Powerset and Partition Creation

Question

Let $A$ be a set with at least three elements.

If $\mathcal P = \{B_1 ,B_2, B_3\} $ is a partition of $A$, is $\{B_1^c , B_2^c,B_3^c\}$ a partition of A?

So I am thinking the answer is no, since there is the possibility that two or more elements of $\mathcal P $ may be equal.

Answer 1

$A=\{1,2,3\}$. $\mathcal{P}=\{\{1\},\{2\},\{3\}\}$. Then $\{1\}^c = \{2,3\}$ and $\{2\}^c=\{1,3\}$, so $\{1\}^c\cap \{2\}^c = \{3\}\neq \emptyset$.

Answer 2

$B_{1}^{c} = B_{2}\cup B_{3}$ and $B_{2}^{c} = B_{1}\cup B_{3}$, if you take any element in $B_{3}$, then it will be in both $B_{1}^{c}$ and $B_{2}^{c}$

Answer 3

There is no possibility that two or more elements of $\mathscr{P}$ are equal: they are distinct by hypothesis.

HINT: $A=B_1\cup B_2\cup B_3$, where the sets $B_1,B_2$, and $B_3$ are pairwise disjoint. What is $B_1^c$ in terms of $B_2$ and $B_3$? What is $B_2^c$ in terms of $B_1$ and $B_3$?