# Is every upper bound of a subset of a partially ordered set also a maximal element of the set?

Here's Definition 4.1-1 on page 210 in Introductory Functional Analysis With Applications by Erwine Kreyszig.

A partially ordered set is a set $$M$$ on which there is defined a partial ordering, that is, a binary relation which is written $$\leqq$$ and satisfies the conditions

(PO1) $$a \leqq a$$ for every $$a \in M$$. (Reflexivity)

(PO2) If $$a \leqq b$$ and $$b \leqq a$$, then $$a = b$$. (Antisymmetry)

(PO3) If $$a \leqq b$$ and $$b \leqq c$$, then $$a \leqq c$$. (Transitivity)

"Partially" emphasizes the fact that $$M$$ may contain elements for which neither $$a \leqq b$$ nor $$b \leqq a$$ holds. Then $$a$$ and $$b$$ are called incomparable elements. In contrast, two elements $$a$$ and $$b$$ are called comparable elements if they satisfy $$a \leqq b$$ or $$b \leqq a$$ (or both).

A totally ordered set or chain is a partially ordered set such that every two elements of the set are comparable. In other words, a chain is a partially ordered set that has no incomparable elements.

An upper bound of a subset $$W$$ of a partially ordered set $$M$$ is an element $$u \in M$$ such that $$x \leqq u \ \ \ \mbox{ for every } \ x \in W.$$

(Depending on $$M$$ and $$W$$, such a $$u$$ may or may not exist.) A maximal element of $$M$$ is an element $$m \in M$$ such that $$m \leqq x \ \mbox{ implies} \ m = x.$$ {Again, $$M$$ may or may not have maximal elements. Note further that a maximal element need not be an upper bound. }

I have copied verbatim what Kreyszig has stated.

Now my question is, if $$S$$ is a non-empty subset of a partially ordered set $$M$$, then (according to this set of definitions) is every upper bound of $$S$$ (in $$M$$) also a maximal element of $$S$$?

An easy counterexample is $M:=\mathbb{Z}$ with partial ordering induced by the absolute value, the subset $S$ being $\{-1,1\}$. Every point of $S$ is an upper bound, but none is maximal.
• Do you mean that for all $x, y \in \mathbb{Z}$, we define $x \leqq y$ to mean that $\vert x \vert \leq \vert y \vert$? Jul 31, 2016 at 10:48