Prove that $\liminf x_n = -\limsup (-x_n)$ How can we prove that $\liminf x_n = -\limsup (-x_n)$?
The definitions we are using are
$\limsup x_n = \lim\limits_{n\to\infty} \sup\{x_k; k\ge n\}$
$\liminf x_n = \lim\limits_{n\to\infty} \inf\{x_k; k\ge n\}$
 A: I'll go with the definition you mention. Defining $\liminf x_n$ as the smallest limit of all converging subsequences in $[-\infty,+\infty)$ and $\limsup$ as the greatest etc... would make the statement slightly easier to check.
Assume you have proved that $-\sup (-A)=\inf A$ for every subset $A$ of $\mathbb{R}$.
Applying this to the set $A=\{x_n\;;\;n>N\}$ yields:
$$
-\sup_{n>N}(-x_n)=\inf_{n>N}x_n\qquad \forall N.
$$
Taking the limit on both sides gives the formula. Even in the case where these sequences are constant equal to $-\infty$, that is $\liminf x_n=-\infty$ and $\limsup (-x_n)=+\infty$.
Now let us prove the set property. Clearly, $A$ is not bounded below if and only if $-A$ is not bounded above. In this case, we get $-(+\infty)=-\infty$ and the property holds. Now assume we are not in the latter case. Take $a\in A$. We have
$$
m\leq a\qquad\iff\qquad -a\leq -m.
$$
Therefore, $m$ is a lower bound for $A$ if and only if $-m$ is an upper bound for $-A$. It follows that $-\inf A$ is an upper bound for $-A$, so $\sup(-A)\leq -\inf A$. And likewise, $-\sup(-A)$ is a lower bound for $A$, so $-\sup(-A)\leq \inf A$. This proves the equality $-\sup(-A)=\inf A$.
