Proposition
Let $\{A_n:n\geq1\}$ be a sequence of subsets of $\Omega$. Show that
$$ \begin{align} &1)\;\text{If}\;A_k\subset A_{k+1}\;\text{for all}\;k\in\mathbb{N},\text{then}\;\lim_{n\rightarrow\infty}A_n=\bigcup_{n=1}^\infty A_n\\ &2)\;\text{If}\;A_k\supset A_{k+1}\;\text{for all}\;k\in\mathbb{N},\text{then}\;\lim_{n\rightarrow\infty}A_n=\bigcap_{n=1}^\infty A_n\\ \end{align} $$
1) If $A_k\subset A_{k+1}$ for all $k\in\mathbb{N}$, then $\{A_n:n\geq1\}\uparrow A$, i.e. the sequence is monotone non-decreasing. In such a case we have that $\bigcap_{m\geq n} A_m \overset{?}{=} A_n$. In other words, if
$$ \begin{align} \lim_{n\rightarrow\infty} A_n&=A_*\\ &=\liminf_{n\rightarrow\infty}A_n\\ &=\bigcup_{n=1}^\infty\bigcap_{m=n}^\infty A_m\\ &=\bigcup_{n=1}^\infty A_n. \end{align} $$