limsup of a sequence vs limsup of a subsequence

Question

Let $X$ be a set and let $(E_n)$ be a sequence of subsets of $X$. The $\limsup E_n$ is defined as follows: $$\limsup E_n = \bigcap^{\infty}_{m=1} \bigcup^{\infty}_{n=m} E_n .$$

Suppose $(E_{n_k})$ is a subsequence of $(E_n)$. Will $\limsup E_{n_k}$ differ from $\limsup E_n$ by a zero set or something larger?

I believe their difference is probably a zero set, but how to actually check this?

EDIT: the claim is wrong in general, but would it be correct if we add Cauchy condition ?

Thanks!

Answer 1

Let

$$E_n=\begin{cases} \Bbb R,&\text{if }n\text{ is even}\\ \varnothing,&\text{if }n\text{ is odd}\;. \end{cases}$$

Then $\limsup_nE_n=\Bbb R$, but $\limsup_nE_{2n+1}=\varnothing$.

Answer 2

One example might get you started on constructing some of your own: Consider $E_n=\{0\}$ if $n$ is even, and $E_n=\{1\}$ if $n$ is odd. Then $\limsup_n E_n=\{0,1\}$, but $\limsup_k E_{2k}=\{0\}$.