# Thread: liminf to limsup transformation question..

1. ## liminf to limsup transformation question..

y_n is a bounded sequence

liminf(-y_n)=-limsup(y_n)
n->infinity

i cant understand how it happens??

2. Originally Posted by transgalactic
y_n is a bounded sequence

liminf(-y_n)=-limsup(y_n)
n->infinity

i cant understand how it happens??
If $A$ be a non-empty bounded subset of $\mathbb{R}$ then $\sup (-A) = -\inf A \text{ and }\inf (-A) = -\sup A$.
This is the key fact that we need. (Notation: $-A = \{ - a|a\in A\}$)

Let $r = \liminf (-y_n)$.
Therefore, $r = \sup \left\{ \inf\{ -y_k |k\geq n\} | k\geq 1\right\}$
But then, $-r = \inf \left\{ \boxed{- \inf \{ -y_k |k\geq n\}} | k\geq 1\right\}$
Apply again to the boxed expression, $-r = \inf \left\{ \sup \{ y_k |k\geq n\} | k\geq 1 \right\}$
Thus, $-r = \limsup (y_n)$.
Putting this together we get, $\liminf (-y_n) = - \limsup (y_n)$.

3. You may want to wait for confirmation on this...it may not be rigorous enough....also how late it is may cause concern for incorrectness.

I think you alternatively to TPH's response may be able to say that since $x_n$ is a bounded sequence that the set of subsequential limits $S$ of $x_n$ is a bounded subset of the reals. This fact lets us apply all the basic "rules and theorems" of subsets of the reals to $S$. So now using TPH's first remark and noting that $\limsup x_n=\sup S$ and $\liminf x_n=\inf S$ we can draw our conclusion.