# equivalent definitions of limit superior

Definition I

Let $\{ s_n \}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} \rightarrow x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+\infty$, $-\infty$.

Then the limit superior $s^*$ is the unique number with the properties:

(a) $s^* \in E$.

(b) If $x> s^*$, there is an integer $N$ such that $n \geq N$ implies $s_n < x$.

Definition II

Let $(a_k)$ be a sequence of real numbers. Define $s_m =\sup\{a_k\mid k\ge m\}.$ Then the limit superior is defined to be $s^*=\lim\limits_{m\rightarrow \infty}s_m.$

I have show these two definitions are equivalent.

Attempt :

Let $(a_k)$ be a bounded sequence of real numbers. Define $s_m =\sup\{a_k\mid k\ge m\}.$ Since $(s_m)$ is monotonic and bounded, $(s_m)$ converges.

Denote $U =\lim\limits_{m\rightarrow \infty}s_m$ so that

$$\forall \epsilon >0, \exists N\in \mathbb{N} \text{ such that } |s_m-U|<\epsilon, \forall m\ge N.$$

If we show that $U$ satisfies the properties mentioned in definition I, we are done.

So, Claim : If $y>U,$ then $\exists m\in \mathbb{N}$ such that $k\ge m$ implies $a_k < y.$

Choose $\epsilon := y-U$

This implies $\exists N\in \mathbb{N}$ such that $|s_m-U|<y-U, \forall m\ge N.$

$$\implies s_m-U<y-U, \forall m\ge N.$$

$$\implies s_m<y, \forall m\ge N.$$

Thus for $k\ge m,$ we have $a_k\le s_m < y, \forall m\ge N.$

Hence, If $y>U,$ then $\exists m\in \mathbb{N}$ such that $k\ge m$ implies $a_k < y$ and the claim is verified.

Question : How do you show that $U \in E$ i.e., there exists a subsequence $(a_{n_k})$ which converges to $U.$

Note that $s_m$ is monotonically decreasing so that $s_m\geq U$ for each $m$.
Based on $s_1\geq U$ we conclude that some $n_1\geq1$ exists with $a_{n_1}\geq U-2^{-1}$.
Then based on $s_{n_1+1}\geq U$ we conclude that some $n_2>n_1$ exists with $a_{n_2}\geq U-2^{-2}$.
Then based on $s_{n_2+1}\geq U$ we conclude that some $n_3>n_2$ exists with $a_{n_3}\geq U-2^{-3}$.
We have $s_{n_k}\geq a_{n_k}\geq U-2^{-k}$ so the sequence $(a_{n_k})_k$ will evidently converge to $U$.