Section 11*: Problem 2 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

(a) Let $\prec$ be a strict partial order on the set $A$ . Define a relation on $A$ by letting $a\preceq b$ if either $a\prec b$ or $a=b$ . Show that this relation has the following properties, which are called the partial order axioms:

(i) $a\preceq a$ for all $a\in A$ .

(ii) $a\preceq b$ and $b\preceq a$ $\Rightarrow$ $a=b$ .

(iii) $a\preceq b$ and $b\preceq c$ $\Rightarrow$ $a\preceq c$ .

(b) Let $P$ be a relation on $A$ that satisfies properties (i)-(iii). Define a relation $S$ on $A$ by letting $aSb$ if $aPb$ and $a\neq b$ . Show that $S$ is a strict partial order on $A$ .

The purpose of the exercise is to show that the two definitions of the partial order, i.e. the one given by the axioms (i)-(iii) and the other one given in the text based on a strict partial order, are equivalent in the sense that a relation satisfies the axioms (i)-(iii) of a partial order if and only if there is a strict partial order inducing the relation.

(a) A strict partial order must satisfy nRT, so that (a) if $a\prec b$ then neither $a=b$ nor $b\prec a$ holds, and (b) if ( $a\prec b$ or $a=b$ ) and ( $b\prec c$ or $b=c$ ) then (

or $a=c$ ). Hence, the induced relation

must satisfy (ii) and (iii). It also satisfies (i) by definition.

(b) $S$ satisfies nR by definition. Now, if $aSb$ and $bSc$ , then $aPb$ and $bPc$ , and, by (iii), $aPc$ . If at the same time $c=a$ , then, by (ii), we would have $b=c$ contradicting $bSc$ . Therefore, $a\neq c$ , and $aSc$ . Hence, $S$ satisfies T.

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Section 11*: Problem 2 Solution

Section 11*: Problem 2 Solution