Maximal Chains in a Poset Let us say for two elements $x$ and $y$ of a poset, $y$ covers $x$ if $x < y$
and there is no $z$ such that $x < z < y$. Show that if $P$ is a finite poset with the longest chain
has length $l$ and if for any pair $y$ covering $x$ in $P$ there exists a chain of length $l$ containing
both $x$ and $y$, then every maximal (by inclusion) chain in $P$ has length $l$. Assume $P$ has
no isolated elements.
I am attempting to solve this problem using contradiction by assuming there is a maximal chain with length less than $l$ and proving that this would yield a chain with length larger than $l$ using the fact that two elements $x$ and $y$ where $y$ covers $x$ is contained in a chain with length $l$.
 A: SKETCH: Suppose that $C$ is a maximal chain in $P$ of length less than $\ell$.


*

*Explain why $C$ has at least two elements.

*Show that if $n=|C|$, we can index $C=\{x_1,\ldots,x_n\}$ so that $x_{k+1}$ covers $x_k$ for $k=1,\ldots,n-1$.


For $k=1,\ldots,n-1$ let $C_k$ be a chain of length $\ell$ containing $x_k$ and $x_{k+1}$, let $$C_k^{-}=\{y\in C_k:y\le x_k\}\;,$$ and let $$C_k^+=\{y\in C_k:x_{k+1}\le y\}\;.$$
Note that $|C_k^-|+|C_k^+|=\ell$.


*

*Use the maximality of $P$ to show that $|C_1^+|=\ell-1$.

*Use this to show that $|C_2^-|\le 2$ and hence that $|C_2^+|\ge\ell-2$. (Assume for now that $\ell$ is large enough for this to make sense.)

*Use this to show that $|C_3^-|\le 3$ and hence that $|C_3^+|\ge\ell-3$. (Again, assume for now that $\ell$ is large enough for this to make sense.)

*Recognize that this can be converted to a proof by induction that $|C_k^+|\ge\ell-k$ for $k=1,\ldots,n-1$, and do the conversion.

*Conclude that $|C_{n-1}^+|\ge\ell-n+1\ge 2$, so that $|C_{n-1}^-|=\ell-|C_{n+1}^-|\le\ell-2$.

*Use the maximality of $P$ to get a contradiction by showing that $|C_{n-1}^-|=\ell-1$.

