$Proof through inclusion (A∆B) \cup A = A \cup B$

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$Proof through inclusion (A∆B) \cup A = A \cup B$

Hey guys,
Maybe someone of you can help. I need to prove the following through inclusion:
(A∆B) ∪ A = A ∪ B

Any suggestion on how to solve this?

Set Theory Proofs

Philippsch3

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Answer 1

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We would like to prove that
\[A\cup B= (A \Delta B)\cup A .\]
Note that $(A \Delta B)\cup A = [(A\cup B) -(A\cap B)]\cup A$. Since
\[A\Delta B \subset A\cup B \text{and} A \subset A\cup B,\]
we have
\[(A\Delta B ) \cup A\subset (A\cup B). (1)\]
Next we prove
\[(A\cup B) \subset (A\Delta B ) \cup A. \]
Let $x \in A\cup B$. We consider two cases:

(i) If $x\in A$, then $x\in (A\Delta B ) \cup A.$

(ii) If $x\not \in A$ then we must have $x\in B$. Hence $x\in (A\cup B)-(A\cap B)=A\Delta B.$ Therefore $x\in (A\Delta B)\cup A$.

Thus $x\in A\cup B$ implies $x \in (A\Delta B)\cup A$. This means
\[(A\cup B) \subset (A\Delta B ) \cup A. (2)\]

Finally (1) and (2) implies
\[(A\cup B) =(A\Delta B ) \cup A. \]

Erdos

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