This is a discrete math question. I am stuck. PLEASE help me. Suppose A,S C U, (<) E LOrd(A), and B := (S). If we define a relation (3) on A × B for(а,b), (аz, bz) € (Ax B):(a1, b1) 3 (a2, b2) ):(а, S a2) л (b, с b2)show that ((A × B), (3)) is an order lattice.

Name: This is a discrete math question. I am stuck. PLEASE help me. Suppose
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: This is a discrete math question. I am stuck. PLEASE help me. Suppose A,S C U, (<) E LOrd(A), and B := (S). If we define a relation (3) on A × B for(а,b), (аz, bz) € (Ax B):(a1, b1) 3 (a2, b2) ):(а, S a2) л (b, с b2)show that ((A × B), (3)) is an ord

Suppose x=a,b∈A×B is an upper bound of the two point set. We have show that…

Answered: Suppose A,S CU, (<) e LOrd(A), and B…

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

Chapter1: Line And Angle Relationships

Section1.4: Relationships: Perpendicular Lines

Problem 17E: Does the relation is a brother of have a reflexive property consider one male? A symmetric property...

See similar textbooks

Related questions

Topic Video

Question

This is a discrete math question. I am stuck. PLEASE help me.

Suppose A,S C U, (<) E LOrd(A), and B := (S). If we define a relation (3) on A × B for
(а,b), (аz, bz) € (Ax B):
(a1, b1) 3 (a2, b2) ):
(а, S a2) л (b, с b2)
show that ((A × B), (3)) is an order lattice.

Expert Solution

Step 1

Suppose $x = (a, b) \in A \times B$ is an upper bound of the two point set.

We have show that

$U = (a_{1} \lor a_{2}, b_{1 \lor} b_{2}) \underset{~}{<} x = (a, b) \Leftrightarrow a_{1} \leq 0 and b_{1} \subseteq b_{2} (a_{2}, b_{2}) \underset{~}{<} (a, b) \Leftrightarrow a_{2} \leq a and b_{2} \leq b Since, a_{1} \leq a and a_{2} \leq a \Rightarrow a is an upper bond of \{a_{1}, a_{2}\} Since, (A, \leq) is Lattice a_{1} \lor a_{2} exists and is least upper bond . \Rightarrow a_{1} \lor a_{2} \leq a Also, b_{1} \subseteq b and b_{2} \subseteq b$

So together we can write,

$(a_{1} \lor a_{2} \subseteq a) and (b_{1} \cup b_{2} \subseteq b) \Leftrightarrow (a_{1} \lor a_{2}, b_{1} \cup b_{2}) \underset{~}{<} (a, b) = x$

So, the two element subset is,

$(a_{1}, b_{1}) \underset{~}{<} (a_{1} \lor a_{2}, b_{1} \cup b_{2}) (a_{2}, b_{2}) \underset{~}{<} (a_{1} \lor a_{2}, b_{1} \cup b_{2}) Also, a_{2} \leq a_{1} \lor a_{2} and b_{2} \subseteq b_{1} \cup b_{2} \Rightarrow (a_{2}, b_{2}) \underset{~}{<} (a_{1} \lor a_{2}, b_{1} \cup b_{2})$

Step 2

To show that,

$(a_{1} \lor a_{2}, b_{1} \cup b_{2}) is least upper bond of \{(a_{1}, b_{1}), (a_{2}, b_{2})\}$

Firstly since $(A, \leq)$ is a Lattice.

$a_{1} \land a_{2}$ exists for two element subset $\{a_{1}, a_{2}\}$ of $A$

Also $a_{1} \land a_{2} \leq a_{2} a n d b_{1} \cap b_{2} \leq b_{2} \Rightarrow (a_{1} \land a_{2}, b_{1} \cap b_{2}) \underset{~}{<} (a_{2}, b_{2}) So, (a_{1} \land a_{2}, b_{1} \cap b_{2}) i s lower bound of \{(a_{1}, b_{1}), (a_{2}, b_{2})\} \subseteq A \times B$

Step by step

Solved in 3 steps

Knowledge Booster

$Research Design Formulation$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions