Filters and antichains in preorders In a preorder $\mathbb{P}$ (reflexive and transitive), what means to say that a subset $C\subset\mathbb{P}$ is not contained in any filter of $\mathbb{P}$? For example, it is obvious that if $C$ contains antichains, then there are no filters containing $C$. Is the converse true?
Alternatively, my question is: is there any characterization to say when a set $C\subset\mathbb{P}$ is contained in a filter?
An answer to this would be interesting to me because I am working with the negation of $\text{MA}_{\aleph_1}$.
Thanks!
 A: In general, there won't be a very good characterization. For example, we can find a poset $\mathbb{P}$ and a set $C\subseteq\mathbb{P}$ such that 


*

*$C$ is countable,

*every finite tuple from $C$ has a common extension in $\mathbb{P}$, but

*$C$ is not contained in any filter.
Note that the second bullet point is a very strong form of not containing an antichain.
Specifically, let $\mathbb{P}=[\omega]^{<\omega}$ ordered as follows: for finite sets of naturals $a, b\in\mathbb{P}$, we have $a\le_\mathbb{P}b$ iff 


*

*$a=b$, or

*$a$ is empty or a singleton and $a\subseteq b$.
Then take $C=\{\{n\}: n\in\omega\}$. Any finitely subset of $C$ has a common extension (just take the union), but $C$ is not contained in any filter.

We can do even worse: let $\mathbb{P}'=[\omega]^{<\omega}\cup\{\omega\setminus \{n\}: n\in\omega\}$, so all the finite sets and the co-singletons, ordered as: $a\le_{\mathbb{P}'}b$ iff 


*

*$a=b$, or

*$a$ is empty or a singleton and $a\subseteq b$.
Then, again taking $C=\{\{n\}: n\in\omega\}$, we still have that $C$ is not contained in any filter; but this time, removing any single element from $C$ yields a set of conditions which is contained in some filter!
