Filter and ultrafilter properties in Boolean algebra

Question

In a Boolean algebra $(\mathbb{B}, \vee, \wedge, 0, 1, \neg)$, a ultrafilter $U \subseteq \mathbb{B}$ satisfies

1. If $a \wedge b \in U \iff a \in U$ and $b \in U$.
2. If $a \vee b \in U \iff a \in U$ or $b \in U$.

The first one is actually true for all filters ($\Rightarrow$ upward closure, $\Leftarrow U$ downward directed). For the second one, $\Leftarrow$ can be proven by upward closure, but how about the $\Rightarrow$ direction which seems to be true only for ultrafilter ?

NOTE: The only definition of ultrafilter I have at this stage is: an ultrafilter $U$ is a filter satisfying $\forall x \not\in U, \neg x \in U$ (i.e. $\forall x \in \mathbb{B}, x \in U \leftrightarrow \neg x \not\in U$).

Answer 1

Suppose that $a\lor b\in U$ but $b\notin U$. Then $\neg b\in U$, so $(a\lor b)\land\neg b\in U$. And

$$(a\lor b)\land\neg b=a\land\neg b\,,$$

so $a\in U$ by upward closure.

Answer 2

To add to Brian’s answer: yes that property implies it is an ultrafilter as any filter contains $1$ (being non-empty and upper-closed) and so for any $a \in \Bbb B$, $a \lor \lnot a = 1 \in U$, so the right to left implication of 2 then implies $a \in \mathcal{U}$ or $\lnot a \in U$. These cannot hold at the same time as $0 = a \land \lnot a \notin U$. This implies that $U$ is an ultrafilter in your definition.