Interpretation of probability statements in Nina Zubrilina's paper In the paper
https://content.sciendo.com/view/journals/dmgt/ahead-of-print/article-10.7151-dmgt.2210/article-10.7151-dmgt.2210.xml?language=en
The main result is

$$\operatorname{edim}(G(n,p)) \leq (1+o(1))\frac{4 \log n}{\log(1/q)},$$ where $q= 1-2p(1-p)^2(2-p)$

My first question is how should I interpret the result, what is $\operatorname{edim}$ of random graph. Should I interpret it as
$$\mathbb{P} \left[ \operatorname{edim}(G(n,p)) \leq (1+o(1))\frac{4 \log n}{\log(1/q)} \right] \rightarrow 1 \text{ as } n \rightarrow \infty \text{ ?}$$
My second question is concerned with how to interpret lemma 2.2, it is stated that

Let $G=G(n,p)$ be the random graph. Let $V,E$ denote the vertex and edge sets. Let $\omega \in \{1,\cdots,n\}$ be such that for any two distinct edges $e_1,e_2$ of $E$, a uniformly random subset $W \subset V$ of size $\omega$ satisfies
$$\mathbb{P}( W \text{ does not distinguish } e_1,e_2) \leq 1/n^4p^2 $$
Then
$$\operatorname{edim}(G) \leq \omega$$

So, firstly how should I understand $E$ as subset of a random graph, and how can I fix two edges of this seemingly random set by saying "for any two distinct edges $e_1,e_2 \in E$". I am confused about how I interpret such statement. Can any one clarify them ?
 A: We could address your second question by saying that $G \sim G(n,p)$ is a random subgraph of the complete graph $K_n$ on some vertex set $\{v_1, \dots, v_n\}$, and considering the event "$W$ distinguises $e_1, e_2$" for $e_1, e_2 \in E(K_n)$ to be true automatically if $e_1, e_2$ are not both edges of $G$. This makes for a very reasonable interpretation. Also, when the diameter of $G(n,p)$ is $2$ (which is true with probability tending to $1$), the presence of edges $e_1, e_2$ in $G(n,p)$ will not affect distances from $W$ to their endpoints, except in the rare case that one of the endpoints is in $W$.
But then the thing that Lemma 2.2 actually proves is that if $G$ and $W$ are both randomly chosen, the expected number of pairs not distinguished is $\le \frac18$. This only proves that $$P(\operatorname{edim}(G) \le \omega) \ge \frac78$$ because it allows for the possibility that, for example, with probability $\frac18$ we get a $G$ such that every $W$ of size $\omega$ distinguishes all but one pair. So, if the hypotheses of Lemma 2.2 are met, then the main theorem should have a "with probability at least $\frac78$" at the end as well.
However, we can strengthen this result by making the analysis slightly more sophisticated. The expected value calculation also implies that
$$
   \mathbb P_{G \sim G(n,p)}(\text{every }W\text{ fails to distinguish }\log\log n\text{ pairs}) \le \frac1{8 \log\log n}
$$
so with probability tending to $1$, $G$ has a set $W$ of this size that distinguishes all but $\log\log n$ pairs of edges. But we can fix this by making $W$ slightly bigger: for example, for every pair $\{e_1, e_2\}$ that's not distinguished, add one of the endpoints of one of the edges to $W$. This bigger set still has size $(1+o(1))\frac{4\log n}{\log 1/q}$, because $\log\log n$ is a much smaller term, so the main theorem now holds with high probability.
