MHB Proving an Infinite $\sigma-$Algebra is Uncountable

[mathmari]

Hey!

Show that an infinite $\sigma-$algebra is uncountable.

Could you give me some hints what I could do??

Do I have to start by supposing that an infinite $\sigma-$algebra is countable??

But how could I get a contradiction?? (Wondering)

 

[Opalg]

Hey!

Show that an infinite $\sigma-$algebra is uncountable.

Could you give me some hints what I could do??

Do I have to start by supposing that an infinite $\sigma-$algebra is countable??

But how could I get a contradiction?? (Wondering)

Here is an outline strategy for a proof. Suppose that $A$ is an infinite $\sigma$-algebra (consisting of subsets of some set $X$).

Case 1. $A$ contains an infinite descending chain of nonempty subsets of $X$. (In other words, there exists an infinite sequence $S_1 \supset S_2 \supset S_3 \supset \ldots$ of nonempty elements of $A$, each one strictly containing the following one.) In that case, let $T_n = S_n \setminus S_{n+1}$ (set-theoretic difference), for $n=1,2,3,\ldots$. Then the sets $T_n$ are nonempty and pairwise disjoint. For each subset $J$ (finite or infinite) of the natural numbers, let $$T_J = \bigcup_{n\in J}T_n.$$ There are uncountably many such sets (because there are uncountably many subsets of the natural numbers), they are all different, and they all belong to $A$. Therefore $A$ is uncountable.

Case 2. There are no infinite descending chains in $A$. In that case, every descending chain $S_1 \supset S_2 \supset S_3 \supset \ldots\supset S_n$ of nonempty elements of $A$ must terminate in a minimal nonempty element $S_n$ of $A$. There must be infinitely many such atoms in $A$ (otherwise $A$ would be finite), and these atoms must be pairwise disjoint. As in Case 1, you can construct uncountably many elements of $A$ by taking unions of the atoms.

 

[mathmari]

Case 2. There are no infinite descending chains in $A$. In that case, every descending chain $S_1 \supset S_2 \supset S_3 \supset \ldots\supset S_n$ of nonempty elements of $A$ must terminate in a minimal nonempty element $S_n$ of $A$. There must be infinitely many such atoms in $A$ (otherwise $A$ would be finite), and these atoms must be pairwise disjoint. As in Case 1, you can construct uncountably many elements of $A$ by taking unions of the atoms.

Could you explain me further the second case?? (Wondering)

What does "There must be infinitely many such atoms in $A$ " mean ?? (Wondering)

 

[johng1]

Hi,
Opalg has given you an excellent outline for a proof. If you still have questions, here is my fleshing out of her proof: