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This is a follow-up to this question by Dominic van der Zypen. For each bijection $f:\mathbb{N}\to\mathbb{N}$, let $$\operatorname{rc}(f) := \liminf_{N\to\infty} \frac{\left|\left\{(m,n)\in\{1,\dots,N …

$\newcommand\N{\mathbb N}$No, it is not possible to have $\mu_{[\N]^2}\big({\operatorname{rev}(f)}\big) = 1$. Given the function $f$, we will say that $n\in\mathbb{N}$ is good if $f(n)<f(k)$ for more …

$[\mathbb{N}]^n$, with edges between $a,b\in[\mathbb{N}]^n$ if $\#(a\cap b)=n-1$, is an infinite graph in which all vertices have infinite degree. Moreover, for any two vertices $a,b$ in $[\mathbb{N}] …

Yes. Consider the countable set $\mathbb{Q}^2$ and for each $x\in\mathbb{R}$ the subset $\{(a,b)\in\mathbb{Q}^2;a\leq x,b\leq-x\}$. Edit: See some easier examples in the comment below by bof. There ar …

One has $\max\big\{\mu\big(\exp(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$, as mentioned by Emil Jeřábek in the comments. We can also create a function $\varphi$ with $\mu\big(\text{shr}(\varp …