Show that AnA'= nullset
If $\displaystyle A=\{ \}$ then proof is complete.Originally Posted by dcapdogg
If $\displaystyle A\not = \{\}$ then $\displaystyle x\in A$
Then, $\displaystyle x\not \in A'$ thus, $\displaystyle A\mbox{ and }A'$ are disjoint thus, $\displaystyle A\cap A'=\{\}$
Hello, dcapdogg!
The proof depends upon what axioms and definitions are established.
Show that $\displaystyle A \cap A' \:=\:\emptyset$
$\displaystyle A \cap A'$ is the set of elements that are in $\displaystyle A$and in $\displaystyle A'$.
By definition of complement: .If $\displaystyle x \in A$, then $\displaystyle x \not\!\!{\in} A'$
. . Hence, there are no elements common to $\displaystyle A$and $\displaystyle A'.$ .They are disjoint.
Therefore: .$\displaystyle A \cap A' \:= \:\emptyset$