I have a partially ordered set , and I'm asked to find the maximal chain and antichain of :
$A=${$1,2,\dots,7$} with a relation $R$ s.t.
$R=I_A \cup ${$(a,b)|a,b \in {1,2,\dots,5} \space and \space a < b $}$\cup${$(6,7)$}
I have a partially ordered set , and I'm asked to find the maximal chain and antichain of :
$A=${$1,2,\dots,7$} with a relation $R$ s.t.
$R=I_A \cup ${$(a,b)|a,b \in {1,2,\dots,5} \space and \space a < b $}$\cup${$(6,7)$}
Note if you look at $A=\{1,2,3,4,5\}$ and $B=\{6,7\}$, none of the elements in $A$ is comparable with any of the elements in $B$, while all elements within $A$ are comparable with each other, and all elements within $B$ are comparable with each other.
Maximal chains: $A$ and $B$. Both are chains, and you cannot extend them further: say you want to add an element $b\not\in A$ into $A$ to get a bigger chain, but then $b\in B$ and is not comparable with any elements from $A$. On the other hand, any chain is fully contained in either $A$ or $B$ (because otherwise it will contain incomparable elements) and so can be extended to either $A$ or $B$.
Maximal antichains: $\{a,b\}$ where $a\in A$ and $b\in B$ (all $10$ of those). These are antichains, and you cannot add any additional elements, because if you try to add another element from $A$, it will be comparable with $a$, and if you try to add another element from $B$, it will be comparable with $b$. On the other hand, as $A$ and $B$ are chains, any antichain can have at most one element from $A$ and at most one element from $B$, and, if not already one of those $10$ antichains above, it can be obviously extended to one of those.