properties of complement

Let $X$ be a set and $A,B$ are subsets of $X$.

1. ${(A^{\mathrm{\complement }})}^{\mathrm{\complement }}=A$.

   Proof.

   $a\in {(A^{\mathrm{\complement }})}^{\mathrm{\complement }}$ iff $a\notin A^{\mathrm{\complement }}$ iff $a\in A$.∎

2. ${\mathrm{\varnothing }}^{\mathrm{\complement }}=X$.

   Proof.

   $a\in {\mathrm{\varnothing }}^{\mathrm{\complement }}$ iff $a\notin \mathrm{\varnothing }$ iff $a\in X$.∎

3. $X^{\mathrm{\complement }}=\mathrm{\varnothing }$.

   Proof.

   $a\in X^{\mathrm{\complement }}$ iff $a\notin X$ iff $a\in \mathrm{\varnothing }$.∎

4. $A\cup A^{\mathrm{\complement }}=X$.

   Proof.

   $a\in A\cup A^{\mathrm{\complement }}$ iff $a\in A$ or $a\in A^{\mathrm{\complement }}$ iff $a\in A$ or $a\notin A$ iff $a\in X$.∎

5. $A\cap A^{\mathrm{\complement }}=\mathrm{\varnothing }$.

   Proof.

   $a\in A\cap A^{\mathrm{\complement }}$ iff $a\in A$ and $a\in A^{\mathrm{\complement }}$ iff $a\in A$ and $a\notin A$ iff $a\in \mathrm{\varnothing }$.∎

6. $A\subseteq B$ iff $B^{\mathrm{\complement }}\subseteq A^{\mathrm{\complement }}$.

   Proof.

   Suppose $A\subseteq B$. If $a\in B^{\mathrm{\complement }}$, then $a\notin B$, so $a\notin A$, or $a\in A^{\mathrm{\complement }}$. This shows that $B^{\mathrm{\complement }}\subseteq A^{\mathrm{\complement }}$. On the other hand, if $B^{\mathrm{\complement }}\subseteq A^{\mathrm{\complement }}$, then by applying what’s just been proved, $A={(A^{\mathrm{\complement }})}^{\mathrm{\complement }}\subseteq {(B^{\mathrm{\complement }})}^{\mathrm{\complement }}=B$.∎

7. $A\cap B=\mathrm{\varnothing }$ iff $A\subseteq B^{\mathrm{\complement }}$.

   Proof.

   Suppose $A\cap B=\mathrm{\varnothing }$. If $a\in A$, then $a\in B^{\mathrm{\complement }}$, or $a\notin B$, which implies that $A\cap B=\mathrm{\varnothing }$. Suppose next that $A\subseteq B^{\mathrm{\complement }}$. If there is $a\in A\cap B$, then $a\in B$ and $a\in A$. But the second containment implies that $a\in B^{\mathrm{\complement }}$, which contradicts the first containment.∎

8. $A\setminus B=A\cap B^{\mathrm{\complement }}$, where the complement is taken in $X$.

   Proof.

   $a\in A\setminus B$ iff $a\in A$ and $a\notin B$ iff $a\in A$ and $a\in B^{\mathrm{\complement }}$ iff $a\in A\cap B^{\mathrm{\complement }}$.∎

9. (de Morgan’s laws) ${(A\cup B)}^{\mathrm{\complement }}=A^{\mathrm{\complement }}\cap B^{\mathrm{\complement }}$ and ${(A\cap B)}^{\mathrm{\complement }}=A^{\mathrm{\complement }}\cup B^{\mathrm{\complement }}$.

   Proof.

   See here (http://planetmath.org/DeMorgansLawsProof).∎