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 \(\displaystyle T_J = \bigcup_{n\in J}T_n.\) 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:

 

[blue_raver22]

To prove that an infinite $\sigma-$algebra is uncountable, we can use a proof by contradiction.

First, assume that the infinite $\sigma-$algebra, denoted by $\mathcal{F}$, is countable. This means that there exists a bijection between the elements of $\mathcal{F}$ and the set of natural numbers, $\mathbb{N}$.

Since $\mathcal{F}$ is a $\sigma-$algebra, it must contain the set of all countable unions of its elements. This means that for any countable collection of sets in $\mathcal{F}$, their union must also be in $\mathcal{F}$.

Now, consider the collection of all singleton sets in $\mathcal{F}$, denoted by $\{A_n\}_{n=1}^{\infty}$, where $A_n$ is the $n$th element in $\mathcal{F}$. Since $\mathcal{F}$ is assumed to be countable, this collection must also be countable.

However, by definition, the union of all singleton sets is the set of all natural numbers, $\mathbb{N}$, which is not a countable union of sets in $\mathcal{F}$. This contradicts the fact that $\mathcal{F}$ is a $\sigma-$algebra.

Therefore, our initial assumption that $\mathcal{F}$ is countable must be false, and thus, an infinite $\sigma-$algebra must be uncountable.