In Zhao's note(page 86) for Noga Alon's Probabilistic Methods:
Remark 8.1.4. When $\mathbb{P}(A_i)=o(1)$, Harris inequality gives us $$ \mathbb{P}(X=0)=\mathbb{P}\left(\bar{A}_{1} \cdots \bar{A}_{k}\right) \geq \mathbb{P}\left(\bar{A}_{1}\right) \cdots \mathbb{P}\left(\bar{A}_{k}\right)=\prod_{i=1}^{k}\left(1-\mathbb{P}\left(A_{i}\right)\right)=e^{-(1+o(1)) \sum_{i-1}^{k} \mathbb{P}\left(A_{i}\right)}=e^{-(1+o(1)) \mu} $$
I was confused by the $o(1)$ notations here:
- What does it mean by $\mathbb{P}(A_i)=o(1)$? I understand it when we say $f(n)=o(1)$ it means that $f(n)\rightarrow 0$ when $n\rightarrow \infty$. But by $\mathbb{P}(A_i)=o(1)$ I see no corresponding variable like $n$.
- How does the second-to-last equation come from? Especially, how to understand the $o(1)$ in $e^{-(1+o(1)) \sum_{i-1}^{k} \mathbb{P}(A_{i})}$. What I know is that $1-x\le e^{-x}$ when $x\in [0,1]$.