Hi I would like to confirm the following ideas regarding filters in order theory:
By definition I have that a filter is a subset of a poset $(P, \leq)$ which satisfies:
- $\mathcal{F}$ is non-empty.
- For any $x,y$ in $\mathcal{F}$, there is some element $z$ in $\mathcal{F}$ such that $z \leq x$ and $z \leq y$ (thus $\mathcal{F}$ is downward directed).
- For every $x$ in $\mathcal{F}$ and $y$ in $P$, $x \leq y$ implies that $y$ is in $\mathcal{F}$ (thus $\mathcal{F}$ is an up-set).
A base for filter $\mathcal{F}$ is a subset $\mathcal{B} \subset \mathcal{F}$ such that finite intersection and supersets of $\mathcal{B}$ is equivalent to $\mathcal{F}$.
A subset $\mathcal{B} \subset P$ is a base for a filter in $P$ iff
- $\mathcal{B}$ is non-empty.
- If $x,y$ in $\mathcal{B}$ then there is a $z$ in $\mathcal{B}$ such that $z \leq x$ and $x \leq y$ (downward directed).
Given any upward directed $B\uparrow$ of poset $P$, we can form a filter base by defining it as follows: $$\mathcal{B} := \big\{ \{ a: a \in B, b \leq a\}: b\in B\big\}$$
We check that this is a filter base as follows:
- $\mathcal{B} \neq \emptyset$ since $b \leq b$, then there must exist some $G \in \mathcal{B}$ such that $\{b\} \subset G$ for every $b \in B$.
- Since $B$ is upward directed it follows that there is some $c \in B$, for every pair $b_{1},b_{2} \in B$, such that $c \geq b_{1}$ and $c \geq b_{2}$, therefore there must exist some $G \subset \{ a: a \in B,~~ a \geq b_{1} \} \cap \{a: a \in b_{2},~~ a \geq b_{2} \}$, where $G \in \mathcal{B}$. Since $B$ is upward directed, for $b_{1}, b_{2} \in B$ there exists some $c \in B$ such that $b_{1} \leq c$ and $b_{2} \leq c$. Thus we can take $G:= \{ a: a \in B,~ a \geq c \}$.
Is all of this fine? Thanks.
The definitions are fine, but the discussion of generating a filter from an upward directed $B\subseteq P$ has some problems. First, it appears that the inequality in the definition of $\mathscr{B}$ goes the wrong way: it should be
$$\mathscr{B}=\big\{\{a\in B:b\le a\}:b\in B\big\}\;.$$
With this change the proof that for any $B_0,B_1\in\mathscr{B}$ there is a $B_2\in\mathscr{B}$ such that $B_2\subseteq B_0\cap B_1$ is correct. The previous bullet point, however, is not: what you need to show is not that $\varnothing\notin\mathscr{B}$, but rather that $\mathscr{B}\ne\varnothing$. And your argument does seem to be an attempt at proving that $\mathscr{B}\ne\varnothing$, but it’s not quite right: $b\in B$ does not imply that $\{b\}\in\mathscr{B}$. However, the existence of a $b\in B$ does mean that $\{a\in B:b\le a\}\in\mathscr{B}$ and hence that $\mathscr{B}$ is non-empty.