Height of a certain order

Question

Let $$\mathcal{M} := \{(a_{n,i})_{(n,i) \in \mathbb{N}^2} \in \mathbb{R}^{\mathbb{N} \times \mathbb{N}} \mid \forall n \in \mathbb{N} \ \forall i \in \mathbb{N} \ 0 \leq a_{n,i} \leq a_{n+1,i}\}$$

Let the quasiorder $\preceq$ on $\mathcal{M}$ be defined via $(a_{n,i})_{(n,i) \in \mathbb{N}^2} \preceq (b_{n,i})_{(n,i) \in \mathbb{N}^2}$ iff $\forall n \in \mathbb{N} \ \exists k \in \mathbb{N} \ \forall i \in \mathbb{N} \ a_{n,i} \leq b_{k,i}$.

Question: Does the first uncountable ordinal $\omega_1$ embed into the poset $(\mathcal{M},\prec)$?

Motivation: I want to compare convergence rates at countably many positions. Intuitively, $a_{n,i}$ is how close to the limit the $n$-th item is at the $i$-th position. The order $\prec$ then captures a kind of converges faster or uniformly not too much slower everywhere.

It seems to me that the answer should be rather straight-forward, but right now I do not even see what it should be.

Edit: I had used $\mathbb{N}$ in place of $\mathbb{R}_{\geq 0}$ first, but then realized that I was not even sure that the two equivalent.

Answer 1

We can find such an embedding based on the (slightly modified) fast-growing hierarchy:

$F_0(n) = 2n$

$F_{\alpha+1}(n) = F_{\alpha}^n(n)$

If $\alpha$ is a limit ordinal, then $F_{\alpha}(n) = \sup_{m \le n}\{F_{\alpha[m]}(n)\}$.

where $\alpha[n]$ is known as the fundamental sequence of $\alpha$. For our purposes, for each countable limit ordinal $\alpha$ we simply choose $\alpha[n]$ to be an arbitrary increasing sequence of ordinals with supremum $\alpha$. We can do this using the axiom of choice.

Next, we will define an embedding $g:\omega_1 \to (\mathcal{M},\prec)$ by:

$g(\alpha)_{n,i} = F_{\alpha}(n+i)$

To prove this is an embedding, we use some easy to prove facts:

1. If $m < n$, $F_\alpha(m) < F_\alpha(n)$.
2. If $\alpha < \beta$, there is an $n \in \mathbb{N}$ such that for all $m \ge n$, $F_\alpha(2m) < F_\beta(m)$.

From 2. we can see that for any $\alpha < \beta < \omega_1$, we can choose an $n \in \mathbb{N}$ such that for all $m \ge n$, $F_\alpha(2m) < F_\beta(m)$, so for any $j \in \mathbb{N}$, letting $k = n+j$ yields $g(\beta)_{k,i} = F_\beta(k+i) > F_\alpha(2k+2i) > F_\alpha(j+i) = g(\alpha)_{n,i}$ for all $i \in \mathbb{N}$. So $g(\alpha) \preceq g(\beta)$. On the other hand, There cannot be a $k$ such that $g(\alpha)_{k,i} = F_\alpha(k+i) \ge F_\beta(i) = g(\beta)_{0,i}$ for all $i \in \mathbb{N}$, since for $i = n+k$ we have $F_\beta(i) > F_\alpha(2i) = F_{\alpha}(n+k+i) > F_\alpha(k+i)$. So $g(\beta) \npreceq g(\alpha)$, as desired.