# proof of alternative characterization of ultrafilter

## Proof that $A\coprod B=X$ implies $A\in\mathcal{U}$ or $B\in\mathcal{U}$

Once we show that $A\notin\mathcal{U}$ implies $B\notin\mathcal{U}$, this result will follow immediately.

On the one hand, suppose that $A\notin\mathcal{U}$ and that there exists a $C\in\mathcal{U}$ such that $A\cap C$ is empty. Then $C\subseteq B$. Since $\mathcal{U}$ is a filter and $C\in\mathcal{U}$, this implies that $B\in\mathcal{U}$.

On the other hand, suppose that $A\notin\mathcal{U}$ and that $A\cap C$ is not empty for any $C$ in $\mathcal{U}$. Then $\{A\}\cup\mathcal{U}$ would be a filter subbasis. The filter which it would generate would be finer than $\mathcal{U}$. The fact that $\mathcal{U}$ is an ultrafilter means that there exists no filter finer than $\mathcal{U}$. This contradiction shows that, if $A\notin\mathcal{U}$, then there exists a $C$ such that $A\cap C$ is empty. But this would imply that $C\subseteq B$ which, in turn would imply that $B\in\mathcal{U}$.

## Proof that $\mathcal{U}$ is an ultrafilter.

Assume that $\mathcal{U}$ is a filter, but not an ultrafilter and that $A\coprod B=X$ implies $A\in\mathcal{U}$ or $B\in\mathcal{U}$. Since $\mathcal{U}$ is not an ultrafilter, there must exist filter $\mathcal{U}^{\prime}$ which is strictly finer. Hence there must exist $A\in\mathcal{U}^{\prime}$ such that $A\notin\mathcal{U}$. Set $B=X\setminus A$. Since $A\coprod B=X$ and $A\notin\mathcal{U}$, it follows that $B\in\mathcal{U}$. Since $\mathcal{U}\subset\mathcal{U}^{\prime}$, it is also the case that $B\in\mathcal{U}^{\prime}$. But $A\in\mathcal{U}^{\prime}$ as well; since $\mathcal{U}^{\prime}$ is a filter, $A\cap B\in\mathcal{U}^{\prime}$. This is impossible because $A\cap B\in\mathcal{U}^{\prime}$ is empty. Therefore, no such filter $\mathcal{U}^{\prime}$ can exist and $\mathcal{U}$ must be an ultrafilter.

## Proof of generalization to $A\cup B=X$

On the one hand, since $A\cup B=X$ implies $A\coprod B=X$, the condition $A\cup B=X\Rightarrow A\in\mathcal{U}\vee B\in\mathcal{U}$ will also imply that $\mathcal{U}$ is an ultrafilter.

On the other hand, if $A\cup B=X$, there must exists $A^{\prime}\subseteq A$ and $B^{\prime}\subseteq B$ such that $A^{\prime}\coprod B^{\prime}=X$. If $\mathcal{U}$ is assumed to be a filter, $A^{\prime}\in\mathcal{U}$ implies that $A\in\mathcal{U}$. Likewise, $B^{\prime}\in\mathcal{U}$ implies that $B\in\mathcal{U}$. Hence, if $\mathcal{U}$ is a filter such that $A\cup B=X$ implies that either $A\in\mathcal{U}$ or $B\in\mathcal{U}$, then $\mathcal{U}$ is an ultrafilter.

## Proof of first proposition regarding finite unions

Let $B_{j}=\coprod_{i=1}^{j}A_{i}$ and let $C_{j}=\coprod_{i=j+1}^{n}A_{i}$. For each $i$ between $1$ and $n-1$, we have $B_{i}\coprod C_{i}=X$. Hence, either $B_{i}\in\mathcal{U}$ or $C_{i}\in\mathcal{U}$ for each $i$ between $1$ and $n-1$. Next, consider three possibilities:

1. 1.

$B_{1}\in\mathcal{U}$: Since $B_{1}=A_{1}$, it follows that $A_{1}\in\mathcal{U}$.

2. 2.

$B_{n-1}\notin\mathcal{U}$: Since $B_{n-1}\coprod C_{n-1}=X$, it follows that $C_{n-1}\in\mathcal{U}$. Because $C_{n-1}=A_{n}$, it follows that $A_{n}\in\mathcal{U}$.

3. 3.

$B_{1}\notin\mathcal{U}$ and $B_{n-1}\in\mathcal{U}$: There must exist an $i\in\{2,\ldots,n-1\}$ such that $B_{i-1}\notin\mathcal{U}$ and $B_{i}\in\mathcal{U}$. Since $B_{i-1}\notin\mathcal{U}$, $C_{i-1}\in\mathcal{U}$. Since $\mathcal{U}$ is a filter, $C_{i-1}\cap B_{i}\in\mathcal{U}$. But also $C_{i-1}\cap B_{i}=A_{i}$ which implies that $A_{i}\in\mathcal{U}$.

This examination of cases shows that if $\coprod_{i=1}^{n}A_{i}=X$, then there must exist an $i$ such that $A_{i}\in\mathcal{U}$. It is also easy to see that this $i$ is unique — If $A_{i}\in\mathcal{U}$ and $A_{j}\in\mathcal{U}$ and $i\neq j$, then $A_{i}\cap A_{j}=\emptyset$, but this cannot be the case since $\mathcal{U}$ is a filter.

## Proof of second proposition regarding finite unions

There exist sets $A^{\prime}_{i}$ such that $A^{\prime}_{i}\subseteq A_{i}$ and $\coprod_{i=1}^{n}A_{i}=X$. By the result just proven, there exists an $i$ such that $A^{\prime}_{i}\in\mathcal{U}$. Since $\mathcal{U}$ is a filter, $A^{\prime}_{i}\in\mathcal{U}$ implies $A_{i}\in\mathcal{U}$. Note that we can no longer assert that $i$ is unique because the $A_{i}$’s no longer are required to be pairwise disjoint.

Title proof of alternative characterization of ultrafilter ProofOfAlternativeCharacterizationOfUltrafilter 2013-03-22 14:42:23 2013-03-22 14:42:23 rspuzio (6075) rspuzio (6075) 16 rspuzio (6075) Proof msc 54A20