How to calculate the $\limsup$ and $\liminf$ of a piece wise sequence of sets I am struggling to understand $\limsup$ and $\liminf$ of a sequence of sets. The following is a question I have made up for the sake of example:
Suppose I want to find the $\limsup_n$ and $\liminf_n$  of a sequence of sets $c_n = (1-1/(n+2), 1 + 1/(n+2))$ when n is odd and $(1/(n+2), 4/5-1/(n+2))$ when n is even.
I am struggling at two levels. First, trying to find $\liminf$ of the subsequence when $n$ is odd and when n is even (So basically, calculating the $\liminf$ and $\limsup$ of a basic sequence of sets). Second, how to calculate the $\liminf$ and $\limsup$ of a sequence whose two important subsequences have different limits.
 A: I find that the easiest way to think about the limits inferior and superior of the sequence $\langle c_n:n\in\Bbb N\rangle$ is that $\liminf_nc_n$ is the set of all real numbers that are in all but finitely many of the sets $c_n$, and $\limsup_nc_n$ is the set of all real numbers that are in infinitely many of the sets $c_n$. Here
$$c_n=\begin{cases}
\left(1-\frac1{n+2},1+\frac1{n+2}\right),&\text{if }n\text{ is odd}\\
\left(\frac1{n+2},\frac45-\frac1{n+2}\right),&\text{if }n\text{ is even,}
\end{cases}$$
so $c_0=\left(\frac12,\frac3{10}\right)$, $c_1=\left(\frac23,\frac43\right)$, $c_2=\left(\frac14,\frac{11}{20}\right)$, and so on. This is easier to understand if we look at the subsequences of terms with odd indices and with even indices. For odd $n$ we get
$$\left(\frac23,\frac43\right),\left(\frac45,\frac65\right),\left(\frac67,\frac87\right),\ldots\,,\tag{1}$$
and for even $n$ we get
$$\left(\frac12,\frac45-\frac12\right),\left(\frac14,\frac45-\frac14\right),\left(\frac16,\frac45-\frac16\right),\ldots\,.\tag{2}$$
The sets in $(1)$ are squeezing down on $1$: their intersection is the singleton $\{1\}$. Thus, the only real number that is in all but finitely many of them is $1$ itself, and it is also the only real number that is in infinitely many of them: $\{1\}$ is the $\liminf$ and the $\limsup$ of this subsequence.
The sets in $(2)$ are expanding: each is contained in the next, and their union is $\left(0,\frac45\right)$. This means that any real number that is in $c_{2n}$ is in $c_{2m}$ for each $m\ge n$ and is therefore in all but finitely many of these sets. In particular, every element of their union $\left(0,\frac45\right)$ is in one of these sets and is therefore in all but finitely many of them (and of course also in infinitely many of them). Here again the $\liminf$ and $\limsup$ are equal, this time to $\left(0,\frac45\right)$.
Now for the original sequence you need to answer two question:

*

*What real numbers are in infinitely many of the sets $c_n$?

*What real numbers are in all but finitely many of the sets $c_n$?

What I’ve done for the subsequences should make both of these questions fairly easy to answer.
