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[SOLVED] (A - B)\cup B = A

dwsmith

Well-known member
Feb 1, 2012
1,673
$(A - B)\cup B = A$ 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.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$(A - B)\cup B = A$ 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.