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 $n1$, 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 $n1$. Next, consider three possibilities:

1.
${B}_{1}\in \mathcal{U}$: Since ${B}_{1}={A}_{1}$, it follows that ${A}_{1}\in \mathcal{U}$.

2.
${B}_{n1}\notin \mathcal{U}$: Since ${B}_{n1}\coprod {C}_{n1}=X$, it follows that ${C}_{n1}\in \mathcal{U}$. Because ${C}_{n1}={A}_{n}$, it follows that ${A}_{n}\in \mathcal{U}$.

3.
${B}_{1}\notin \mathcal{U}$ and ${B}_{n1}\in \mathcal{U}$: There must exist an $i\in \{2,\mathrm{\dots},n1\}$ such that ${B}_{i1}\notin \mathcal{U}$ and ${B}_{i}\in \mathcal{U}$. Since ${B}_{i1}\notin \mathcal{U}$, ${C}_{i1}\in \mathcal{U}$. Since $\mathcal{U}$ is a filter, ${C}_{i1}\cap {B}_{i}\in \mathcal{U}$. But also ${C}_{i1}\cap {B}_{i}={A}_{i}$ 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 $i\ne j$, then ${A}_{i}\cap {A}_{j}=\mathrm{\varnothing}$, but this cannot be the case since $\mathcal{U}$ 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 ${A}_{i}^{\prime}\in \mathcal{U}$. Since $\mathcal{U}$ is a filter, ${A}_{i}^{\prime}\in \mathcal{U}$ 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 

Canonical name  ProofOfAlternativeCharacterizationOfUltrafilter 
Date of creation  20130322 14:42:23 
Last modified on  20130322 14:42:23 
Owner  rspuzio (6075) 
Last modified by  rspuzio (6075) 
Numerical id  16 
Author  rspuzio (6075) 
Entry type  Proof 
Classification  msc 54A20 