I am asked to prove that if a family $\mathcal{A}$ of subsets of X is $\textit{weakly partition regular}$, then it contains an ultrafilter.
The definition of "weakly partition regular" I'm working with is "a family $\mathcal{A}$ such that for every partition $X = C_1 \sqcup ... \sqcup C_r $ ($r$ might vary) of the entire set there exists an $i$ such that $C_i \in \mathcal{A}$.
The hint I am given is to consider the family $\{ A \mid A^c \notin \mathcal{A}\}$.
I really can't do it and I am starting to think that some hypotheses might be missing.
As an alternative attempt to solve it I'm wondering if a minimal weakly partition regular family must be an ultrafilter (I believe I've proven the converse).
You are correct, there is a missing assumption that $\mathcal{A}$ is closed upwards (which, from what I know, is usually part of the definition of weakly partition regular). Assuming that:
An ultrafilter (on $X$) is a filter such that $B∈\mathcal{F}⇔X\setminus B∉\mathcal{F}$.
So given $C=\{ A \mid X\setminus A \notin \mathcal{A}\}$ we can see:
Let $A,B∈C$ then $A\cap B∪(X \setminus A)∪((X\setminus B)\cap A)=X⇒ A\cap B\in \mathcal{A}\quad(*)$
This follows from the fact that $(X\setminus B)∩ A∈\mathcal{A}⇒X\setminus B∈\mathcal{A}$. Similar deconstruction of $X$ using any finite number of $A_i∈C$ will also work.
Let's assume that $∅∉\mathcal{A}$(as it is trivial if it is), then $(*)$ proves that $C$ has the finite intersection property. Extending $C$ into a filter $C⊆\mathcal{F}_C⊆\mathcal{A}$ can be easily done by noting that $\mathcal{A}$ is closed by supset, so every supset of finite intersection of elements of $C$ is in $\mathcal{A}$.
Extend $\mathcal{F}_C$ into $\mathcal{U}$ an ultrafilter, assume that $B∈\mathcal{U}$ but $B∉\mathcal{A}$, then by definition $X\setminus B∈C⊆\mathcal{F}_C⊆\mathcal{U}$
Unfortunately I don't have any counter examples in mind that shows that the extra assumption is necessary
As an alternative attempt to solve it I'm wondering if a minimal weakly partition regular family must be an ultrafilter (I believe I've proven the converse).Also note that that if $\mathcal{A}$ is weakly partition regular, then $\mathcal{A}∪\{∅\}$ is also weakly partition regular, so this is wrong
Sorry, I didn't notice the "minimal" part.