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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 $(a_n)$?

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

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

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  • $\begingroup$ 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 $A$? $\endgroup$ – dove Dec 17 '18 at 18:26
  • $\begingroup$ 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. $\endgroup$ – Peter Szilas Dec 17 '18 at 19:22

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