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

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.

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 $$y, for otherwise such $$y$$ would be a strictly smaller upper bound for $$\{x_n:n\geq 1\}$$. Hence $$\mathcal{F}_x=\cup_{y, 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.

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.

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 $$x$$.

• Thank you for your answer. I tried your approach by showing that $x\leq y$ implies $\mathcal{F}_x\subset\mathcal{F}_y$ below. Do you think its ok? Apr 29, 2021 at 1:35
• Troposphere? Sorry for disturbing you your feedback is very appreciated. Apr 30, 2021 at 13:00

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 $$x and $$z\neq \min \Omega$$. If $$z$$ has no immediate predecessor, then $$\mathcal{F}_z=\cup_{y, and since $$x we have $$\mathcal{F}_x\subset \mathcal{F}_z$$. If $$x$$ has an immediate predecessor $$y, 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 $$J_x=\Omega$$.