Show that $\cup_{x\in\Omega}\mathcal{F}_x$ is a $\sigma$-algebra

Question

Let $X$ be an arbitrary set and let $\Omega$ denote the minimal uncountable well-ordered set. Given $\mathcal{E}\subset \mathcal{P}(X) $ with $\emptyset\in \mathcal{E}$ we define a collection of subsets of $\mathcal{P}(X)$ by transfinite recursion as follows:

Define $\mathcal{E}^c:=\{E^c:E\in\mathcal{E}\}$ and $\mathcal{E}_{\sigma}:=\{\cup_{n=1}^{\infty}E_n:(E_n)\subset\mathcal{E}\}$. Define $1:=\min \Omega$ and set $\mathcal{F}_1:=\mathcal{E}\cup \mathcal{E}^c$. If $x\in\Omega$ and if $x$ has an immediate predecessor $y$ then we set $\mathcal{F}_x:=(\mathcal{F}_y)_{\sigma}\cup ((\mathcal{F}_y)_{\sigma})^c$, and if $x$ has no immediate predecessor then we set $\mathcal{F}_x:=\cup_{y<x}\mathcal{F_y}$.

The goal is to show that $\cup_{x\in\Omega}\mathcal{F}_x$ is a $\sigma$-algebra in $X$. Am stuck on proving closure under countable unions:

Let $(E_n)\subset \cup_{x\in\Omega}\mathcal{F}_x$. For each $n$ choose $x_n\in\Omega$ such that $E_n\in \mathcal{F}_{x_n}$ (axiom of countable choice). Since the set $\{x_n:n\geq 1\}$ is a countable subset of $\Omega$, it has an upper bound. Let $x$ denote the smallest upper bound. If $x\neq x_n$ for each $n$, then $x$ has no immediate predecessor $y<x$, for otherwise such $y$ would be a strictly smaller upper bound for $\{x_n:n\geq 1\}$. Hence $\mathcal{F}_x=\cup_{y<x}\mathcal{F_y}\supset \cup_{n\geq 1}\mathcal{F}_{x_n}$, and since $\Omega$ has no largest element it follows that $\cup_{n\geq 1} E_n \in (\mathcal{F_x})_{\sigma}\subset \mathcal{F_{x+1}} $, where $x+1$ denotes the immediate sucessor of $x$ in $\Omega$.

But what about the case where $x=x_n$ for some $n$?

EDIT: I tried to follow Troposhere's approach below. Any feedback is very appreciated.

Answer 1

The basic fact you seem to have overlooked is that $\mathcal E \subseteq \mathcal E_\sigma$. (Just take all the $E_n$s in the definition to be identical).

From this fact you can prove that $\mathcal F_\alpha \subseteq \mathcal F_{\alpha+1}$ and in general (by induction on $\beta$) that $\alpha\le\beta \Rightarrow \mathcal F_\alpha \subseteq \mathcal F_\beta$ even when $\beta$ is a successor.

Once you know this, the argument you've presented for $x\ne x_n$ actually works no matter there is a maximal $x_n$ or not.

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Alternatively you could drop your requirement that $x$ must be a smallest upper bound. Simply choosing $x$ as some upper bound of the $x_n$ will suffice for your argument. And it is true in general that every $x\in\Omega$ sits below a limit ordinal that is also in $\Omega$, namely the least upper bound of $\{x+n\mid n\in\mathbb N\}$.

However, even though this alternative works, it loses out on the fundamental insight that $\mathcal F_x$ is an increasing function of $x$.

Answer 2

I will try to follow Troposhere's approach by showing that $x\leq y$ implies $\mathcal{F}_x\subset\mathcal{F}_y$.

Let $x\in \Omega$ and let $J_x:=\{y\in \Omega : x\leq y\Rightarrow \mathcal{F}_x \subset \mathcal{F}_y\}$. It suffices to show that $J_x=\Omega$. We use transfinite induction. Let $z\in \Omega$ and suppose $S_z\subset J_x$. Assume $x\leq z$. If $x=z$, then $\mathcal{F}_x=\mathcal{F}_z$. Otherwise $x<z$ and $z\neq \min \Omega$. If $z$ has no immediate predecessor, then $\mathcal{F}_z=\cup_{y<z}\mathcal{F_y}$, and since $x<z$ we have $\mathcal{F}_x\subset \mathcal{F}_z$. If $x$ has an immediate predecessor $y<z$, then $x\leq y$, and since $y\in S_z\subset J_x$ we have $\mathcal{F}_x\subset \mathcal{F}_y$. It then follows from $\mathcal{F}_z=(\mathcal{F}_y)_{\sigma}\cup ((\mathcal{F}_y)_{\sigma})^c$ that $\mathcal{F}_x\subset\mathcal{F}_z$. In any case $x\leq z$ implies $\mathcal{F}_x\subset\mathcal{F}_z$, and so $z\in J_x$. By transfinite induction we conclude that $J_x=\Omega$.