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August 17th 2006, 02:46 AM
dcapdogg

sets

Show that AnA'= nullset
August 17th 2006, 04:34 AM
ThePerfectHacker

Quote:

Originally Posted by dcapdogg

Show that AnA'= nullset

If $A=\{ \}$ then proof is complete.
If $A\not = \{\}$ then $x\in A$
Then, $x\not \in A'$ thus, $A\mbox{ and }A'$ are disjoint thus, $A\cap A'=\{\}$
August 17th 2006, 06:30 AM
Soroban

Hello, dcapdogg!

The proof depends upon what axioms and definitions are established.

Quote:

Show that $A \cap A' \:=\:\emptyset$

$A \cap A'$ is the set of elements that are in $A$ and in $A'$ .

By definition of complement: .If $x \in A$ , then $x \not\!\!{\in} A'$
. . Hence, there are no elements common to $A$ and $A'.$ .They are disjoint.

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

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