$\liminf a_n = \sup_{n \in \mathbb{N}}(\inf_{k \ge n}a_k)$ How to prove these equalities? Here $a_n$ is a sequence in $ \mathbb{R} \cup \{- \infty, + \infty \}$
$\liminf_{n\to\infty} a_n = \sup_{n \in \mathbb{N}}(\inf_{k \ge n}a_k)$
and
$\limsup_{n\to\infty} a_n = \inf_{n \in \mathbb{N}}(\sup_{k \ge n}a_k)$
I know that $\ \limsup_{n\to\infty} a_n = \sup\{ \text{ all  limit  points  of } a_n \} $ and similarly $\liminf_{n\to\infty} a_n $ is the infimum of that set. I feel that it should be somehow used in proving the above equalities but I'm not sure how to do that exactly.
 A: $\inf\limits_{k\geq n} a_k$ is an increasing sequence, thus $\liminf\limits_{n\to\infty} := \lim\limits_{n\to\infty}\inf\limits_{k \geq n} a_k = \sup\limits_{n\in\mathbb N}\inf\limits_{k \geq n} a_k $ You can change the limit to a supremum, because the sequence  $\inf\limits_{k\geq n} a_k$ is increasing.
A: Hint: you can use the fact that 
$$\lim\; \inf\; a_n = \lim_{n\to\infty} \inf(\bigcup_{k=n}^\infty a_k)$$
and $\displaystyle A_n=\inf(\bigcup_{k=n}^\infty a_k)$ is a monotone sequence.
A: I will help you with the limsup part so you get the idea. I am also assuming that you have a bounded sequence $(a_{n})_{n=1}^{\infty}$, because otherwise the claim would be $0=\infty$ if we take $a_{n}=0$ for odd $n$ and $a_{n}=n$ for even $n$, using the provided definition for limsup.
Let $A$ denote the set of all accumulation points of the sequence $(a_{n})_{n=1}^{\infty}$. Your definition is that $\limsup a_{n}=\sup A$. You want to show that $a:=\inf_{n\in\mathbb{N}}(\sup_{k\geq n} a_{k})\in A$ and $c\leq a$ for all $c\in A$, as these two conditions imply the given result.
First fix $\varepsilon>0$. Since $(a_{n})_{n=1}^{\infty}$ is bounded and $(\sup_{k\geq n} a_{k})_{n=1}^{\infty}$ is nonincreasing, then the limit $\lim_{n\to\infty}(\sup_{k\geq n} a_{k})$ exists and it equals $a$. Hence by definition of convergence there exists $n_{\varepsilon}\in\mathbb{N}$ so that $\sup_{k\geq n} a_{k}\in B(a,\varepsilon)$ for all $n\geq n_{\varepsilon}$ $(*)$, where $B(a,\varepsilon)$ is the $\varepsilon$-radius ball around $a$. In particular we find $n_{1}\in\mathbb{N}$ so that $a_{n_{1}}\in B(a,\varepsilon)$, because if $a_{k}\notin B(a,\varepsilon)$ for all $k$, then $\sup_{n\geq k}a_{n}\notin B(a,\varepsilon)$ for all $k$ since $B(a,\varepsilon)^{c}$ is a closed set, which is a contradiction with $(*)$. Moreover by $(*)$ we find $n_{2}>n_{1}$ so that $a_{n_{2}}\in B(a,\varepsilon)$, and thus inductively we find an increasing sequence of indices $n_{1},n_{2},n_{3},...$ so that $a_{n_{k}}\in B(a,\varepsilon)$ for all $k\in\mathbb{N}$. Since $\varepsilon>0$ was arbitrary, then this shows that $a$ is an accumulation point of the sequence $(a_{n})_{n=1}^{\infty}$, i.e. $a\in A$. 
So it remains to show that for any $c\in A$ we have $c\leq a$. Choose $c\in A$. For each $k\in\mathbb{N}$ take $a_{n_{k}}\in B(c,\frac{1}{k})$ with $n_{k}\geq k$ (such elements exists, since $c\in A$) to obtain a subsequence $(a_{n_{k}})_{k=1}^{\infty}$ converging to $c$. I will leave it for you to verify that now also $\inf_{j\in\mathbb{N}}\,(\sup_{k\geq j}a_{n_{k}})=c$. Hence, since $n_{k}\geq k$ for all $k$, then
\begin{align*}
c &=\lim_{k\to\infty}a_{n_{k}}=\inf_{j\in\mathbb{N}}\,(\sup_{k\geq j}a_{n_{k}})\leq \inf_{j\in\mathbb{N}}\,(\sup_{k\geq j}a_{k})=a,
\end{align*}
i.e. $c\leq a$. So this shows the claim, because we have shown that $a=\sup A$‚ and the right hand side is exactly your definition of $\limsup a_{n}$.
