Let $(a_n)$ be a bounded sequence of real numbers, and define $$\beta_n = \sup \{ a_k : k \geq n \}. $$ This sequence converges to a limit,
$$\lim_{n \to \infty} \beta_n = \limsup a_n.$$
I'm interested in proving the existence of a subsequence of $(a_n)$ that converges to $\limsup a_n.$ I've read over the other posts concerning this and gave a satisfactory proof similar to these arguments, but I'm wondering if the following argument could be used as well. Note that in my study of analysis I've already proven the Bolzano-Weirstass theorem in $\mathbb{R}$. Here goes.
If $\beta_n$ is a bounded real sequence, Bolzano-Weierstrass implies that it has a convergent subsequence $(\beta_{n_j})$. Since $\lim \beta_n$ exists and is equal to $\limsup a_n,$ then $(\beta_{n_j})$ converges to $\limsup a_n.$
I'm not certain if this argument works because I want a subsequence of $(a_n).$ Is there a way to argue that $(\beta_{n_j})$ is a subsequence of $(a_n)$?
Best Answer
No, it doesn't work because $\beta_n$ needs not be an element of $(a_k)$.
We need a slight modification: for $n$, by the definition of supremum, there's an element $a_{k_n}>\beta_n-\frac1n$.
Then we get $L\leftarrow \ \beta_n-\frac1n<a_{k_n}\le\beta_n\ \to L$, where $L=\lim\beta_n$.