## You are here

Homeproof of alternative characterization of ultrafilter

## Primary tabs

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

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. $B_{1}\notin\mathcal{U}$ 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.

## Mathematics Subject Classification

54A20*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections