0
$\begingroup$

I have been trying to prove that the two definitions of $\limsup$ are equivalent. I would appreciate it if someone could verify my attempt! Thanks in advance!


Here are the two definitions:

  1. For any bounded sequence $(x_n)$, $\displaystyle \limsup_{n\to\infty} x_n$ is defined to be $\displaystyle\limsup_{n\to\infty}x_n\colon = \lim_{n\to\infty} \sup \{ x_{n}, x_{n+1} , x_{n+2} , \ldots \}$.

  2. Let $(x_n)$ be a bounded sequence. Let $T$ be the set of all cluster points of $(x_n)$. Then we define $\displaystyle\limsup_{n\to\infty} x_n=\sup T$.


My attempt:

Let $(x_n)$ be bounded sequence. We define $(y_n)$ to be the sequence $y_n = \sup \{ x_{n}, x_{n+1} , x_{n+2} , \ldots \} $ and $\alpha = \lim_{n\to\infty} y_n$ and $\beta = \sup T$ where $T$ is the set of all cluster points of $(x_n)$. We'll be done if we show that $\alpha = \beta $.

$\beta$ is a cluster point of $(x_n)$. Thus there must be a subsequence $(x_{n_k})$ of $(x_n)$ such that $\lim_{k\to \infty} x_{n_{k}} = \beta$. For each $k \in \mathbb{N}$, $n_k \ge k$ and thus $x_{n_k} \le y_k = \sup \{ x_{k}, x_{k+1} , x_{k+2} , \ldots \} $. Thus, taking limits both sides of the previous inequality, we have that $\beta \le \alpha$.

To prove $\alpha \le \beta$, we show that $\alpha$ is a cluster point of $(x_n)$ then we will be done. We will construct a subsequence of $(x_n)$ which converges to $\alpha$. We use the fact that $\alpha = \lim_{n\to\infty} y_n$ to achieve this. Let $\varepsilon =1$. Then there exists $N\in\mathbb{N}$ such that $\alpha -1 < \sup \{ x_{n}, x_{n+1} , \ldots \} < \alpha + 1 $ for all $n\ge N$. Let $N_1=N$. Hence, $\alpha -1 < \sup \{ x_{N_1}, x_{N_1+1} , \ldots \} < \alpha + 1 $. Since $\alpha -1$ is not an upper bound for the set $\{ x_{N_1}, x_{N_1+1} , \ldots \}$, there exists $n_1 \ge N_1$ such that $\alpha -1 < x_{n_1} \le \sup \{ x_{N_1}, x_{N_1+1} , \ldots \} < \alpha + 1 $. Now, we do it for $\varepsilon = 1/2$. Then yet again there exists $N\in\mathbb{N}$ such that $\alpha -\frac{1}{2} < \sup \{ x_{n}, x_{n+1} , \ldots \} < \alpha + \frac{1}{2} $ for all $n\ge N$. Let $N_2 =\max \{ N, n_1 \}$. Hence, $\alpha -\frac{1}{2} < \sup \{ x_{N_2}, x_{N_2 +1} , \ldots \} < \alpha + \frac{1}{2} $. Since $\alpha -1/2$ is not an upper bound for the set $\{ x_{N_2}, x_{N_2+1} , \ldots \}$, there exists $n_2 \ge N_2$ such that $\alpha -1/2 < x_{n_2} \le \sup \{ x_{N_2}, x_{N_2+1} , \ldots \} < \alpha + 1/2 $. By induction, for each $k \in \mathbb{N}$, we pick $x_{n_k}$ such that $n_{k+1} > n_{k}$ and $|x_{n_{k}}-\alpha | < \frac{1}{k}$ . By our construction $\lim_{k \to \infty} x_{n_k} = \alpha$. Thus, $\alpha$ is a cluster point of $(x_n)$ and hence $\alpha \le \beta$. Thus we are done!


(Do not mark this post as duplicate as I do not seek for solutions!)

$\endgroup$
1
$\begingroup$

This looks valid to me! You show that the $\lim \limits_{n \to \infty}$sup{${x_n, x_{n+1},...}$} is equal to sup(T). This shows the two definitions are equal.

Your steps look good. You show that $\beta \leq \alpha$ by establishing a cutoff point after which $x_{n_k} \leq\space $sup{$x_k, x_{k+1}, ...$}. You then take the limit of each side of this inequality. This should be valid since the left side is less than the right for all $x_{n_k}$.

You then show that $\alpha \leq \beta$ by constructing a subsequence of $x_n$ that converges to $\alpha$ [which establishes $\alpha$ as a cluster point of $(x_n)$] and then using the fact that $\beta$ is the supremum of all cluster points of $(x_n)$.

I was unable to find any flaws in these steps.

$\endgroup$

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.