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I asked this question on Math.stackexchange but got no answer.

In the paper Zubrilina - Asymptotic behavior of the edge metric dimension of the random graph (MR, 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 $$\mathbb{P}\left[\operatorname{edim}(G(n,p)) \leq (1+o(1))\frac{4 \log n}{\log(1/q)}\right] \rightarrow 1 \text{ as $n \rightarrow \infty$?} $$

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 $\omega \in \{1,\dotsc,n\}$ 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 $$\mathbb{P}( \text{$W$ does not distinguish $e_1$, $e_2$}) \leq 1/n^4p^2. $$ 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?

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    $\begingroup$ You should define terms in your question (like $\mathrm{edim}$, which is surely not a ubiquitous notion). $\endgroup$ Jun 25 '20 at 15:51
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    $\begingroup$ Also, you should link to your MSE question to avoid duplication of effort. Anyways, it seems your concerns are about basic shorthands when dealing with random objects, which are standard but I agree can be confusing when not spelled out. $\endgroup$ Jun 25 '20 at 16:02
  • $\begingroup$ The original link went to a page with only generic content for me ("0 Articles Found"). I pasted in the DOI link, which takes me to an articles-to-appear page which doesn't seem to list Zubrilina's. So I pasted in the name of the article and a link to the MR. I hope that was all all right. $\endgroup$
    – LSpice
    Apr 21 at 23:20
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I just skimmed the paper. When she writes

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

she means this inequality holds "asymptotically almost surely", i.e., the probability that this inequality holds goes to 1 as $n\to \infty$ (note: this is for a fixed $p$). This matches what you thought, and it's clear from her use of "a.a.s" in Lemma 2.3 and in Section 3. I previously misread your interpretation to think you were suggesting the result was a bound on $P(edim(G(n,p)))$ which of course makes no sense.

Similarly, in Lemma 2.2, edim$(G)$ refers to the family of graphs, and it's enough to prove the inequality holds a.a.s. as $n\to \infty$. So, when she says "Let $G = G(n,p)$ be the random graph" she is saying $G$ represents the family of graphs drawn from this random variable. That's confirmed in the proof, where she talks about the expected size of the edge set, $E$ -- it's a random variable. I think the definition of $\omega$ is fine as stated. She's choosing a number to guarantee a probabilistic inequality. It would be like if I said, "Consider an $n$-sided die, and let $n$ be a number such that the probability of rolling two consecutive 1s is less than 0.01." That statement does define a number. Of course, there is a probability on a given draw of $G = G(n,p)$ that you get a graph with no edges, or only 1 edge, and maybe she could have spelled out how to handle those cases. But I think they are vacuous, because the definition of "edge metric dimension" says "for any distinct $e_1,e_2\in E$". Also, the probability of this situation occurring goes to zero as $n\to \infty$.

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  • $\begingroup$ I don't understand what you mean by "the edge metric dimension of the family of random graphs." This notion is just defined for a graph. I think that mahmoud's interpretation of the inequality holding a.a.s. is correct. For instance, the paper cites arxiv.org/abs/1208.3801 which proves a similar thing but for metric dimension instead of edge metric dimension (and states the a.a.s. bit in their theorem more clearly). $\endgroup$ Jun 25 '20 at 16:19
  • $\begingroup$ @SamHopkins: I think you're right. I'll edit. $\endgroup$ Jun 25 '20 at 16:32
  • $\begingroup$ The formal interpretation of Lemma 2.2 should also involve some kind of a.a.s. statement, but there's a little more unwinding to do there. $\endgroup$ Jun 25 '20 at 16:36
  • $\begingroup$ I edited a second time. I just jumped the gun on the first edit. $\endgroup$ Jun 25 '20 at 16:44
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    $\begingroup$ For the first case: I think that for many readers, it would still be unclear what it means to say that "the probability that $X_n\leq (1+o(1))f(n)$ goes to $1$ as $n\to\infty$". The "$o()$" notation already includes an $n\to\infty$ limit, which makes it hard to understand that statement. To interpret it really clearly one could write: for any $\epsilon>0$, the probability that $X_n\leq (1+\epsilon)f(n)$ goes to $1$ as $n\to\infty$. $\endgroup$ Jun 26 '20 at 9:45

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