Restricted Tukey-Teichmüller Theorem/Strong Form

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Theorem

Let $X$ be a set.

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

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


Let $\mathcal A$ have finite character.

For all $A \in \mathcal A$ and all $x \in X$, let either:

$A \cup \left\{ {x}\right\} \in \mathcal A$

or:

$A \cup \left\{ {x'}\right\} \in \mathcal A$


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

Let $A \in \mathcal A$.

Let:

$\mathcal B = \left\{ {B: B \subseteq X \text{ and } A \cup B \in \mathcal A}\right\}$


It is to be shown that $\mathcal B$ 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 $F \subseteq B \subseteq X$:

$F \subseteq X$

It remains to be shown 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 $\mathcal B$:

$\left({F \cap B}\right) \cup A \in \mathcal A$

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

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:

$A \cup B \in \mathcal A$

$\Box$


Let $B \in \mathcal B$ and $x \in X$.

Then:

$B \cup A \in \mathcal A$

so either $B \cup A \cup \left\{ {x}\right\}$ or $B \cup A \cup \left\{ {x'}\right\}$ is in $\mathcal A$.

But then $B \cup \left\{ {x}\right\}$ or $B \cup \left\{ {x'}\right\}$ is in $\mathcal B$ by the definition of $\mathcal B$.

Thus $\mathcal B$ satisfies the premises of the weak form of the Restricted Tukey-Teichmüller Theorem.

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 $B \in \mathcal B$:

$C = A \cup B \in \mathcal A$

$\blacksquare$


Source of Name

This entry was named for John Wilder Tukey and Oswald Teichmüller.


Sources

  • 2005: R.E. HodelRestricted versions of the Tukey-Teichmuller Theorem that are equivalent to the Boolean prime ideal theorem (Arch. Math. Logic Vol. 44: pp. 459 – 472)