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!)

  • $\begingroup$ What if $(x_n)$ is not bounded below? Can one still be able to show the result? $\endgroup$
    – Steve
    Nov 28, 2021 at 9:27

1 Answer 1


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.


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