Problem about $\lim \sup C_{n}$ and $\lim \inf C_{n}$ of sequence of sets. 
Problem: Let $$\lim \sup A_{n}=\lim_{n \to \infty}\sup A_{n}:=\bigcap_{n\in \mathbb{N}}\bigcup_{k\geq n}A_{k} \quad \text{and} \quad \lim \inf A_{n}= \lim_{n\to \infty}\inf A_{n}:=\bigcup_{n \in \mathbb{N}}\bigcap_{k\geq n}A_{k}$$
Calculate $\lim \sup C_{n}$ and $\lim \inf C_{n}$,
$$C_{n}=(-\infty, c_{n}], \quad c_{2n}=1+\frac{1}{2n}, \quad c_{2n+1}=-1-\frac{1}{2n+1}$$

My attempt: By, definition we can see that
\begin{eqnarray}
\lim \inf C_{n
}=\lim_{n \to \infty} \inf C_{n}=\bigcup_{n\in \mathbb{N}}\bigcap_{k\geq n}C_{k}=\bigcup_{n\in \mathbb{N}}\bigcap_{k\geq n}(-\infty,c_{k}]
\end{eqnarray}
Now, for $c_{2k}$ we have
\begin{eqnarray}
\lim \inf C_{n}=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}\left(-\infty,1+\frac{1}{2n}\right]=\bigcup_{n=1}^{\infty}\left[\left(-\infty,1+\frac{1}{2n}\right]\cap \left(-\infty,1+\frac{1}{2(n+1)}\right]\cap \left(-\infty,1+\frac{1}{2(n+2)}\right]\cap \cdots \right]
\end{eqnarray}
and for $c_{2k+1}$ we have
\begin{eqnarray}
\lim \inf C_{n}=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}\left(-\infty,-1-\frac{1}{2n+1}\right]=\bigcup_{n=1}^{\infty}\left[\left(-\infty,-1-\frac{1}{2(n+1)+1}\right]\cap \left(-\infty,-1-\frac{1}{2(n+2)+1}\right]\cap \cdots \right]
\end{eqnarray}
but I don't know how to continue. Is it correct? How can I continue from here? Is there other form to approach or solve this types of problems?
 A: You cannot look only at even indices or only at odd indices in computing $$\bigcup_{n\ge 1}\bigcap_{k\ge n}(\leftarrow,c_k]\,;$$ you have to look at all indices $k\ge n$.
Notice first that the sequence $\langle c_{2n}:n\ge 1\rangle$ is decreasing, while $\langle c_{2n+1}:n\ge 1\rangle$ is increasing. Thus, the sets $(\leftarrow,c_{2n}]$ are getting smaller as $n$ increases, while the sets $(\leftarrow,c_{2n+1}]$ are getting bigger as $n$ increases. Finally, note that for any $n,m\ge 1$, $(\leftarrow,c_{2n+1}]\subsetneqq(\leftarrow,c_{2m}]$. It follows that
$$\bigcap_{k\ge 2n}C_k=C_{2n+1}=\bigcap_{k\ge 2n+1}C_k$$
for each $n\ge 1$. Thus,
$$\liminf_nC_n=\bigcup_{n\ge 1}C_{2n+1}=\bigcup_{n\ge 1}\left(\leftarrow,-1-\frac1{2n+1}\right]\,,$$
and I’ll leave it to you to finish the calculation.
To find $\limsup_nC_n$ you’ll need first to compute $\bigcup_{k\ge n}C_k$ for each $n\ge 1$. For each $n$ this union will be one of the sets $C_{2k}$ with even subscript, because those properly contain all of the sets with odd subscripts; you just have to work out which one of these sets it is.
I find that the most intuitive way to think about the $\limsup$ and $\liminf$ of a sequence $\langle A_n:n\in\Bbb N\rangle$ of sets is this: $\limsup_nA_n$ is the set of points that are in infinitely many of the sets $A_n$, and $\liminf_nA_n$ is the set of points that are in all of the sets $A_n$ from some point on, i.e., the points that are in all but finitely many of the sets $A_n$.
A: I think it's easier to see if we translate the math with more words:
$\lim_{n \to \infty}\sup C_{n}:=\bigcap_{n\in \mathbb{N}}\bigcup_{k\geq n}C_{k}$ is a set $S$. The meaning of this is that $x$ will get into $S$ if and only if no matter which integer $n\ge 1$ we choose, we can find an integer $k\ge n$ such that $x\in C_k$.
And since $C_k$ are given explicitly, we can easily test this condition, splitting $\mathbb R$ into intervals.
Hints:
$x>1:\ x\notin C_{2n+1}$ clearly, for any integer $n$ and since we can choose $N$ large enough that $1+\frac{1}{2N}<x$, then $x\notin C_{2n}$ for $\textit{any}\ n\ge N.$
$x\le 1:\ x\in C_{2n}$ for $\textit{any}$ integer $n$.
Similarly, you can test for the $\liminf$.
