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  1. If $p_n$ is a bounded sequence and a subsequence $p_{n_k}$ converges to a limit q.

Show that $q \le \limsup_{n\to \infty}p_n$.

  1. If $p_n$ is a bounded sequence and $\limsup_{n\to \infty}p_n = p$, show that there is a subsequence $p_{n_k}$ that also converges to p.

I think the two questions are related but I couldn't figure out how to prove the first one.

So I will start with no. 2.

For 2,

Since $p_n$ is a bounded sequence, a subsequence $p_{n_k}$ converges. Thus, $\lim_{n\to \infty} p_{n_k} = q$. So all I need to do is to show $p=q$ since $\limsup_{n\to \infty}p_n = p$.

Assume $p \neq q$, and let $\varepsilon = \frac {|q-p|}{2}$

Also, $P_n = \sup \{p_m : m\ge n\}$ and $\limsup_{n\to \infty} p_n = \lim_{n\to \infty} P_n$

Since $\limsup_{n\to \infty}p_n = \lim_{n\to \infty} P_n= p$ and $\lim_{n\to \infty} p_{n_k} = q$, for $\varepsilon \gt 0 $,

there exists $N_1 \in \Bbb N: n \gt N_1 \Rightarrow |p_{n_k}-q| \lt \varepsilon$

Also there exists $N_2 \in \Bbb N: n \gt N_2 \Rightarrow |P_n-p| \lt \varepsilon$

Choose $N = max\{N_1, N_2\}$

$|q-p| = |q- p_{n_k} + p_{n_k} -p| \le |p_{n_k} -q| + |p_{n_k} -p|$

but $p_{n_k} \le P_n$. Thus, $|q-p|\le |p_{n_k} -q| + |P_n -p| \lt \varepsilon + \varepsilon = 2\varepsilon = |q-p| $, which is a contradiction.

Hence $p=q$ and $\lim_{n\to \infty}$ sup $p_n = \lim_{n\to \infty} p_{n_k} =p$

For 1,

$\lim_{n\to \infty} p_{n_k}=q$ and $\lim_{n\to \infty}$sup $p_n = \lim_{n\to \infty} P_n$ where $P_n = sup\{p_m: m\ge n\}$

And that is all I got. I don't know where to go from there.

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    $\begingroup$ Use \limsup_{n \to \infty} instead of \lim_{n \to \infty} \sup. $\endgroup$ – Henricus V. Mar 24 '16 at 17:19
  • $\begingroup$ I'd say your argument for part 2 is inaccurate. It looks as if you try to show that for a bounded sequence $(p_n)$ s.t. $\limsup p_n=p$, every convergent subsequence converges to $p$ (a counterexample is the sequence $p_n=(-1)^n$). Technically, the problem in your proof is where you say "but $p_{n_k}≤P_n$. Thus, $|q−p|≤|p_{n_k}−q|+|P_n−p|<ε+ε=2ε=|q−p|$". This is not necessarily true. The distance between $p_{n_k}$ and $p$ might be greater than the distance between $P_n$ and $p$. $\endgroup$ – Nate River Mar 24 '16 at 20:48
  • $\begingroup$ @Nate River. Ok. do you have any suggestion how to improve it? $\endgroup$ – abuchay Mar 24 '16 at 21:04
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For 1, notice

$$ \limsup_{n} p_n=\inf_{n} \sup_{k\geq n} p_k $$ while $$ q=\lim_{n} p_{l_n}=\limsup_{n} p_{l_n}=\inf_{n} \sup_{k\geq n} p_{l_k} $$

To conclude, we observe that, for each $n$, $$ \sup_{k\geq n} p_{l_k}\leq \sup_{k\geq n} p_k $$ hence $$ q=\lim_{n} p_{l_n}=\limsup_{n} p_{l_n}=\inf_{n}\sup_{k\geq n} p_{l_k}\leq \inf_{n}\sup_{k\geq n} p_k=\limsup_n p_n $$

For 2, for simplicity let's denote $P_n=\sup_{k\geq n}$. A possible proof, using the fact that the sequence $P_n$ is decreasing, is this: (I guess there are better, simpler ones, but here we go anyway):

Case 1: For all $n$ there is $k>n$ s.t. $P_k<P_n$. This implies that there is a decreasing subsequence $(p_{k_n})_n$ s.t. for all $n$, $P_{k_n}=p_{k_n}$. This implies $p=\lim_n p_{k_n}$.

Case 2: There is $n_0$ s.t. for all $n\geq n_0,\: P_n=P_{n_0}=p$. This means, for all $n\geq n_0$, $p=\sup_{k\geq n} p_k$. Using the definition of supremum, we can find a subsequence $p_{k_n}$ s.t. for all $n$, $0\leq p-p_{k_n}\leq 1/n$.

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