Im reading Friedmans modern analysis, and im having some flaws in my logical thinking... the question is to prove that $\left(\limsup\limits_{n}E_n\right)^c = \liminf\limits_{n}E_n^c$. My first attempt was this:
$\left(\limsup\limits_{n}E_n\right)^c = \{x\in X \mid x \notin \limsup\limits_{n}E_n\} = \{x \in X \mid x \notin \{x\in X \mid x \in E_n \text{ for infinitely many n}\}\} = \{x \in X \mid x \in E_n^c \text{ for infinitely many n}\} = \limsup\limits_{n}E_n^c$, which is clearly wrong. The second attempt was:
$\left(\limsup\limits_{n}E_n\right)^c = \{x\in X \mid x \notin \limsup\limits_{n}E_n\} = \{x \in X \mid x \notin \{x\in X \mid x \in E_n \text{ for infinitely many n}\}\} = \{x \in X \mid x\in E_n \text{ for at most finitely many n} \} = \{x \in X \mid x\in E_n^c \text{ for all but finitely many n}\} = \liminf\limits_{n}E_n^c$. The second attempt gives the right answer, but im not sure its correct. Also, whats the main difference between these two "solutions"? I believe the hard part is to find "the complement of infinitely many"? If anyone could guide me some in my thinking it would be much appreciated.
The second attempt is correct. The first attempt contains no mistakes, it is just that you can say more.
If $x\in E_n$ does not hold for infinitely many $n$, then you can infere that $x\in E_n^C$ for infinitely many $n$. But you can infere something even stronger too: $x\in E_n^C$ for all but finitely many $n$.
You have to make use of the stronger conclusion to prove what you want to prove.