1
$\begingroup$

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 $S$?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you mean that for all $x, y \in \mathbb{Z}$, we define $x \leqq y$ to mean that $\vert x \vert \leq \vert y \vert$? $\endgroup$ – Saaqib Mahmood Jul 31 '16 at 10:48
  • $\begingroup$ Yes, exactly like that $\endgroup$ – b00n heT Jul 31 '16 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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