MHB Show existence of subsequences

[evinda]

Hello! (Wave)

Let $(a_n)$ be a bounded sequence such that $\inf_{\ell} a_{\ell}<a_n< \sup_{\ell} a_{\ell}$ for each $n=1,2, \dots$

I want to show that there are subsequences $(a_{k_n})$ increasing and $(a_{m_n})$ decreasing such that $a_{k_n} \to \sup_{\ell} a_{\ell}$ and $a_{m_n} \to \inf_{\ell} a_{\ell}$.Could you give me a hint how we could find the desired subsequences? (Thinking)

 

[I like Serena]

Hey evinda!

I don't think it's true. (Worried)

Suppose we take the sequence $(a_n) = (0,2,1,1,1,1,...)$.
Then it's bounded and its supremum is $2$, but there is no subsequence that converge to $2$ is there? (Wondering)

 

[Olinguito]

Hey evinda!

I don't think it's true. (Worried)

Suppose we take the sequence $(a_n) = (0,2,1,1,1,1,...)$.
Then it's bounded and its supremum is $2$, but there is no subsequence that converge to $2$ is there? (Wondering)

That sequence attains its supremum (and infimum). evinda is looking at bounded sequences that do not attain their least upper or greatest lower bounds.

@evinda:
This is my suggestion. Let $s=\sup_la_l$. Try and construct a subsequence $\left(a_{k_n}\right)$ as follows. Let $a_{k_1}=a_1$. Then there exists $a_{n_1}$ such that $a_{k_1}<\dfrac{a_{k_1}+s}2<a_{n_1}<s$ since $s$ is the least upper bound. Let $a_{k_2}=a_{n_1}$. Now $\exists a_{n_2}$ such that $a_{k_2}<\dfrac{a_{k_2}+s}2<a_{n_2}<s$; let $a_{k_3}=a_{n_2}$. Continue this way to get the subsequence $\left(a_{k_n}\right)$ where at the $i$th stage $\exists a_{n_i}$ such that $a_{k_i}<\dfrac{a_{k_i}+s}2<a_{n_i}<s$ and you let $a_{k_{i+1}}=a_{n_i}$. Clearly $\left(a_{k_n}\right)$ is increasing and $\dfrac{a_{k_1}+\left(2^n-1\right)s}{2^n}<a_{k_{n+1}}<s$ (easily proved by induction); since $\dfrac{a_{k_1}+\left(2^n-1\right)s}{2^n}\to s$ as $n\to\infty$, $\displaystyle\lim_{n\to\infty}a_{k_n}=s$ by the squeeze theorem.

Similarly for the other subsequence $\left(a_{m_n}\right)$.

 

[GJA]

Hi Everyone,

Wanted to politely jump into say that the construction above needs a slight modification since it does not ensure that the subsequence we select converges to the supremum. Ignoring the infimum stuff, consider the sequence defined by: $a_{2n+1}=-\dfrac{1}{2n+1}$ and $a_{2n}=1-\dfrac{1}{2n}.$ Then it is possible to have selected the subsequence $-1, -1/3, -1/5, \ldots,$ which does not converge to the supremum value of 1.

Note: I admit this example does not possesses the strict inequality property on the infimum, but that is independent of what we are considering here.

 

[Olinguito]

Hi Everyone,

Wanted to politely jump into say that the construction above needs a slight modification since it does not ensure that the subsequence we select converges to the supremum. Ignoring the infimum stuff, consider the sequence defined by: $a_{2n+1}=-\dfrac{1}{2n+1}$ and $a_{2n}=1-\dfrac{1}{2n}.$ Then it is possible to have selected the subsequence $-1, -1/3, -1/5, \ldots,$ which does not converge to the supremum value of 1.

Note: I admit this example does not possesses the strict inequality property on the infimum, but that is independent of what we are considering here.

Thanks GJA. I also noticed the same thing and was editing my post as you posted; I think the edited version will work now. (Nod)