Convergent sequence of sets How to show that $\{E_n\}$ is a convergent sequence if and only if there is no point $x\in X$ such that $x\in E_n$, $x\in X - E_m$ hold for infinitely many $n$ and infinitely many $m$.
 A: Use the ideas at the beginning of Henno’s answer to this earlier question of yours. If $x\in E_n$ for infinitely many $n$, then $x\in\limsup_nE_n$. If $x\in X\setminus E_n$ for infinitely many $n$, then it is not true that $x\in E_n$ for all but finitely many $n$, so $x\notin \liminf_nE_n$. Thus, if $x\in E_n$ for infinitely many $n$ and $x\in X\setminus E_n$ for infinitely many $n$, then $$x\in\limsup_nE_n\setminus\liminf_nE_n\;,$$ so $\liminf_nE_n\ne\limsup_nE_n$, and $\lim_nE_n$ does not exist.
Now you just need to check that all of the implications reverse.
A: I'm going to assume that you define convergence of an arbitrary family of sets in terms of $\limsup$ and $\liminf$; recall that
$$\begin{align*}
\liminf_{n\to\infty}A_n &= \bigcup_{m=1}^{\infty}\bigcap_{n=m}^{\infty}A_n,\\
\limsup_{n\to\infty}A_n &= \bigcap_{m=1}^{\infty}\bigcup_{n=m}^{\infty}A_n.
\end{align*}$$
Now, what you want to show is that:

Proposition. Let $\{A_n\}$ be a family of sets. Then:
  
  
*
  
*$x\in\liminf\limits_{n\to\infty} A_n$ if and only if there exists $N\gt 0$ such that $x\in A_n$ for all $n\geq N$; if and only if $x$ is, eventually, in all $A_n$; if and only if $x$ is in $X-A_n$ for only finitely many $n$ (where $X=\cup A_n$). 
  
*$x\in\limsup\limits_{n\to\infty} A_n$ if and only if $x$ is in infinitely many $A_n$. 
  

Now, $\liminf E_n$ is always contained in $\limsup E_n$. So in order for equality to hold, you need every element of $\limsup E_n$ to be in $\liminf E_n$. Verify that this holds precisely when your condition holds.
