0

Let $s_n$ be a sequence such that $-5\leq s_n \leq 22$ for all $n\in \mathbb{N}$. Prove $\exists$ a sequence of positive integers $n_k \in \mathbb{N}$ with 3 properties:

  • $n_1<n_2<...<n_k<n_{k+1}<...$ (i.e., $n_k$ < $n_{k+1}$, for all $k \in \mathbb{N}$)
  • $\lim_{k\to \infty} s_{n_{k}}$ exists; and
  • $-5 \leq \lim_{k\to \infty} s_{n_{k}} \leq 22$.

Okay bear with me here. I feel like this problem is fairly simple, but I'm just not getting it. From what I'm seeing, I'm supposed to find an $s_n$ that satisfies these 3 listed properties? So, $s_n$ has to be increasing, the limit of it's subsequence has to exist, and fall between -5 and 22? I'm just not sure how I go about proving all of that at once.

sfgiantsfan19
  • 445
  • 1
    If you know the theorem : Every bounded sequence has a convergent subsequence. It is called Bolzano-Weierstrass. Then your question is quite immediate. – Both Htob Mar 18 '15 at 01:41
  • 1
    Note that the third condition is fulfilled automatically if the subsequence has a limit at all. You don’t need the subsequence to be increasing, but if you can get a monotone subsequence (increasing or decreasing), you’re pretty much done. See this question. – Brian M. Scott Mar 18 '15 at 01:42
  • 1
    @BothHtob: Since the result is trivial with the BW theorem, it’s likely that the exercise precedes that theorem. – Brian M. Scott Mar 18 '15 at 01:43
  • @BothHtob Your comment needs to be edited slightly. But I'm a bit confused now. So, the statement is true by B-W, then I prove using MCT? – sfgiantsfan19 Mar 18 '15 at 02:00
  • 1
    @sfgiantsfan19 I am sorry about making you confused. Actually, if you can apply B-W, you will immediately get the result. However, I suppose that you cannot use B-W. You should go by proving the statement instead : Every sequence has a monotone subsequence. After that, you should proceed by showing that that monotone subsequence, say $(s_{n_k})$, will either converge to $\sup {s_{n_k}}$ or $\inf {s_{n_k}}$. (Particularly, to $\sup$ if monotone increasing, to $\inf$ if monotone decreasing.) Hope this help. – Both Htob Mar 18 '15 at 02:50
  • @BothHtob Not at all, I'm just not quite proficient in this field. But thank you for clearing that up! I am allowed to just use B-W, but it's good to know other ways as well. I'm pretty sure I can formulate an answer. It sounds like I just need to brush up on my theorems some more and I'll get it. – sfgiantsfan19 Mar 18 '15 at 03:27
  • 1
    @sfgiantsfan19 If you can use B-W, then I think that everything is quite immediate, possibly except the last condition. But, again, try using monotone convergence theorem and the fact that $$-22 \leq \inf {s_{n_k}} \leq \sup {s_{n_k}} \leq 5$$, the third condition follows immediately. – Both Htob Mar 18 '15 at 03:44

1 Answers1

1

Since for all $n\in \mathbb{N}$, $-5\le s_n\le 22$, $(s_n)$ is bounded. Then by Bolzano-Weierstrass Theorem, $(s_n)$ has a convergent sub sequence $(s_{n_k})$. That is there exists a sequence of positive integers $(n_k)$ such that $n_1<n_2<...<n_k<...$ and $\lim\limits_{k\to \infty}S_{n_k}=l$ for some $l\in \mathbb{R}$. Now we need to show $-5\le l\le 22$. Since $\lim\limits_{k\to \infty}S_{n_k}=l$, for all $\epsilon>0$, there exists $k_\epsilon\in \mathbb{N}$ such that for all $k>k_\epsilon$, $|l-S_{n_k}|<\epsilon$.

Therefore for all $k>k_\epsilon$, $S_{n_k}-\epsilon <l<S_{n_k}+\epsilon \Rightarrow -5-\epsilon <l<22+\epsilon $.

Therefore for all $\epsilon>0$, $-5-\epsilon <l<22+\epsilon \Rightarrow -5\le l\le 22$.

ASB
  • 4,139