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Jech gives the following definitions in Chapters 7 and 14 of Set Theory (2002)

Given a set $S$, a set filter $F$ on $S$ is a subset of $\mathcal{P}(S)$ satisfying the following properties:

  1. $S \in F$ and $\varnothing \not\in F$;
  2. if $X, Y \in F$, then $X \cap Y \in F$;
  3. if $X \subseteq Y$ and $X \in F$, then $Y \in F$.

Given a poset $(P, \leq)$, a poset filter $F$ on $P$ is a subset of $P$ satisfying the following properties:

  1. $F$ is not empty;
  2. if $X, Y \in F$, then there exists $Z \in F$ such that $Z \leq X$ and $Z \leq Y$;
  3. if $X \leq Y$ and $X \in F$, then $Y \in F$.

I'm wondering why there is an asymmetry between these definitions. In particular, I notice that if $F$ is a set filter on $S$, then $F$ is a poset filer on $(\mathcal{P}(S), \subseteq)$. However, if $F$ is a poset filter on $(\mathcal{P}(S), \subseteq)$, then $F$ is a set filter on $S$ if and only if $\varnothing \not\in F$.

Why do we allow poset filters to contain a minimum element but do not allow the same for set filters?

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First off, many authors do allow $\emptyset \in F$, but then you need to exclude the improper filter $P(S)$ in a lot of cases. This is sort of like in ring theory when we talk about ideals, and we need to specify frequently that the ideal is proper, e.g. when we talk about maximal ideals. So excluding that case from the definition of filter is a reasonable convention, arguably the best convention, but a convention nonetheless.

For posets, one thing to keep in mind is that a poset might not have a least element, so it's sort of awkward to say that something that might not exist needs to be excluded. However, we could get around this by instead stipulating that $F\ne P,$ so we could just as well ask why that isn't part of the definition.

Aside from the reality that we can't reasonably expect all conventions will be in perfect harmony with one another, the reason Jech chose these is probably reflective of how these concepts are used in set theory.

Set filters are mostly pertaining to the theory of ultrafilters and related topics. And this is just like (in fact in a sense dual to) the ring theory example I mentioned... ultrafilters are maximal proper filters, and it just makes sense to assume all filters are proper in this context.

On the other hand, poset filters are mostly for forcing, where we mostly care about generic filters. If a poset happens to have a least element, then the singleton containing the least element is dense, so the least element needs to be in the generic filter and the generic filter is improper. So if we disallowed proper filters, we would have to caveat, e.g. the theorem that any poset in a countable model has a filter generic over the model.

In practice that's not all that important, since the case where there is a least element and consequently whole poset is the only generic filter results in a trivial forcing extension anyway. So posets that have a least element, or even atomic posets (or even non-atomless posets) generally are avoided in practice for reasons of triviality. For instance, later in Jech, when you view a Boolean algebra as a forcing poset, the natural and right thing to do is not to use the partial order on $B$ as is, but to chop off the least element and use the partial order on the rest.

But it might be enough to explain the choice of convention.

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    $\begingroup$ Chopping of the least element of a Boolean algebra makes the two notions of filter match up as the OP wants. A set filter on a set $S$ is a poset filter in $\mathcal P(S)\setminus\{\varnothing\}$. $\endgroup$ Commented Jul 12 at 0:54

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