# Component of Finite Union in Ultrafilter

## Theorem

Let $S$ be a non-empty set.

Let $\mathcal U$ be a ultrafilter on $S$.

Let $\left\{{Y_1, \dots, Y_n}\right\}$ be a pairwise disjoint set of subsets of $S$ such that $S = Y_1 \cup \cdots \cup Y_n$.

Then there is a unique $k \in \left\{{1, \dots, n}\right\}$ such that $Y_k \in \mathcal U$.

## Proof

Assume that none of the $Y_k$ are empty.

If not, then any empty ones can simply be removed.

### Uniqueness

Aiming for a contradiction, suppose that $Y_j, Y_k \in \mathcal U$ with $j \ne k$.

Then since $Y_1, \dots, Y_n$ are a pairwise disjoint:

$Y_j \cap Y_k = \varnothing$

But by the definition of an ultrafilter, $\mathcal U$ has the finite intersection property.

It follows from this contradiction that no such $Y_j, Y_k \in \mathcal U$ with $j \ne k$ exists.

$\Box$

### Existence

Aiming for a contradiction, suppose that $Y_1, \dots, Y_n \notin \mathcal U$.

Since the $Y_k$ are pairwise disjoint:

$\displaystyle Y_i^c = \bigcup \left\{{Y_k: k \ne i}\right\}$

Then by the definition of an ultrafilter:

$\displaystyle \forall i: Y_i^c = \bigcup \left\{{Y_k: k \ne i}\right\} \in \mathcal U$

But:

$\displaystyle \bigcap_{i \mathop = 1}^n \bigcup \left\{{Y_k: k \ne i}\right\} = \varnothing$

contradicting the fact that $\mathcal U$ has the finite intersection property.

Thus $Y_k \in \mathcal U$ for some $k$.

$\blacksquare$

## 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)