proof of alternative characterization of ultrafilter

Proof that $A\coprod B=X$ implies $A\in 𝒰$ or $B\in 𝒰$

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

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

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

Proof that $𝒰$ is an ultrafilter.

Assume that $𝒰$ is a filter, but not an ultrafilter and that $A\coprod B=X$ implies $A\in 𝒰$ or $B\in 𝒰$. Since $𝒰$ is not an ultrafilter, there must exist filter ${𝒰}^{\prime }$ which is strictly finer. Hence there must exist $A\in {𝒰}^{\prime }$ such that $A\notin 𝒰$. Set $B=X\setminus A$. Since $A\coprod B=X$ and $A\notin 𝒰$, it follows that $B\in 𝒰$. Since $𝒰\subset {𝒰}^{\prime }$, it is also the case that $B\in {𝒰}^{\prime }$. But $A\in {𝒰}^{\prime }$ as well; since ${𝒰}^{\prime }$ is a filter, $A\cap B\in {𝒰}^{\prime }$. This is impossible because $A\cap B\in {𝒰}^{\prime }$ is empty. Therefore, no such filter ${𝒰}^{\prime }$ can exist and $𝒰$ 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 $A\cup B=X\Rightarrow A\in 𝒰\vee B\in 𝒰$ will also imply that $𝒰$ 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 $𝒰$ is assumed to be a filter, $A^{\prime }\in 𝒰$ implies that $A\in 𝒰$. Likewise, $B^{\prime }\in 𝒰$ implies that $B\in 𝒰$. Hence, if $𝒰$ is a filter such that $A\cup B=X$ implies that either $A\in 𝒰$ or $B\in 𝒰$, then $𝒰$ 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 $B_i\in 𝒰$ or $C_i\in 𝒰$ for each $i$ between $1$ and $n-1$. Next, consider three possibilities:

1. $B_1\in 𝒰$: Since $B_1=A_1$, it follows that $A_1\in 𝒰$.

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

3. $B_1\notin 𝒰$ and $B_{n-1}\in 𝒰$: There must exist an $i\in \{2,\mathrm{\dots },n-1\}$ such that $B_{i-1}\notin 𝒰$ and $B_i\in 𝒰$. Since $B_{i-1}\notin 𝒰$, $C_{i-1}\in 𝒰$. Since $𝒰$ is a filter, $C_{i-1}\cap B_i\in 𝒰$. But also $C_{i-1}\cap B_i=A_i$ which implies that $A_i\in 𝒰$.

This examination of cases shows that if ${\coprod }_{i=1}^n{A}_i=X$, then there must exist an $i$ such that $A_i\in 𝒰$. It is also easy to see that this $i$ is unique — If $A_i\in 𝒰$ and $A_j\in 𝒰$ and $i\ne j$, then $A_i\cap A_j=\mathrm{\varnothing }$, but this cannot be the case since $𝒰$ is a filter.

Proof of second proposition regarding finite unions

There exist sets $A_i^{\prime }$ such that $A_i^{\prime }\subseteq A_i$ and ${\coprod }_{i=1}^n{A}_i=X$. By the result just proven, there exists an $i$ such that $A_i^{\prime }\in 𝒰$. Since $𝒰$ is a filter, $A_i^{\prime }\in 𝒰$ implies $A_i\in 𝒰$. 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
Canonical name	ProofOfAlternativeCharacterizationOfUltrafilter
Date of creation	2013-03-22 14:42:23
Last modified on	2013-03-22 14:42:23
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	16
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 54A20