representing an uncountable free group as a union of an increasing sequence of countable subgroups

Question

Let $(G_\alpha)$ and $(K_\alpha)$ $(\alpha<\aleph_1)$ be strictly increasing chains of countable sets such that if $\alpha$ is a limit, then $G_\alpha=\bigcup_{\beta<\alpha}G_\beta$ and $K_\alpha=\bigcup_{\beta<\alpha}K_\beta$. Assume $\bigcup_{\alpha<\aleph_1}G_\alpha=\bigcup_{\alpha<\aleph_1}K_\alpha$. Does there exist a club $C$ such that $K_\gamma=G_\gamma$ for all $\gamma\in C$?

In the paper ``The Abelianization of Almost Free Groups" (the end of the proof of Lemma 2.5 on page 1801), this assertion is made where the $G_\alpha$ and $K_\alpha$ are subgroups of a free group with some additional properties, but the author makes the assertion I want without any explanation, so I assume the reason must be simple and may not depend on the group theory.

Answer 1

The set $C$ of $\gamma$ such that $K_\gamma = G_\gamma$ is closed, because if $\alpha$ is a limit point of $C$, then $$G_\alpha = \bigcup_{\beta < \alpha } G_\beta = \bigcup_{\beta <\alpha, \beta \in C} G_{\beta} = \bigcup_{\beta <\alpha, \beta \in C} K_{\beta} = \bigcup_{\beta<\alpha} K_\beta= K_\alpha.$$

So it suffices to show that $C$ is unbounded. Given any $\alpha_1$, because $K_{\alpha_1}$ is countable and is contained in $\bigcup_{\alpha<\aleph_1} G_\alpha$, we have $K_{\alpha_1} \subseteq G_{\alpha_2}$ for some $\alpha_2$. Similarly we have $G_{\alpha_2}\subseteq K_{\alpha_3} \subseteq G_{\alpha_4} \subseteq \dots$ so $$G_{\lim_{n \to \infty} }\alpha_n = K_{\lim_{n \to \infty} } \alpha_n$$ and thus $C$ contains an element greater than $\alpha_1$. So $C$ is unbounded.