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 $S$?


1 Answer 1


I don't think so: why do you think it should?

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.

  • $\begingroup$ Do you mean that for all $x, y \in \mathbb{Z}$, we define $x \leqq y$ to mean that $\vert x \vert \leq \vert y \vert$? $\endgroup$ Jul 31, 2016 at 10:48
  • $\begingroup$ Yes, exactly like that $\endgroup$
    – b00n heT
    Jul 31, 2016 at 10:59

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