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 $m$.
Convergent sequence of sets
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0Thanks guys i 'll try using these tools for my future posts. – 2012-02-12
2 Answers
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.
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.