How does this statement on posets follow from Zorn's lemma?

Question

Let $P$ be a poset with a maximal element $r$, i.e. an element $r\in P$ such that $r\geq x$ for all $x\in P$. One can think of the poset as a tree with root $r$.

I want to show that there exists a ''smallest maximal element'', i.e. an element $m\in P$ such that $m\geq x$ for all $x\in P$ and if $m'$ is another element with that property then $m'\geq m$. This is somehow the first element where the tree branches.

All looks much like Zorn's lemma but I don't see how it can be helpful. How can I proof the proposition?

Answer 1

First let us remember the definitions for maximum and maximal elements.

Let $(P,<)$ be a partially ordered set, $x\in P$ is called maximal if for all $y\in P$ we have $y\ge x\rightarrow y=x$. In addition to that, $x$ is called a maximum if $x\ge y$ for all $y\in P$.

Note that being a maximum is stronger than being the unique maximal element, and one can easily concoct a partial order without a maximum, but with a unique maximal element.

Zorn's lemma: Let $(P,<)$ be a partially ordered set. If every chain is bounded, then there exists a maximal element. (It need not be a maximum.)

It seems that you try to show that if we have a maximum in $P$, and we remove it then we have a new maximal element (not necessarily a maximum though).

It would seem to me that you example you have in mind is a partially-well ordered set. However this is not always the case.

For example, $P=[0,1]$ with $<$ as the usual ordering of real numbers. We have that $1$ is the maximum element, however there is no "smallest" maximal element in $[0,1)$ under the same ordering.

This is because you have to require that every chain is bounded, if we have an unbounded chain whose only bound is the maximum then by removing it the partial order no longer fulfills the needed assumption of Zorn's lemma, and so it does no longer guarantee us a maximal element.

Answer 2

Your definition of maximal element implies that it is unique: if $m$ and $m'$ are both maximal, then $m \leq m'$ and $m' \leq m$ which implies that $m = m'$. I'm not an expert on poset terminology, but what you call a maximal element I would refer to as a 'top element' or a '1'. This is the terminology from Stanley's Enumerative Combinatorics, IIRC.

In particular, what you intuitively are calling the "smallest maximal element" as the place where the tree first branches is not a maximal element at all in your definition, unless the tree branches immediately at its root.

The standard way to define a maximal element is as an element $m$ with the property that if $m \leq m'$, then $m' = m$. This allows for more than one maximal element to exist, although it is not hard to come up with examples that show a smallest maximal element does not have to exist (using this definition of maximal). The simplest would be two elements $x$ and $y$ that are not related. Or, if you want the Hasse diagram to be connected, you have could have $x$, $y$, and $z$ with the relations $x \geq z$ and $y \geq z$.

As is often the case in mathematics, when you run into a stumbling block, go back and examine whether you are asking the right question. My suspicion is that you are trying to ask another question and, hopefully, when you have formulated it correctly, the proof will become evident.

Edit: Cliff notes -- As others stated much more succinctly than I, your definition of a maximal element is that of a maximum element. Maximum elements are unique and if a maximum exists than it is the only maximal element of the poset. In an arbitrary poset the only way a minimum element in the subposet of maximal elements (i.e. a "smallest maximal element") can exist is if it is the only maximal element. But as Asaf pointed out in the comments below, a unique maximal element need not be a maximum element.