Is it possible for a sequence of sets $A_{1}$, $A_{2}$, $A_{3}, ... \in \mathscr{F} $ to all be disjoint, AND all be countably infinite sets?

Question

The original problem:

Let $\Omega$ be a countably infinite set, and let $\mathscr{F}$ be the field consisting of all finite subsets of $\Omega$ and their complements.

If $A$ is finite, set $\mu(A)=0$, and if $A^c$ is finite, set $\mu(A)=1$. Show that $\mu$ is finitely additive, but not countably additive.

I've proven the cases:

• All $A_{n}$ are finite.

• There exists an $A_{*}^{c}$ that is finite.

The last case I thought to include:

• All $A^{c}_{n}$ are finite.

Therefore, each $A_n$ is countably infinite. However, we assume that all $A$ are disjoint. Is this possible?

In other words, would there be some element that at least two sets have in common as $n \rightarrow \infty$ ?

Answer 1

Take $S_p = \{p,p^2,p^3,....\}$ where $p$ is a prime number.

It is clear that $S_{p_1} \cap S_{p_2} = \emptyset$ for $p_1 \neq p_2$ but each $S_p$ is countably infinite.

Answer 2

Write all positive integers in an array thus: $$\matrix{1\cr 2&3\cr 4&5&6\cr 7&8&9&10\cr 11&12&13&14&15\cr \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\cr}$$ and let $A_k$ be the set whose elements are in column $k$. That is, $$A_1=\{1,2,4,7,11,\ldots\}\ ,\quad A_2=\{3,5,8,12,\ldots\}$$ and so on.