Show that AnA'= nullset

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- Aug 17th 2006, 02:46 AMdcapdoggsets
Show that AnA'= nullset

- Aug 17th 2006, 04:34 AMThePerfectHackerQuote:

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, $\displaystyle A\cap A'=\{\}$ - Aug 17th 2006, 06:30 AMSoroban
Hello, dcapdogg!

The proof depends upon what axioms and definitions are established.

Quote:

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: .$\displaystyle A \cap A' \:= \:\emptyset$