1
$\begingroup$

There is this question here where there's a definition of $\lim \sup S_n$ and $\lim \inf S_n$ where $S_n$ is a sequence of sets, specifically, a sequence of subsets of a given set.

The defintion is then given as a union of intersections, and intersection of unions, respectively.

I want to get a better handle on this definition.

I now want to construct a sequence of real intervals $S_n$ (closed or open) such that neither $\lim \sup S_n$ nor $\lim \inf S_n$ is empty, and such that they are not equal.

I keep trying but my $\lim \inf S_n$ is empty, as soon as I make my $\lim \sup S_n$ not empty and not equal to the $\lim \inf S_n$.

Any hints?

Thanks in advance

$\endgroup$
1
$\begingroup$

Similar to how oscillating functions might not have limits (say, $\sin(x)$), we can define $$ S_n = \begin{cases} [0,1] &\text{if $n$ even}\\ [-1,0] &\text{if $n$ odd} \end{cases}$$ Then $\lim \inf S_n = \{0\}$ and $\lim \sup S_n = [-1,1]$.

$\endgroup$
  • $\begingroup$ Is it always the case that $\lim \inf \subseteq \lim \sup$? $\endgroup$ – JuliusL33t Aug 3 '17 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.