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\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

Suppose we have an increasing sequence of sets $A_1\subseteq
A_2\subseteq\cdots$

Let $\ds A_0=\bigcup_{n=1}^\infty A_n$.  Let $D_n=A_{n+1}-A_n$.
Prove the following identities

\begin{itemize}
\item[i)]
$\ds
A_{2n}=A_1\cup\bigcup_{k=1}^{n-1}D_{2k}\cup\bigcup_{k=0}^{n-1}D_{2k+1}$

\item[ii)]
$\ds A_0=A_{2n}\cup\bigcup_{k=n}^\infty
D_{2k}\cup\bigcup_{k=n}^\infty D_{2k+1}$

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
$\ds
A_{2n}=A_1\cup\bigcup_{k=1}^{n-1}D_{2k}\cup\bigcup_{k=0}^{n-1}D_{2k+1}$

$n=2 \,\,\,\, A_1\cup D_2\cup D_1\cup D_3$

$=A_1\cup(A_2-A_1)\cup(A_3-A_2)\cup(A_4-A_3)=A_4$

since $A_1\subseteq A_2\subseteq A_3\subseteq A_4$

$\ds
A_1\cup\bigcup_{k=1}^{n}D_{2k}\cup\bigcup_{k=0}^{n}D_{2k+1}=A_{2n}\cup
D_{2n}\cup D_{2n+1}$

$\ds =A_{2n}\cup(A_{2n+1}-A_{2n})\cup(A_{2n+2}-A_{2n+1})=A_{2n+2}$

Hence the result by induction.


\item[ii)]
$\ds A_{2n}\cup\bigcup_{k=n}^\infty D_{2k}\cup\bigcup_{k=n}^\infty
D_{2k+1}$

$\ds
=A_1\cup\bigcup_{k=1}^{n-1}D_{2k}\cup\bigcup_{k=0}^{n-1}D_{2k+1}
\cup\bigcup_{k=n}^\infty D_{2k}\cup\bigcup_{k=n}^\infty D_{2k+1}$

$\ds =A_1\cup\bigcup_{i=1}^\infty D_i=\bigcup_{n=1}^\infty
A_i=A_0$ since $A_i\subseteq A_{i-1}\cdots$

\end{itemize}



\end{document}
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