I'm reading something that defines ultrafilters like this.
An $\text{ultrafilter}$ on $\mathbb{N}$ is a subset $\omega$ of $2^{\mathbb{N}} - \{\varnothing\}$ so that
- (Completeness) For any $A \subset \mathbb{N}$, exactly one of $A$ or $A^c$ is in $\omega$.
- (Consistency) If $A \subset B$ and $A \in \omega$, then $B \in \omega$.
Right after the definition there are two exercises.
- If $\mathbb{N} = A_1 \sqcup \cdots \sqcup A_n$ is a finite partition of $\mathbb{N}$, then exactly one $i$ has $A_i \in \omega$.
- If $A,B \in \omega$, then $A \cap B \in \omega$.
In $1$, I can see that at most $1$ of the $A_i$ can be in $\omega$, but I can't see why at least $1$ should be in there. If I could use the second exercise, I would have it, but the order of the exercises seems to imply I don't need it yet.
In $2$, it seems like the strategy should be to assume not and get a contradiction, since our only tools to showing $A\cap B \in \omega$ would be to show the complement can't be in $\omega$, or exhibit a subset of $A \cap B$ which is in $\omega$ and use consistency. The assumption that $(A\cap B)^c \in \omega$ doesn't seem to buy me anything though.
All the other resources on ultrafilters I can find are all pretty intense set theory stuff, and this is in notes about groups acting on cube complexes, so it seems like it should be simple. Any suggestions?
I'm writing an answer here just to close the question.
Bof has a clear counterexample, and after speaking to a professor it seems like the most reasonable correction is to just add exercise $2$ as an assumption, so that exercise $1$ follows easily.