Does $\liminf_{n\to \infty} -a_{n}= -\limsup_{n\to \infty}a_{n}$?

Question

Let $(a_{n})$ be a bounded sequence.
How to prove $$\displaystyle\liminf_{n\to \infty} -a_{n}= -\displaystyle\limsup_{n\to \infty}a_{n}$$
I don't how formally prove this..can someone guide me? tnx!

Answer 1

Show that $-\textrm{sup}\{a_m: m \geq n \} = \textrm{inf}\{-a_m: m \geq n \}$.

Answer 2

Using the relations $\sup(-A)=- \inf(A),\inf(-A)=-\sup(A)$ and the definitions of limes inferior/superior:

$$\liminf (-a_n)=\sup_k \inf_{n \geq k} (-a_n)=\sup_k (- \sup_{n \geq k} a_n)=-\inf_k \sup_{n \geq k} a_k=-\limsup (a_n). $$