Do the complements of a partition of a set into two give a partition?

Question

Let $A$ be a nonempty set. Let $\{A_1,A_2\}$ be a partition of $A$. Consider the collection of set difference $\{A_1',A_2'\}$ = $\{A\setminus A_1, A\setminus A_2\}$. Is this a partition of $A$?

Answer 1

Hint: Show that $A\setminus A_1=A_2$ and $A\setminus A_2=A_1$ using the properties of a partition.

Answer 2

If $\{A_1,A_2\}$ is a partition of $A$ then the following hold:

1. $A=A_1\cup A_2$.
2. $A_1\cap A_2=\varnothing$.

In particular $A_1=A\setminus A_2$ and $A_2=A\setminus A_1$. Therefore the two partitions that you have described are the same partition.

You can verify that if we take a partition into three parts, then this is no longer true.

Answer 3

Hints:

$$\begin{align*}(1)&\;\;A=A_1\cup A_2\implies (A-A_1)\cup(A-A_2)\stackrel{\text{de Morgan}}=A-(A_1\cap A_2)=A-\emptyset=A\\ (2)&\;\; (A-A_1)\cap(A-A_2)\stackrel{\text{de Morgan}}=A-(A_1\cup A_2)=A-A=\emptyset\implies\end{align*}$$

De above shows the answer to your question is yes.