0
$\begingroup$

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 $U.$

$\endgroup$
2
$\begingroup$

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 $a_{n_3}\geq U-2^{-3}$.

Et cetera.

We have $s_{n_k}\geq a_{n_k}\geq U-2^{-k}$ so the sequence $(a_{n_k})_k$ will evidently converge to $U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.