If $c$ is any limit point of $(a_{n})_{n=m}^{\infty}$, then we have that $L^{-}\leq c\leq L^{+}$.

Question

Let $(a_{n})_{n=m}^{\infty}$ be a sequence of real numbers, let $L^{+}$ be the limit superior of this sequence, and let $L^{-}$ be the limit inferior of this sequence (thus both $L^{+}$ and $L^{-}$ are extended real numbers).

Proposition

If $c$ is any limit point of $(a_{n})_{n=m}^{\infty}$, then we have that $L^{-}\leq c\leq L^{+}$.

Proof

Let us recall the definitions involed in the proposition.

We shall start with the definition of limit point:

For every $\varepsilon > 0$ and every $N\geq m$ there is a natural number $n\geq N$ such that $|a_{n} - c| \leq \varepsilon$.

On the other hand, $\displaystyle L^{+} = \limsup_{n\rightarrow\infty}a_{n} = \inf(a^{+}_{N})_{N=m}^{\infty}$, where $a^{+}_{N} = \sup(a_{n})_{n=N}^{\infty}$.

Similarly, $\displaystyle L^{-} = \liminf_{n\rightarrow\infty}a_{n} = \sup(a^{-}_{N})_{N=m}^{\infty}$, where $a^{-}_{N} = \inf(a_{n})_{n=N}^{\infty}$.

Unfortunately, I am not able to manipulate these definitions in order to obtain the desired result.

Could someone please help me with this? This is not a homework. It is part of my self-study.

Answer 1

Assume $c \gt L^{+}$ and choose $\epsilon \in (0,c - L^{+})$. By definition of limit point, $|a_n - c| \lt \epsilon/2$ for infinitely many $n$. But then $a_n \gt L^{+} + \epsilon/2$ for infinitely many $n$, a contradiction. Similar argument for the case $c \lt L^{-}$.