Solution to Consider the following relation on set B = {a, b, {a}, {b}, {a, b}}: P … - Sikademy
Author Image

Archangel Macsika

Consider the following relation on set B = {a, b, {a}, {b}, {a, b}}: P = {(a, b), (b, {a, b}), ({a, b}, a), ({b}, a), (a, {a})}. Which one of the following sets is a partition S of B = {a, b, {a}, {b}, {a, b}}? 1. {{a, b, {a}, {b}}, {{a, b}}} 2. {{a}, {b}, {a, b}} 3. {{a, b, {a}}, {{a}, {b}, {a, b}}} 4. {a, b, {a}, {b}, {a, b}} (A partition of the given set B can be defined as a set S = {S1, S2, S3, …}. The members of S are subsets of B (each set Si is called a part of S) such that a. for all i, Si =/ 0/ (that is, each part is nonempty), b. for all i and j, if Si =/ Sj, then Si  Sj = 0/ (that is, different parts have nothing in common), and c. S1  S2  S3  … = B (that is, every element in B is in some part Si). It is possible to form different partitions of B depending on which subsets of B are formed to be elements of S.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

Let us find out which one of the following sets is a partition S of the set B = \{a, b,\{a\}, \{b\}, \{a, b\}\}.


1. For the family S=\{\{a, b, \{a\}, \{b\}\}, \{\{a, b\}\}\} we have that


a. \{a, b, \{a\}, \{b\}\}\ne\emptyset,\ \{\{a, b\}\}\ne\emptyset

b. \{a, b, \{a\}, \{b\}\}\cap \{\{a, b\}\}=\emptyset

c. \{a, b, \{a\}, \{b\}\}\cup\{\{a, b\}\}=B


Therefore, S is partition of B.


2. For the family \{\{a\}, \{b\}, \{a, b\}\} we have that \{b\}\cap\{a, b\}=\{b\}\ne\emptyset, and hence this family is not a partition of B.


3. For the family \{\{a, b, \{a\}\}, \{\{a\}, \{b\}, \{a, b\}\}\} we have that \{a, b, \{a\}\}\cap\{\{a\}, \{b\}, \{a, b\}\}=\{\{a\}\}\ne\emptyset, and hence this family is not a partition of B.


4. For that set \{a, b, \{a\}, \{b\}, \{a, b\}\} we have that \{a\}\cap \{a, b\}=\{a\}\ne\emptyset, and hence this family is not a partition of B.

Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-2964-qpid-1663