# Existence of a subsequence converging to limsup

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 $$(a_n)$$?

• No because $(\beta _{n_j})$ is not a subsequence of $(a_n)$. Moreover, why taking a subsequence of $(\beta _n)$ since it already converge ? – Surb Dec 16 '18 at 21:47
• math.stackexchange.com/questions/581128/… – parsiad Dec 16 '18 at 22:09

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 $$L\leftarrow \ \beta_n-\frac1n, where $$L=\lim\beta_n$$.

• Great, thanks, I thought so! Appreciate the speedy response. – dove Dec 17 '18 at 0:23

In context:

$$b_n= \sup (a_k| k \ge n)$$

$$A:= \lim_{n \rightarrow \infty} b_n=\lim \sup a_n.$$

Show that $$A$$ is a limit point of $$(a_n)_{n \in \mathbb{N}}$$.

1)Let $$\epsilon >0$$ be given.

There is a $$N$$ such that for $$n \ge N$$

$$|b_n -A| \lt \epsilon/2.$$

2) By definition of $$b_n = \sup (a_k|k \ge n)$$ there is a

$$a_{n_k}$$ such that for $$k \ge k_0$$

$$|a_{n_k}-b_n| \lt \epsilon/2.$$

For $$n \ge N$$, and $$k \ge k_0$$

3) $$|a_{n_k}-A| \le$$

$$|a_{n_k}-b_n| +|b_n-A| \lt \epsilon$$, i.e.

$$A$$ is a limit point of $$(a_n)_{n \in \mathbb{N}}$$, and $$a_{n_k}$$ converges to.$$A$$.

• Thank you! I haven't been formally introduced to the notion of limit points yet, but this makes things clear. I assume this follows from a proof that if $A$ is a limit point of $(a_n)$ then there is a subsequence of $(a_n)$ converging to $A$? – dove Dec 17 '18 at 18:26
• dove.Welcome. Yes.Hopefully adds a bit of context.I myself am not very much at ease with limsup and liminf. Bounded a_n has limit points, a subsequence converges to every limit point.Shown above: limsup is a imit point A of (a_n), hence a subsequence converges to A.Greetings. – Peter Szilas Dec 17 '18 at 19:22