I am currently working on my discrete math problem. Let's define a consecutive chain $C = (a_1, a_2, \cdots)$ of poset $(P, \leq)$ as a chain with the property that $\not\exists z (a_i \leq z \leq a_{i+1})$, where $z \neq a_i, a_{i+1}$. The poset I am considering is $(\mathcal{F}, \subseteq)$ where $\mathcal{F} \subseteq 2^{[n]}$ with the following property.
If $A, C \in \mathcal{F}$ and $A \subseteq B \subseteq C$, then $B \in \mathcal{F}$.
If I can prove the following version of Dilworth's theorem on $(\mathcal{F},\subseteq)$, I will be done.
Let $r$ be the length of the largest antichain of $\mathcal{F}.$ Then $\mathcal{F}$ can be partitioned into $r$ consecutive chains.
I have read the original proof of Dilworth's theorem by Galvin. It was by the induction on the size of the poset. However I have no idea how to modify it the prove my claim. Is that generalized version of Dilworth' theorem false? Even when restricted to the case that the poset is $(\mathcal{F}, \subseteq)$?
Thanks in advance for any form of help, hint, or solution.