Set of Finite Character with Countable Union is Type M

Theorem

Let $S$ be a set of sets of finite character.

Let its union $\bigcup S$ be countable.

Then $S$ is of type $M$.

That is:

every element of $S$ is a subset of a maximal element of $S$ under the subset relation.

Proof 1

By Countable Set has Choice Function, $S$ has a choice function.

The result follows from Set of Finite Character with Choice Function is Type $M$.

$\blacksquare$

Proof 2

Let $S$ be a set of sets of finite character whose union $\bigcup S$ is countable.

Let $D := \bigcup S$ be enumerated as:

$D = \set {d_1, d_2, \ldots, d_n, d_{n + 1}, \ldots}$

Let $b \in S$ be arbitrary.

From the Principle of Recursive Definition, we can generate a countable sequence of elements of $S$ as follows:

Let $b_0 := b$.

Let it be assumed that $b_n$ has been defined.

Then we take:

$b_{n + 1} := \begin{cases} b_n \cup \set {d_n} & : \text {if $b_n \cup \set {d_n}$ is an element of $S$} \\ b_n & : \text {otherwise} \end{cases}$

Let $C$ be the set defined as

$C := \set {b_0, b_1, \ldots, b_n, b_{n + 1}, \ldots}$

From Increasing Sequence of Sets forms Nest, $C$ is a nest which, as $C$ is a set, is in fact a chain.

Hence $\bigcup C \in S$.

From Union of Chain in Set of Finite Character with Countable Union is Maximal Element, $\bigcup C$ is a maximal element of $S$ under the subset relation.

Thus $b$ is a subset of a maximal element of $S$ under the subset relation.

As $b$ is arbitrary, the result follows.

$\blacksquare$