Restricted Tukey's Theorem/Strong Form

Theorem
Let $X$ be a set.

Let $\mathcal A$ be a non-empty set of subsets of $X$.

Let $'$ be a unary operation on $X$.

Suppose that:


 * $(1)\quad$ $\mathcal A$ has finite character.


 * $(2)\quad$ For all $A \in \mathcal A$ and all $x \in X$, either $A \cup \{ x \} \in \mathcal A$ or $A \cup \{ x' \} \in \mathcal A$

Then:

For each $A \in \mathcal A$ there exists a $C \in \mathcal A$ such that $A \subseteq C$ and for all $x \in X$, either $x \in C$ or $x' \in C$.

Proof from the weak form
Let $A \in \mathcal A$.

Let $\mathcal B = \{ B: B \subseteq X \text{ and } A \cup B \in \mathcal A \}$.

$\mathcal B$ has finite character:

First suppose that $B \in \mathcal B$ and $F$ is a finite subset of $B$.

Then since $B \in \mathcal B$, $B \subseteq X$ and $A \cup B \in \mathcal A \}$.

We wish to show that $F \in \mathcal B$.

Since $F \subseteq B \subseteq X$, $F \subseteq X$, so it remains to show that $A \cup F \in \mathcal A$.

$A \cup F \subseteq A \cup B$.

Let $G$ be a finite subset of $A \cup F$.

Then $G$ is a finite subset of $A \cup B$.

Since $A \cup B \in \mathcal A$ and $\mathcal A$ has finite character, $G \in \mathcal A$.

Thus every finite subset of $A \cup F$ is in $\mathcal A$.

Since $\mathcal A$ has finite character, $A \cup F \in \mathcal A$.

Thus $F \in \mathcal B$.

Suppose instead that $B \subseteq X$ and every finite subset of $B$ is an element of $\mathcal B$.

We wish to show that $B \in \mathcal B$.

In order to do this, we must show that $A \cup B \in \mathcal A$.

Let $F$ be a finite subset of $A \cup B$.

$\mathcal A$ has finite character, $F \cap A \in \mathcal A$.

Since $F \cap B$ is a finite subset of $B$, $F \cap B \in \mathcal B$ by assumption.

Then by the definition of $\mathcal B$, $(F \cap B) \cup A \in \mathcal A$.

But $F$ is a finite subset of $(F \cap B) \cup A \in \mathcal A$.

Since $\mathcal A$ has finite character, $F \in \mathcal A$.

As this holds for all finite subsets of $A \cup B$ and $\mathcal A$ has finite character, $A \cup B \in \mathcal A$.

If $B \in \mathcal B$ and $x \in X$, then $B \cup A \in \mathcal A$, so either $B \cup A \cup \{ x \}$ or $B \cup A \cup \{ x' \}$ is in $\mathcal A$. But then $B \cup \{x\}$ or $B \cup \{x'\}$ is in $\mathcal B$ by the definition of $\mathcal B$.

Thus $\mathcal B$ satisfies the premises of the Restricted Tukey-Teichmüller Theorem/Weak Form.

Thus there is a $B \in \mathcal B$ such that for all $x \in X$, either $x \in B$ or $x' \in B$.

Let $C = A \cup B$.

Then since $B \subseteq C$, if $x \in X$ then either $x \in C$ or $x' \in C$.

But since $B \in \mathcal B$, $C = A \cup B \in \mathcal A$.