I'm getting in a muddle trying to understand statements which include both Landau symbols and the notion of "with high probability" (w.h.p.) in random graphs.
Setting/Background
Reading about Erdos-Renyi random graphs, I come across statements like
(1) "with high probability, the largest component $C_1$ has size $O(\log(n))$
and, in a separate situation,
(2) "with high probability $\text{diam}(G) = (3 + o(1))(1/\varepsilon)\log(\varepsilon^3n)$"
In the context of Erdos-Renyi random graphs, I understand "$G(n,p)$ has property A w.h.p." to mean that $\mathbb{P}(G\text{ has A}) \to 1$ as $n \to \infty$
Furthermore, I understand $f(n) = O(g(n))$ (in a more general setting) to mean that there are constants $n_0$ and $M$ such that for all $n\geq n_0$, $|f(n)|\leq M.|g(n)|$.
Finally, I understand $f(n) = o(g(n)))$ to mean $|f(n)/g(n)|\to 0$ as $n \to \infty$
Question
The only sensible way I can think of to parse statement (1) is as follows:
(1') There exists constant $M$ such that $\mathbb{P}(|C_1| \leq M\log(n)) \to 1$ as $n \to \infty$
Is this correct?
On the other hand, the best I can come up with for (2) is
(2') $$\mathbb{P}\left(\left| \frac{\text{diam}(G)\varepsilon}{\log(\varepsilon^3 n)}-3 \right| \to 0 \right)=1$$
which I'm significantly less convinced by. This is based on the idea that perhaps
$$f(G(n,p)) = o(g(n)) \text{ w.h.p.}$$
should be interpreted to mean $$\mathbb{P}\left(\left| \frac{f(G(n,p))}{g(n)} \right| \to 0 \right)=1$$
(NB: As this is my first question on here, please feel free to give pointers on how I can better match community expectations for questions.)
Your interpretation of (1) is spot on. An intuitive way to think of it is that there is a constant $M$ so that the probability of any connected component having more than $M\log(n)$ vertices is essentially zero (probability getting ever closer to zero as the size of the graph, $n$, increases).
For (2), think of $o(1)$ as a vanishing function (i.e. some $f(n)\to 0$). Thus, saying w.h.p $\text{diam}(G)=(3+o(1))(1/\epsilon)\log(\epsilon^{2}n)$ just means that as $n\to\infty$, w.h.p $\text{diam}(G)\sim \frac{3}{\epsilon}\log(\epsilon^{2}n)$; in other words, the diameter of $G$ (w.h.p) is asymptotic to $\frac{3}{\epsilon}\log(\epsilon^{2}n)$. What the $(3+o(1))$ tells you is that the constant out front is typically larger than $3$ (but negligibly larger). This is stronger than saying $\text{diam}(G)=\Theta(\frac{1}{\epsilon}\log(\epsilon^{2}n))$ because you can explicitly say what the best constant is (i.e. the $3$).