[SOLVED] (A - B)\cup B = A

[dwsmith]

$(A - B)\cup B = A$ 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]

$(A - B)\cup B = A$ 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.