# An infinite $\sigma$-algebra contains a infinite sequence of disjoint nonempty sets, my attempt.

As stated in the title the question is "Show that an infinite $$\sigma$$-algebra contains a infinite sequence of disjoint nonempty sets". I've seen this question asked a lot here, many with different answers. I'd like to know if my approach to it is correct.

Consider $$\mathcal{A}$$ an infinite $$\sigma$$-algebra on $$X$$ and define $$\mathfrak{C}=\{\mathcal{C}\subset \mathcal{A}-\{\emptyset\},\, \mathcal{C}\text{ is a disjoint collection of subsets} \}$$ Let show that there exists finite $$\mathcal{C}\in\mathfrak{C}$$ of arbitrarily large size. If this wasn't the case, take $$N=\max\{|\mathcal{C}|,\,\mathcal{C}\in \mathfrak{C}\}$$, there exists $$\mathcal{C}_0$$ such that $$|\mathcal{C}_0|=N$$ and therefore $$\mathcal{C}_0=\{E_1,\dots,E_N\}$$.

If $$\bigcup_{n=1}^N E_n \neq X$$ then $$\mathcal{C}_0$$ would not be maximal as $$\mathcal{C}_0\cup \{X-\bigcup_{n=1}^N E_n\}$$ would be a collection of $$N+1$$ disjoint subsets, therefore we have $$\bigcup_{n=1}^N E_n = X$$. Now consider $$\mathcal{A}'$$, the $$\sigma$$-algebra generated by $$\mathcal{C}_0$$, $$\mathcal{A}'$$ will be finite due to $$\mathcal{C}_0$$ being finite.

Since $$\mathcal{A}$$ is infinite there exists $$E\in \mathcal{A}-\mathcal{A}'$$. For some $$n_0$$ we must have $$E_{n_0}\not\subset E$$ and $$E_{n_0}\cap E\neq \emptyset$$, otherwise we would have that for all $$n$$ either $$E_n\subset E$$ or $$E_n\cap E=\emptyset$$ meaning $$E=\bigcup_{E_n\cap E\neq \emptyset} E_n\in \mathcal{A}'$$ Now, set $$E'_{n_0}:=E_{n_0}\cap E\in \mathcal{A}-\{\emptyset\}$$ and $$E_{N+1}:=E_{n_0}-E'_{n_0}\in \mathcal{A}-\{\emptyset\}$$. We have $$E_{n_0}=E'_{n_0}\cup E_{N+1}$$ where the union is disjoint, moreover $$\mathcal{C}:=\{E_1,\dots,E'_{n_0},\dots,E_N,E_{N+1}\}\in \mathfrak{C}$$ and $$|\mathcal{C}|=N+1$$. This contradicts the maximality of $$N$$, therefore there exists finite $$\mathcal{C}\in\mathfrak{C}$$ of arbitrarily large size.

EDIT: As detailed in the comments there was a mistake, I'll try to see if I can salvage part of this proof.

• Every set in $\mathfrak C$ been finite doesn't imply existence of $N$ such that all set is less then $N$. – mihaild Apr 12 at 15:40
• Woops, you're right. Still I feel I can apply my construction below to increase any finite set to arbitrary size, am I right? – Julio Cáceres Apr 12 at 15:50
• To amend it partially order $\mathfrak{C}$ by inclusion and look at maximal elements. – user647486 Apr 12 at 15:54
• Yeah, I was thinking about that. Although I'm not sure if the inclusion is the right partial order to use. – Julio Cáceres Apr 12 at 15:57
• @JulioCáceres Well, just do it. You will be sure at the end. – user647486 Apr 12 at 16:04