Ultrafilters form a base for a topology Let $UF(\mathbb{N})$ be the set of ultrafilters over $\mathbb{N}$. For $A \subseteq \mathbb{N}$ consider $B=\{b \in UF(\mathbb{N}): A \in b\}$
I am trying to prove $\{B:A \subseteq \mathbb{N}\}$ forms a base for a topology over $UF(\mathbb{N})$.
I know it is a base if it covers $X$, and I already proved that, but I don't know how to prove the intersection of two elements of the base contains an element of the base.
 A: I assume what is meant is that that defining $B_A = \{U \in UF(\mathbb{N}) \mid A \in U\}$, the set $\mathscr{B} = \{B_A \mid A \subseteq \mathbb{N}\}$ is a base of a topology.
To prove this, we must show that any finite intersection of elements of $\mathscr{B}$ can be written as the union of elements of $\mathscr{B}$.
A standard result shows we must prove two things. First, we must show that $UF(\mathbb{N})$ can be written as a union of elements of $\mathscr{B}$. This is trivial, since $UF(\mathbb{N}) = B_\mathbb{N}$.
Second, we must show that given $C, D \in \mathscr{B}$, we can write $C \cap D$ as a union of elements of $\mathscr{B}$. This is also straightforward. For write $C = B_{A_1}$, $D = B_{A_2}$ for some $A, B$. Then $B_{A_1} \cap B_{A_2} = B_{A_1 \cap A_2}$.
So in fact, not only is $\mathscr{B}$ a base, it is a basis.
Your misunderstanding is when you talked about needing to take the intersection of two ultrafilters. This is never done. Instead, we are taking the intersection of two sets of ultrafilters.
Finally, note that there is no need to only consider ultrafilters on the Boolean algebra $P(\mathbb{N})$. This proof works for ultrafilters over any Boolean algebra and is part of the Stone duality.
