This exploration will give you an opportunity to practice two common set proof scenarios: proving a statement about the empty set by contradiction and proving two sets are equal by showing each set is a subset of the other.
Theorem 1: Show that $\overline{U} = \emptyset$.
We need to show that $\overline{U}$ has no elements. Start this off by contradiction.
Assume that $\overline{U}$ isn't empty, i.e. that has at least one element. Call that element x.
Since that element is in $\overline{U}$, where do you know it isn't.
Since $x\in\overline{U}$, we know that it isn't in U by the definitionof the complement of a set.
Is that possible?
But this is a contradiction since U contains all elements. Therefore, $\overline{U}$ is a set with no elements. The empty set is the only such set so $\overline{U}=\emptyset$.
The Double Complement Law: $\overline{\overline{A}} = A$.
This rule says that complements (in a way) act like negative signs. They "reverse" the elements of a set like a negative sign "reverses" the direction of a number. Just like negative signs, two of them will cancel each other out.
First, we need to show that $\overline{\overline{A}} \subseteq A$. Starting by selecting an element to work with.
Let $x\in\overline{\overline{A}}$.
What does it mean for x to be in $\overline{B}$ where $B=\overline{A}$?
Since $x\in\overline{\left(\overline{A}\right)}$, i.e. the complement of $\overline{A}$, x is not in $\overline{A}$.
What does that tell you about where x does have to be?
Since x is not in $\overline{A}$, i.e. it isn't in the complement of A, it must be in A.
What can we then conclude about the subset relationship?
Because every x in $\overline{\overline{A}}$ is also in A, we have $\overline{\overline{A}} \subseteq A$.
Now pick an element of A.
Now let $x\in A\$.
What do you know about the relationship between x and $\overline{A}$?
Since $x\in A$ we know that it can't be in A's complement.
Since $x\notin \overline{A}$ what do you know about its relationship with that set's complement?
Since $x\notin \overline{A}$, it must be in that set's complement, i.e. $x\in\overline{\overline{A}}$.
What can you conclude from that?
Since every $x\in A$ is also in $\overline{\overline{A}}$ we must have $A\subseteq\overline{\overline{A}}$. Since each set is a subset of the other, we must have $\overline{\overline{A}}=A$.
