Example of an inverse system which suddenly "jumps" in size in a specific "controlled" way? I'm looking for an inverse system $(X_\alpha)_{\alpha < \omega_1}$ of vector spaces (EDIT: over a finite field) such that, for some $\lambda \geq 2$ with $\lambda < \lambda^{\omega_1}$ (I believe the case where $\lambda = \kappa^\omega$ for some $\kappa \geq 2$ is particularly interesting), the following conditions hold:

*

*The transition maps $X_\alpha \to X_\beta$, and the maps $\varprojlim_{\alpha < \omega_1} X_\alpha \to X_\beta$, are all surjective.


*For $\alpha < \omega_1$ a limit ordinal, the map $X_\alpha \to \varprojlim_{\beta < \alpha} X_\beta$ is injective.


*For $\alpha < \beta < \omega_1$ and $\gamma < \omega_1$ a limit ordinal, $\lambda = \dim X_\alpha = \dim \ker(X_\beta \to X_\alpha) = \dim ((\varprojlim_{\delta < \gamma} X_\delta ) / X_\gamma)$.


*$\dim(\varprojlim_{\alpha < \omega_1} X_\alpha) = \lambda^{\omega_1}$.
Conditions (1) and (2) are basic structural conditions. Condition (3), in addition to stipulating that $X_\alpha$ remain "small" below $\omega_1$, also stipulates that the "rate of change" of $X_\alpha$ be constant, at $\lambda$. The last part of (3) also rules out simply taking $X_\alpha = V^\alpha$ for some $V$ of dimension $\lambda = \kappa^\omega$. The point of the cardinal $\lambda^{\omega_1}$ in condition (4) is that it is the dimension of $\prod_{\alpha < \omega_1} X_\alpha$, and thus the obvious upper bound subject to the other conditions.

For comparison, consider an inverse system $(X_n)_{n \leq \omega}$ with $X_\omega = \varprojlim_{n < \omega} X_n$ with surjective transition maps such that $1 < |X_n| < \omega$ for all $n < \omega$. In this case, if $X_\omega = \varprojlim_{n <\omega} X_n$, then there is a "jump" in cardinality at $\omega$, all the way up to the continuum. Part of the motivation of the above conditions is to look for similar "jumps" in cardinality at the bigger ordinal $\omega_1$.
In fact, the full motivation of the above conditions is more specific, but in some sense it does boil down to asking for the growth to be "controlled" below $\omega_1$, and suddenly "jump" at $\omega_1$.
 A: Here is my argument, which assumes $|2^\omega|<|2^{\omega_1}|$ and uses an infinite base field. See Tim Campion's answer https://mathoverflow.net/a/376790/164965 for the general case.
Take the base field to be $\mathbb Q,$ set $\lambda=2^\omega=\mathfrak c,$ and $X_{\alpha}=\ell^\infty(\alpha)$: the bounded functions $\alpha\to\mathbb R.$
(If you prefer, you could use simple functions instead of all bounded functions.) The transition map $X_\alpha\to X_\beta$ is the restriction to $\beta.$
For each $\alpha>0,$ the space $X_\alpha$ has $\mathbb Q$-dimension $\mathfrak c.$ If $\dim X_0$ matters, take the example $X'_\alpha=X_{1+\alpha}$ instead.
For $\alpha<\beta<\omega_1$ the transition map has a kernel of $\mathbb R$-dimension at least one, so $\mathbb Q$-dimension $\mathfrak c.$
The limit $\varprojlim_{\delta<\gamma} X_\delta$ at countable limit ordinals $\gamma$ is the set of functions $\gamma\to\mathbb R$ that are bounded when restricted to $\delta$ for any $\delta<\gamma.$ This set contains $X_\gamma$ as a proper $\mathbb R$-subspace, so the $\mathbb Q$-dimension of $(\varprojlim_{\delta<\gamma} X_\delta)/X_\gamma$ is $\mathfrak c.$
The big limit $\varprojlim_{\alpha < \omega_1} X_\alpha$ is just $\ell^\infty(\omega_1)$ because any unbounded function on $\omega_1$ would be unbounded on some $\alpha<\omega_1.$ The dimension of this space is its cardinality, $2^{\omega_1}.$
A: Here is a way to adapt Harry West's answer to the case where the base field $k = \mathbb F_q$ is finite. Write $k_n = \mathbb F_{q^n}$ and $\bar k = \cup_n k_n$. Let $V$ be a $k$-vector space of dimension $\lambda$, where $\lambda = \lambda^\omega < \lambda^{\omega_1}$  (e.g. we could have $\lambda = \mathfrak c$) . Write $V_n = k_n \otimes_k V$ and $\bar V = \bar k \otimes_k V$. Then
$$X(\alpha) = ``\ell^\infty(\alpha; \bar V)":= \cup_n V_n^\alpha \subseteq \bar V^\alpha$$
with the transition maps being restriction. The idea is that the "size" of $v \in \bar V$ is the minimal $n$ such that $v \in V_n$, so $\ell^\infty(\alpha; \bar V)$ comprises the "bounded" $\bar V$-valued functions on $\alpha$.
Then just as in Harry West's answer, we have $\varprojlim_{\alpha < \omega_1} \ell^\infty(\alpha;\bar V) = \ell^\infty(\omega_1; \bar V)$, because if a function is bounded on all $\alpha < \omega_1$, then it's bounded on $\omega_1$. This space contains $V^{\omega_1}$ as a subspace, so it has dimension $\lambda^{\omega_1}$. But for $\alpha < \omega_1$, $\ell^\infty(\alpha;\bar V)$ is a proper subspace of $\varprojlim_{\beta < \alpha} \ell^\infty(\beta;\bar V)$ -- just take any sequence $\alpha_0, \alpha_1, \dots \to \alpha$ and look at functions taking values in $V_n \setminus V_{n-1}$ on $\alpha_n$. The cokernel is at least as big as $V$, so has dimension $\lambda$. All of the other conditions are clear, just as in Harry West's answer.

Notice that the field extensions weren't really used qua field extensions. We might as well simply define $V_\beta = V^{(\beta)}$, where this denotes the $\beta$-fold direct sum of copies of $V$. Then we can take
$$ X(\alpha) = \cup_{\beta < \omega} V_\beta^\alpha \subseteq V_\omega^\alpha $$
The advantage of this formulation, besides working over an arbitrary field, is that it suggests how to generalize beyond $\omega_1$. That is, if we want an inverse system which keeps growing slowly and then suddenly jumps in size at height $\mu$, then we can take, for an appropriate choice of $\lambda = \dim(V)$,
$$X(\alpha) = (\cup_{\beta < \omega} V_\beta^\alpha) \oplus (\cup_{\beta < \omega_1} V_\beta^\alpha) \oplus \cdots$$
where there will be one summand for each regular cardinal $<\mu$. Then at a given limit ordinal $\alpha$, the summand corresponding to $\operatorname{cf}(\alpha)$ will inject non-surjectively into the limit at $\alpha$. If there are infinitely many summands, then something funny happens at cofinality $\omega$, but I think this only increases the non-surjectivity.
