopoulos lecture notes [https://dms.umontreal.ca/~koukoulo/documents/notes/sievemethods.pdf]
$$\#\{n \leq x : p|n \Rightarrow p > \sqrt{x}\} = \frac{(2e^{- \gamma}+ o(1))x}{\log x} + O(4^{(1+o(1))\frac{\sqrt{x}}{\log x}})$$
In the error term I can see that we are using the Prime Number Theorem, but i don't see how we get the $1+o(1)$ or how to get the $o(1)$ in the main term
We start with $\#\{n \leq x: p|n \Rightarrow p > \sqrt{x}\} = \lfloor{x}\rfloor - \sum_{i=1}^{r}\lfloor{\frac{x}{p_{i}}\rfloor}\pm\dots$ by an inclusion-exclusion argument.
We have the estimate that $\lfloor{x}\rfloor = x + O(1).$ Therefore we obtain $\#\{n \leq x: p|n \Rightarrow p > \sqrt{x}\} = x \prod_{p \leq \sqrt{x}}\Big(1 - \frac{1}{p}\Big).$
I know here I am missing the error term. I am not sure how we incorporate this error term going forward, is it just a $O(1)$ error term for every term in the product?
Using Merten's theorem that for $z \geq 2$ we have $\prod_{p \leq z}\Big(1 - \frac{1}{p}\Big) = \frac{e^{- \gamma}}{\log z}\Big(1 + O\Big(\frac{1}{\log z}\Big)\Big),$ gives
$$ \#\{n \leq x: p|n \Rightarrow p > \sqrt{x}\} = \frac{xe^{- \gamma}}{\log{\sqrt{x}}}\Big(1 + O\Big(\frac{1}{\log{\sqrt{x}}}\Big)\Big) = \frac{2xe^{-\gamma}}{\log x}\Big(1 + O\Big(\frac{2}{\log x}\Big)\Big).$$
I can rewrite the inclusion-exclusion argument in terms of the Mobius function as is done in equation 1.1.6 (page 20), but then I don't see how we get the overall result.
It's mostly the asymptotic notation that I'm stuck on as I'm not too familiar with it yet, particularly using "little oh" notation inside of "big oh" notation.
Thanks.