Existence of $\alpha$-sequence of infinitesimal sequences with $\alpha>\omega_1$ We are in ZFC & CH. Given family $Y=\{y_\alpha\}_{\alpha<\omega_1}$ of infinitesimal $\omega$-sequences (i.e. $\lim_{n\to\infty}y_{\alpha n}=0$) of rational numbers with the property:  $\forall\alpha<\beta: \lim_{n\to\infty}\frac{y_{\beta n}}{y_{\alpha n}}=0$. Can we prove that for any infinitesimal sequence $y$ there exists $y_\alpha$ such that $\lim_{n\to\infty}\frac{y_{\alpha n}}{y_{n}}=0$?
 A: The answer is no. First, let be construct a family of $\omega_1$ functions $f_\alpha:\omega\to\omega$ with the following three properties: each $f_\alpha$ tends to infinity, $f_\alpha\ll f_\beta$ whenever $\alpha<\beta$ (where $\ll$ is eventual domination: $f_\alpha(n)<f_\beta(n)$ for large enough $n$) and $f_\alpha(n)\ll n-k$ for all $k>0$.
We start with, for example, $f_0(n)=\lfloor\sqrt{n}\rfloor$. For any $\alpha$, we let $f_{\alpha+1}(n)=f_\alpha(n)+1$. For $\alpha$ a countable limit, pick a sequence $\alpha_k,k\in\omega$ such that $\alpha_k<\alpha$ and $\alpha_k\to\alpha$ (a so-called fundamental sequence for $\alpha$). For any $k$, pick $n_k$ large enough so that $f_{\alpha_k}(n)<n-k$ for $n>n_k$. Define
$$f_{\alpha}(n)=f_{\alpha_k}(n)+1 \text{ for the largest $k$ such that } n\geq k,n>n_k$$
Clearly, for any $k$ we have $f_\alpha\gg f_{\alpha_k}$. Since for any $\beta<\alpha$ there is $k$ with $\alpha_k>\beta$, it follows $f_\alpha\gg f_{\alpha_k}\gg f_\beta$, so $f_\alpha\gg f_\beta$. It follows $f_\alpha$ tends to infinity. Moreover, for any $k$ and  any $n\geq k,n>n_k$ we easily see $f_\alpha<n-k+1$, so $f_\alpha$ has the desired property. By transfinite recursion we get $f_\alpha$ for all $\alpha$.
Now define $y_{\alpha n}=1/f_\alpha(n)!$. Then $y_{\alpha n}\to 0$ (which is what, I think, it means for the sequence to be infinitesimal) and $y_{\beta n}/y_{\alpha n}=f_\alpha(n)!/f_\beta(n)!\leq 1/f_\beta(n)\to 0$ for $\beta>\alpha$, since eventually $f_\beta(n)\geq f_\alpha(n)+1$. Finally, since for each $\alpha$ we eventually have $f_\alpha(n)<n$, we have $y_{\alpha n}>1/n!$, so letting $y_n=1/n!$ we see $y_{\alpha n}/y_n\geq 1$ and doesn't tend to zero.
A: The answer is yes, if you assume the $y_\alpha$ decrease fast enough. Let me elaborate:
Let $(f_\alpha)_{\alpha < \omega_1}$ be a $\omega_1$-scale, i.e. a subset of $\omega ^\omega$ such that $\forall \alpha \,  \forall \beta < \alpha \colon f_\beta < f_\alpha$, where $<$ means eventually dominates, and $\forall f \, \exists \alpha \colon f < f_\alpha$.
Modify the scale as follows: Define $g_0= f_0$ and $g_{\alpha+1} (n) = n \cdot g_\alpha(n)$. For a limit $\lambda$ notice that the family $(g_\alpha)_{\alpha < \lambda}$ is countable, so there is $f_{\alpha'}$ which dominates all $(g_\alpha)_{\alpha < \lambda}$. Set $g_\lambda (n) =n \cdot f_{\alpha'}(n)$.
So $(g_\alpha)_{\alpha < \omega_1}$ is a new $\omega_1$-scale, and if you set $y_\alpha(n)= \dfrac{1}{g_\alpha (n)}$, you get the desired object.
