Answer:
a) Yes
b)No
c) Yes
d) No
Step-by-step explanation:
Remember, a collection [tex]\mathcal{B}[/tex] of subsets of a set [tex]B[/tex] is a partition of [tex]B[/tex] if the union of the subsets is [tex]B[/tex], that is, [tex]\cup \mathcal{B}=B[/tex] and the elements of [tex]\mathcal{B}[/tex] are disjoints.
Let [tex]B=\{-3,-2,-1, 0, 1, 2, 3\}[/tex]
Then
a) [tex]\{-3,-1, 1, 3\}\cup \{-2, 0, 2\} =B[/tex] and [tex]\{-3,-1, 1, 3\}\cap \{-2, 0, 2\}=\emptyset[/tex].
Then the collection [tex]\mathcal{B}=\{\{-3,-1, 1, 3\}, \{-2, 0, 2\}\}[/tex] is a partition of B.
b) [tex]\{-3,-2,-1, 0\}\cup \{0, 1, 2, 3\}=B[/tex] and [tex]\{-3,-2,-1, 0\}\cap \{0, 1, 2, 3\}=\{0\}[/tex]
Since the sets [tex]\{-3,-2,-1, 0\},\{0, 1, 2, 3\}[/tex] aren't disjoints then they aren't a partition of B.
c) [tex]\{-3, 3\}\cup\{-2,2\}\cup\{-1,1\}\cup\{0\}=B[/tex]
and
[tex]\{-3, 3\}\cap\{-2, 2\}=\emptyset\\\{-3, 3\}\cap\{-1, 1\}=\emptyset\\\{-3, 3\}\cap\{0\}=\emptyset\\\{-2, 2\}\cap\{-1, 1\}=\emptyset\\\{-2, 2\}\cap\{0\}=\emptyset\\\{-1, 1\}\cap\{0\}=\emptyset[/tex]
Then the elements of the collection [tex]\mathcal{B}=\{\{-3, 3\},\{-2, 2\},\{-1, 1\},\{0\}\}[/tex] are disjoints.
Therefore, [tex]\mathcal{B}[/tex] is a partition of B.
d) [tex]\{-3,-2, 2, 3\}\cup \{-1, 1\}\neq B[/tex] because [tex]0\in \{-3,-2, 2, 3\}\cup \{-1, 1\}[/tex]. Then the collection [tex]\mathcal{B}=\{\{-3,-2, 2, 3\},\{-1, 1\}\}[/tex] isn't a partition of B.