3236.sagews

\textbf{Theorem 6b }- Suppose that A and B are subsets of U. Then, $A\cup (A\setminus) = U$


& \text{Proof:} 

&& \text{  i. }$\emptyset \cup \emptyset\setminus = U$

                && $A\subset U \iff (A\setminus) \subset U$

                && \text{By Lemma 2, } $A\cup (A\setminus)\subset U$

&& \text{      ii.  By Remark 1, we may assume U is non-empty}

&& \text{           Let } $x \in U$.

&& \text{     } $x\in A$ or $x \notin A$

&& \text{     If} $x\notin A$, \text{then} $x\in A\cup (A\setminus)$

&& \text{     If} $x\in A$ \text{then} $x\in A\cup (A\setminus)$

Therefore,  $A\cup (A\setminus) = U$
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