Does "antichain" mean something different in set-forcing than in lattice theory? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:14:00Z http://mathoverflow.net/feeds/question/47057 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47057/does-antichain-mean-something-different-in-set-forcing-than-in-lattice-theory Does "antichain" mean something different in set-forcing than in lattice theory? Adam 2010-11-23T06:15:09Z 2010-11-23T06:23:12Z <p>On page 3 of <em>Introduction to Lattices and Order</em>, Davey and Priestley define an antichain in a poset $\langle P,\leq\rangle$ as a set of pairwise <strong>incomparable</strong> elements:</p> <blockquote> <p>The ordered set P is an antichain if $x\leq y$ in P only if $x=y$</p> </blockquote> <p>Gratzer's definition is equivalent, but stated in a manner which is difficult to excerpt.</p> <p>On page 53 of <em>Set Theory, an Introduction to Independence Proofs</em>, Kunen defines an antichain in $\langle P,\leq\rangle$ as a set of pairwise <strong>incompatible</strong> elements, saying that two elements $p$ and $q$:</p> <blockquote> <p>are <em>incompatible</em> ($p\bot q$) iff $\neg\exists r\in P(r\leq p\wedge r\leq q)$. An <em>antichain</em> in $P$ is a subset $A\subset P$ such that $\forall p,q\in A(p\neq q\rightarrow p\bot q)$.</p> </blockquote> <p>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.</p> <p><strong>Question</strong>: is it in fact true that "antichain in a poset" means something different to set theorists than to lattice theorists?</p> http://mathoverflow.net/questions/47057/does-antichain-mean-something-different-in-set-forcing-than-in-lattice-theory/47059#47059 Answer by Andres Caicedo for Does "antichain" mean something different in set-forcing than in lattice theory? Andres Caicedo 2010-11-23T06:23:11Z 2010-11-23T06:23:11Z <p>Adam:</p> <p>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? </p> <p>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."</p> <p>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. </p>