Answered: 6.3. Let (A,

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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6.3. Let (A, <A) and (B,<B) be two partially ordered sets. One can define on
the cartesian product A x B a relation < by setting (a1, b1) < (a2, b2) if and
only if a1 <A az and b1 < b2. (a) Show that (A x B,<) is a partially ordered
set known as the product order. (b) When is the product order a total order?

Expert Solution

Step 1

Given: $(A, ⩽_{A})$ and $(B, ⩽_{B})$ are partially ordered sets.

A Cartesian product A x B is defined with relation $⩽$ by setting $(a_{1}, b_{1}) ⩽ (a_{2}, b_{2})$ if and only if $a_{1} ⩽_{A} a_{2}$ and $b_{1} ⩽_{B} b_{2}$ .

Part a:

As $(A, ⩽_{A})$ and $(B, ⩽_{B})$ are partially ordered sets. Therefore, $⩽_{A}$ and $⩽_{B}$ are reflexive, antisymmetric and transitive.

consider an element $(a_{1}, b_{1})$ of A x B.

(i) As A and B are posets, therefore, $a_{1} ⩽_{A} a_{1}$ and $b_{1} ⩽_{B} b_{1}$ .

Therefore, $(a_{1}, b_{1}) ⩽ (a_{1}, b_{1})$ .

Thus the relation $⩽$ is reflexive relation on A x B.

(ii) Now consider $(a_{1}, b_{1}) ⩽ (a_{2}, b_{2})$ , A and B are antisymmetric.

Therefore, $a_{1} ⩽_{A} a_{2}$ implies $a_{1} = a_{2}$ and $b_{1} ⩽_{B} b_{2}$ implies that $b_{1} = b_{2}$ .

Hence the relation $⩽$ is antisymmetric relation on A x B.

(iii) If $(a_{1}, b_{1}) ⩽ (a_{2}, b_{2})$ and $(a_{2}, b_{2}) ⩽ (a_{3}, b_{3})$ are defined on A x B.

As A and B are transitive, therefore,

$a_{1} ⩽_{A} a_{2}$ and $a_{2} ⩽_{A} a_{3}$ implies that $a_{1} ⩽_{A} a_{3}$ .

Also $b_{1} ⩽_{B} b_{2}$ and $b_{2} ⩽_{B} b_{3}$ implies that $b_{1} ⩽_{B} b_{3}$ .

That is $(a_{1}, b_{1}) ⩽ (a_{2}, b_{2})$ and $(a_{2}, b_{2}) ⩽ (a_{3}, b_{3})$ implies that $(a_{1}, b_{1}) ⩽ (a_{3}, b_{3})$ .

Thus $⩽$ is transitive relation on A x B.

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