Question about additive combinatorics proof regarding probability estimate

Solution 1:

Yes, the $O(1)$ in "$\log \log n+O(1)$" stands for some function which is $O(1)$. This means a function $f(x)$ such that there exists $x_0$ so that $|f(x)|\le C\cdot 1$ for all $x\ge x_0$. This function, $f(x)$, can be negative, it just must be bounded from below and above in the long run. Really, the notation $$ s(x)=t(x)+O(r(x)) $$ is a shorthand (or abuse of notation) for $$ s(x)-t(x)\qquad \text{is}\qquad O(r(x)) $$

On the other hand, a function $g(x)$ is $o(1)$ if $\lim_{x\to\infty} g(x)=0$. This is a stronger condition than $O(1)$. Therefore, when you say the probability is $1-o(1)$, this means that the probability is $1$ minus a function going to zero, which is equivalent to saying that the probability approaches one. $1+o(1)$ would mean the same thing.