This is NOT a duplicate of Characterization of lim sup, lim inf. I am asking for a solution verification.
I have finished my first mathematical analysis course last year, but this came to my mind today suddenly. The question was stated below.
Kaczor-Nowak II. 2.4.13(a)(ii). Assume that for a sequence $(a_n)$, the limit superior $L=\displaystyle\varlimsup_{n\to\infty}a_n$ is finite, prove that for every $\varepsilon > 0$ and $k\in\mathbb{N}$ there is $n_k > k$ such that $L-\varepsilon < a_{n_k}$.
Solution. Suppose not. Then there are $\varepsilon > 0$ and $k\in\mathbb{N}$ such that $$a_n \leq L - \varepsilon \quad\text{for all}\quad n>k.$$ By Problem 2.4.9, $\sup S = \varlimsup_{n\to\infty}a_n = L\in S$ where $S$ is the set of limit points of $(a_n)$. It follows from the definition of a limit point that there exists a subsequence $(a_{n_k})$ converging to $L$. Therefore, for any $\varepsilon' > 0$, there is $k_0\in\mathbb{N}$ suhc that $$L - \varepsilon' < a_{n_k} \leq L - \varepsilon \quad\text{for all}\quad k>k_0.$$ Since $\varepsilon'$ is arbitrary, we get $L \leq L - \varepsilon$ contradiction, which contradicts our hypothesis.
I have checked multiple sources and all of them prove this question using contradiction, but when I solved this question, I did it in the following way:
My attempt Let $\varepsilon>0,k\in\mathbb{N}$. By definition, note that $$\sup_{n>k}a_n \geq \inf_{k'\in\mathbb{N}}\sup_{n>k'}a_n = L > L - \varepsilon,$$ we know $L - \varepsilon$ is not an upper bound for $\{a_n : n>k\}$, then there exists $n_k > k$ such that $a_{n_k} > L - \varepsilon$.
In my opinion, my proof is much, much simpler, but I don't see people using this, why?
