(Proof verification) Showing that the two $\limsup$ definitions are equivalent

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!)

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

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 $$(x_n)$$.