- $A\subset B$
- $A \cap B^{c} = \emptyset$
- $A^{c} \cup B = U$
where set $^c$ is the complement of the set
What I have so far is this. The statements are equivalent if they imply each other.
$1\Rightarrow2$: assuming 1 is true, $x \in A$ implies that $x\in B$ which in turn implies that $x\notin B^{c}$. So, by definition the set $A \cap B^{c}$ is always empty and $A\cap B^{c} = \emptyset$
$2\Rightarrow 3$: for this one, I was wondering If I could just say that applying de morgan's laws shows that $(A \cap B^{c})^{c} = \emptyset^{c}$ is logically equivalent to $A^{c} \cup B = U$ showing that 2 implies 3.
If this is incorrect, how would I go about showing that 2 implies 3?
$3\Rightarrow 1$: Assuming that 3 is true, 3 implies that either $x \in B$ or $x \in A^{c}$ and by definition, $x \in A^{c}$ implies that $x \in A $ and so, this implies that $A \subset B$.
Best Answer
$1\Rightarrow 2$ and $2\Rightarrow 3$ seems fine...for $3\Rightarrow 1$ let $x\in A$ then $x\notin A^{c}$ but $A^{c}\cup B=U$ hence $x\in B$ and so $A\subset B$.