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# height of an element in a poset

Let $P$ be a poset. Given any $a\in P$, the lower set $\downarrow\!\!a$ of $a$ is a subposet of $P$. Call the height of $\downarrow\!\!a$ less 1 the *height* of $a$. Let’s denote $h(a)$ the height of $a$, so

$h(a)=\operatorname{height}(\downarrow\!\!a)-1.$ |

From this definition, we see that $h(a)=0$ iff $a$ is minimal and $h(a)=1$ iff $a$ is an atom. Also, $h$ partitions $P$ into equivalence classes, so that $a$ is equivalent to $b$ in $P$ iff $h(a)=h(b)$. Two distinct elements in the same equivalence class are necessarily incomparable. In other words, the equivalence classes are antichains. Furthermore, given any two equivalence classes $[a],[b]$, set $[a]\leq[b]$ iff $h(a)\leq h(b)$, then the set of equivalence classes form a chain.

The height function of a poset $P$ looks remarkably like the rank function of a graded poset: $h$ is constant on the set of all minimal elements, and $h$ is isotone (preserves order). When is $h$ a rank function (the additional condition being the preservation of the covering relation)? The answer is given by a chain condition imposed on $P$, called the *Jordan-Dedekind chain condition*:

(*) In a poset, the cardinalities of two maximal chains between common end points must be the same.

Suppose for each $a\in P$, $h(a)$ is finite and $P$ has a unique minimal element $0$. Then $P$ can be graded by $h$ iff (*) is satisfied. More generally, if we drop the assumption of the uniqueness of a minimal element, then $P$ can be graded by $h$ iff any two maximal chains ending at the same end point have the same length.

## Mathematics Subject Classification

06A06*no label found*

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