Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
$r$ is a maximum, while indeed it is still a maximal, it is also the unique maximal element and it is bigger than any other element. The definition for a maximal element is just that no one is bigger. –  Asaf Karagila Oct 20 '11 at 14:11
If $r\ge x$ for all $x\in P$, then what you have is not only a maximal element, but actually a greatest element. In that case $r$ is the only maximal element. –  Henning Makholm Oct 20 '11 at 14:12
Furthermore: there is not always a "first element where the tree branches". Consider $P=\{1/n\mid n\in\mathbb Z, n\ne 0\}$ with the usual ordering, and add a short one-element branch from each negative $1/n$. Then the branch points have no supremum in $P$. –  Henning Makholm Oct 20 '11 at 14:17
add comment

2 Answers

up vote 4 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
In the poset $\mathbb Z\cup\{a\}$ with the order defined as $<_\mathbb Z\cup\{\langle a,a\rangle\}$ there exists a unique maximal element, but there is no maximum. Being a maximum is a lot stronger than being the unique maximal element. –  Asaf Karagila Oct 20 '11 at 14:27
Thanks! I edited the offending statement. –  Michael Joyce Oct 20 '11 at 14:35
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.