Suppose $ X$ is a separable metric space and ($U_\alpha : \alpha < \gamma$) is an increasing sequence of open sets (i.e. $U_\alpha \subseteq U_\beta$ for $\alpha < \beta$). Show that there is a countable $γ_0$ such that $U_\alpha= U_\beta$ for all $ \alpha, \beta > \gamma_0$.
I'm really not sure where to begin here. Any help would be appreciated.
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EDIT: My previous proof was wrong. Here's a new one. A separable metric space has a countable basis $\mathcal{B}=\{B_m\}$ for its topology, as $m$ runs over $\mathbb{N}$. So every open set $U\subset X$ can be written as a union of these $B_m$. Thus if $U\subsetneq V$, then there's some $B_m\subset V$ but not contained in $U$. (My previous solution claimed there was such a $B_m$ disjoint from $U$, which need not hold.) Finally if we have a chain $\{U_i\}_{i\in I}$ with $\sup_{i<j}U_{i}\subsetneq U_{j}$, there is induced an injection $I\to \mathcal{B}$ by sending $i$ to some $B_{m(i)}$ with $i$ the least index with $B_{m(i)}\subset U_i$. (We're making arbitrary choices here, which could be avoided if necessary.) So $I$ is at most countable. I've slightly reformulated your condition: to get yours back just take $I$ to be the least ordinal such that $U_\alpha\subsetneq U_\beta$ implies $\beta<I$, and assume that an initial segment of the inclusions $\cup_{\alpha<\beta} U_\alpha\subset U_\beta$ are strict.
Note first that $X$, being a separable metric space, is second countable. Let $Y=\bigcup_{\alpha<\gamma}U_\alpha$; then $Y$ is also second countable and therefore Lindelöf. $\{U_\alpha:\alpha<\gamma\}$ is an open cover of $Y$, so it has a countable subcover $\mathscr{U}$. Let $\gamma_0=\sup\{\alpha<\gamma:U_\alpha\in\mathscr{U}\}$. Now show that $U_\alpha=Y$ for $\gamma_0\le\alpha<\gamma$.