This is the proof I am talking about.
http://www.math.cmu.edu/~af1p/Teaching/Combinatorics/F03/Class14.pdf
It is given that : P⁻∩ P⁺=A Otherwise there exists x,i, j such that ai < x < aj and so A is not an anti-chain.
Please could anybody explain this to me ? Why should the intersection of P⁻ and P⁺ be equal to A ?
Since $A$ is obviously included in $P^-$ and in $P^+$, we have $A\subseteq P^-\cap P^+$. The problem is to prove the reverse inclusion. So consider any element $x\in P^-\cap P^+$; I need to show that $x\in A$. By definition of $P^-$ there is an $i$ such that $x\leq a_i$. If $x=a_i$, that means $x\in A$, so I'm done. It remains to consider the situation when $x<a_i$. Similarly, by definition of $P^+$, $x\geq a_j$ for some $j$, and I'm done if equality holds, so I need only consider the situation when $x>a_j$. But then $a_j<x<a_i$, which means that two different elements of $A$ are comparable: $a_j<a_i$. That contradicts the fact that $A$ is an antichain.