On page 3 of Introduction to Lattices and Order, Davey and Priestley define an antichain in a poset $\langle P,\leq\rangle$ as a set of pairwise incomparable elements:

The ordered set P is an antichain if $x\leq y$ in P only if $x=y$

Gratzer's definition is equivalent, but stated in a manner which is difficult to excerpt.

On page 53 of Set Theory, an Introduction to Independence Proofs, Kunen defines an antichain in $\langle P,\leq\rangle$ as a set of pairwise incompatible elements, saying that two elements $p$ and $q$:

are incompatible ($p\bot q$) iff $\neg\exists r\in P(r\leq p\wedge r\leq q)$. An antichain in $P$ is a subset $A\subset P$ such that $\forall p,q\in A(p\neq q\rightarrow p\bot q)$.

So, given a three-element partially ordered set $\{0,a,b\}$ with $0\leq a$, $0\leq b$ the only (non-reflexive) related pairs in the partial order, it appears that $\{a,b\}$ is an antichain in the lattice sense but not in the forcing sense.

Question: is it in fact true that "antichain in a poset" means something different to set theorists than to lattice theorists?



Yes, the notions are different, but I believe the ambiguity is older than forcing; doesn't Halmos use "antichain" for the forcing notion in his book on Boolean algebras?

Typically, when the need arises of distinguishing both notions, I've seen used (and used myself) "$A$ is a weak antichain" for "the elements of $A$ are pairwise incomparable", while "$A$ is a strong antichain" is reserved for the forcing version, "the elements of $A$ are pairwise incompatible."

Usually context suffices to know which version is used. In combinatorial contexts I would think using "antichain" for the "weak" version is more common. Certainly whenever forcing is used, it is the "strong", Boolean- (or forcing-)theoretic version that is used. In any paper where ambiguity could be an issue, I've seen at least a remark.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.