Answer to Question #198876 in Discrete Mathematics for Vina

Question

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.  

Answer 1

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 $B.$