Question:
Show that
$\limsup A_n -\liminf A_n = \limsup(A_n A^c_{n+1}) =\limsup (A^c_n A_{n+1})$
the thing I understand from this queston is the following;
$$\bigcap_{n=1}^\infty \bigcup_{n=k}^\infty A_k \setminus \bigcup_{n=1}^\infty\bigcap_{n=k}^\infty A_k =(\bigcap_{n=1}^\infty \bigcup_{n=k}^\infty A_k) \cap (\bigcup_{n=1}^\infty\bigcap_{n=k}^\infty A_k )^c = \bigcap_{n=1}^\infty \bigcup_{n=k}^\infty ( A_kA_k^c) $$
and I know that $A\setminus B = A\cap B^c $
Probably this way is incorrect.I'm not sure. Please show me the correct solution way. Thank you for helping.
It turns out that the claim in the question is incorrect. The equality is actually $$\limsup_{n\to\infty} A_n \setminus\liminf_{n\to\infty} A_n = \limsup_{n\to\infty} (A_n\triangle A_{n+1}). $$ We have $$\begin{align*} \limsup_{n\to\infty} A_n \setminus\liminf_{n\to\infty} A_n &= \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k\right)\setminus\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k\right)\\ &= \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k\right)\cap\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k\right)^c\\ &= \left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k\right)\cap\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty A_k^c\right)\\ &= \bigcap_{n=1}^\infty\left(\bigcup_{k=n}^\infty A_k \cap \bigcup_{k=n}^\infty A_k^c \right)\\ &= \bigcap_{n=1}^\infty\left(\left(\bigcup_{k=n}^\infty A_k \cap \bigcup_{k=n}^\infty A_{k+1}^c\right) \cup \left(\bigcup_{k=n}^\infty A_k^c\cap\bigcup_{k=n}^\infty A_{k+1}\right) \right)\\ &= \bigcap_{n=1}^\infty\left(\bigcup_{k=n}^\infty (A_k\cap A_{k+1}^c) \cup \bigcup_{k=n}^\infty (A_k^c\cap A_{k+1}) \right)\\ &= \bigcap_{n=1}^\infty\left(\bigcup_{k=n}^\infty (A_k\cap A_{k+1}^c)\cup (A_k^c\cap A_{k+1}) \right)\\ &= \bigcap_{n=1}^\infty\left(\bigcup_{k=n}^\infty A_k\triangle A_{k+1}\right)\\ &= \limsup_{n\to\infty} (A_n\triangle A_{n+1}). \end{align*}$$