Proof: $(A - B)\cup B = A$ iff $B\subseteq A$

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SUMMARY

The equation $(A - B)\cup B = A$ holds true if and only if $B\subseteq A$. The proof begins with the assumption that $B$ is a subset of $A$, leading to the transformation of $(A - B)\cup B$ into $(A\cap B^c)\cup B$. This expression simplifies to $(A\cup B)\cap (B^c\cup B)$, which ultimately results in $A$. The confusion arises when interpreting the final result, where some participants mistakenly conclude that the expression equals $B$ instead of $A$.

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  • Understanding of set theory concepts, including set difference and union.
  • Familiarity with notation such as $A - B$, $B^c$, and subset relations.
  • Basic knowledge of logical equivalences in mathematical proofs.
  • Ability to manipulate and simplify set expressions using algebraic techniques.
NEXT STEPS
  • Study the properties of set operations, particularly focusing on union and intersection.
  • Learn about the universal set and its implications in set theory.
  • Explore more complex proofs involving subsets and set identities.
  • Review logical equivalences and their applications in mathematical reasoning.
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Mathematicians, students of set theory, and anyone interested in understanding the foundational principles of mathematical proofs involving sets.

Dustinsfl
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$(A - B)\cup B = A$ iff $B\subseteq A$.Suppose $B\subseteq A$.
$$
\begin{array}{lcl}
(A - B)\cup B & = & (A\cap B^c)\cup B\\
& = & (A\cup B)\cap (B^c\cup B)\\
& = & A\cap B\\
& = & B
\end{array}
$$

I keep getting = B not A.
 
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dwsmith said:
$(A - B)\cup B = A$ iff $B\subseteq A$.Suppose $B\subseteq A$.
$$
\begin{array}{lcl}
(A - B)\cup B & = & (A\cap B^c)\cup B\\
& = & (A\cup B)\cap (B^c\cup B)\\
& = & A\cap B\\
& = & B
\end{array}
$$

I keep getting = B not A.

Hi dwsmith, :)

Note that, \[(A\cup B)\cap (B^c\cup B)=(A\cup B)\cap V=(A\cup B)\] where \(V\) is the universal set.

Kind Regards,
Sudharaka.
 

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