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@AlexRavsky: OK. I think $\mathrm{cov}(\mathcal N)$ is true (and can be proved as I outlined in my comments on the question), but I don't think I'll have any time to write the details down today.
I think the same idea can be used to show $\mathfrak{x} \leq \mathfrak{r}$. Roughly: instead of finding a convergent subsequence (given to you by $\mathfrak{r}_\sigma$), you just need to find a subsequence such that all subsequential limits are trapped within $\varepsilon/2$ of each other (which $\mathfrak{r}$ does). Then taking differences as in your previous comment, you get a subsequence where all subsequential limits are within $\varepsilon$ of $1$. This allows you to dodge any remote set. (Sorry for all the commenting -- if I have time today, I'll try to turn some of this into an answer.)
Perhaps I'm missing something, but I don't see why $\mathrm{cov}(\mathcal{A}(\mathbb T)) \leq \mathfrak{r}_\sigma$. The characterization of $\mathfrak{r}_\sigma$ you mention gives you an increasing sequence $(u_n)_{n \in \mathbb N}$ such that $(z^{u_n})_{n \in \mathbb N}$ converges, but you have no control over what this sequence converges to. Your definition of $\mathrm{cov}(\mathcal{A}(\mathbb T))$ requires that it converge to $1$. How do you do this?
This shows (modulo my claim about Cohen-generic $z$'s) that $\mathfrak{x} \geq \mathrm{cov}(\mathcal M)$. Similarly, if you can show that a "random" $z$ has the same property, then this would show $\mathrm{cov}(\mathcal N) \leq \mathfrak{x}$ also.
My first thought is to consider what a "generic" $z$ does. For each $z \in \mathbb T$, define $R_z = \{n :\, |z^n-1| \geq 1/2 \}$. (Of course the "1/2" could be made smaller, but I doubt it matters.) I don't have a proof right now, but it seems like a Cohen-generic $z$ should have the property that $R_z \cap A$ is infinite for any infinite $A$ in the ground model. In other words: given $A$, there are only a meager set of $z$'s with $R_z \cap A$ finite. This means you need at least $\mathrm{cov}(\mathcal M)$ $A$'s to get a sufficiently large collection $\mathcal F$. . . .
